{"id":28383,"date":"2024-10-23T07:03:38","date_gmt":"2024-10-23T01:33:38","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=28383"},"modified":"2024-10-24T10:02:22","modified_gmt":"2024-10-24T04:32:22","slug":"samacheer-kalvi-12th-physics-guide-chapter-2","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-physics-guide-chapter-2\/","title":{"rendered":"Samacheer Kalvi 12th Physics Guide Chapter 2 Current Electricity"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide<\/a> Pdf Chapter 2 Current Electricity Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity<\/h2>\n

12th Physics Guide Current Electricity Text Book Back Questions and Answers<\/h3>\n

Part – 1<\/span><\/p>\n

Text Book Evaluation<\/span><\/p>\n

I. Multiple choice questions:<\/span><\/p>\n

Question 1.
\nThe following graph shows the current versus voltage values of some unknown conductor. What is the resistance of this conductor?
\n\"Samacheer
\na) 2 ohm
\nb) 4 ohm
\nc) 8 ohm
\nd) 1 ohm
\nAnswer:
\na) 2 ohm
\nSolution:
\nV = IR
\nR = \\(\\frac{\\mathrm{V}}{\\mathrm{I}}\\)
\nFrom graph,
\n\\(\\text { Slope }=R=\\frac{\\Delta V}{\\Delta I}=\\frac{(4-0)}{(2-0)}=\\frac{4}{2}=2 \\Omega\\)<\/p>\n

Question 2.
\nA wire of resistance 2 ohms per meter is bent to form a circle of radius 1m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is
\n\"Samacheer
\na) \u03c0\u2126
\nb) \\(\\frac{\\pi}{2} \\Omega\\)
\nc) 2\u03c0\u2126
\nd) \\(\\frac{\\pi}{4} \\Omega\\)
\nAnswer:
\na) \u03c0\u2126
\nSolution:
\nCircumference of circle = 2\u03c0r = 2 \u00d7 \u03c0 \u00d7 1 = 2\u03c0
\nResistance of wire = 2 \u00d7 2\u03c0
\nResistance of each section = \\(\\frac{4 \\pi}{2}\\) = 2\u03c0
\nEquivalent Resistance = \\(\\begin{array}{l}
\n\\frac{2 \\pi \\times 2 \\pi}{2 \\pi+2 \\pi} \\\\
\n\\frac{4 \\pi^{2}}{4 \\pi}=\\pi \\text { ohm }
\n\\end{array}\\)<\/p>\n

Question 3.
\nA toaster operating at 240 V has a resistance of 120 \u2126. The power is
\na) 400 W
\nb) 2 W
\nc) 480 W
\nd) 240 W
\nAns :
\nc) 480 W
\nSolution:
\nP = VI
\n= \\(\\frac{V^{2}}{R}=\\frac{240 \\times 24 \\not b}{12 \\phi}\\)
\n= 240 \u00d7 2 = 480W
\nP = 480W<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nA carbon resistor of (47 \u00b1 4.7 ) k\u2126 to be marked with rings of different colours for its identification. The colour code sequence will be
\na) Yellow \u2014 Green \u2014 Violet \u2014 Gold
\nb) Yellow\u00a0 \u2014 Violet \u2014 Orange \u2014 Silver
\nc) Violet \u2014 Yellow \u2014 Orange \u2014 Silver
\nd) Green \u2014 Orange \u2014 Violet \u2014 Gold
\nAnswer:
\nb) Yellow \u2014 Violet \u2014 Orange \u2014 Silver
\nSolution:
\nColour code:
\nB B R O Y G B V G W
\n0 1 2 3 4 5 6 7 8 9
\n4 \u2014 yellow
\n7 \u2014 violet
\n1k\u2126 = 103 = Orange
\nb) is correct answer.<\/p>\n

Question 5.
\nWhat is the value of resistance of the following resistor?
\n\"Samacheer
\na) 100 k\u2126
\nb) 10 k\u2126
\nc) 1 k\u2126
\nd) 1000 k\u2126
\nAnswer:
\na) 100 k\u2126
\nSolution:
\nBrown – 1
\nBlack – 0
\nYellow – 104<\/sup>
\n10 \u00d7 104<\/sup> = 100 \u00d7 103<\/sup> = 100 k\u2126<\/p>\n

Question 6.
\nTwo wires of A and B with circular cross-section are made up of the same material with equal lengths. Suppose RA<\/sub> = 3RB<\/sub>, then what is the ratio of radius of wire A to that of B?
\na) 3
\nb) \u221a3
\nc) \\(\\frac{1}{\\sqrt{3}}\\)
\nd) \\(\\frac{1}{2}\\)
\nAnswer:
\nc) \\(\\frac{1}{\\sqrt{3}}\\)
\nSolution:
\nGiven RA<\/sub> = 3B<\/sub>
\n\\(\\begin{array}{l}
\n\\qquad \\rho=\\frac{\\pi r^{2} R}{l} \\\\
\nR \\alpha \\frac{1}{r^{2}} \\\\
\nR_{A}=\\frac{1}{r_{1}^{2}} \\quad R_{B}=\\frac{1}{r_{2}^{2}} \\\\
\n\\frac{1}{r_{1}^{2}}=3 \\frac{1}{r_{2}^{2}} \\Rightarrow \\frac{r_{1}^{2}}{r_{2}^{2}}=\\frac{1}{3} \\\\
\n\\frac{r_{1}}{r_{2}}=\\frac{1}{\\sqrt{3}}
\n\\end{array}\\)<\/p>\n

Question 7.
\nA wire connected to a power supply of 230 V has power dissipation P1<\/sub>. Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case, power dissipation is P2<\/sub>. The ratio of \\(\\frac{P_{2}}{P_{1}}\\) is
\na) 1
\nb) 2
\nc) 3
\nd) 4
\nAnswer:
\nd) 4
\nSolution:
\n\"Samacheer<\/p>\n

Question 8.
\nIn India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in the USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in the USA will be
\na) R
\nb) 2R
\nc) \\(\\frac{\\mathrm{R}}{4}\\)
\nd) \\(\\frac{R}{2}\\)
\nAnswer:
\nc) \\(\\frac{\\mathrm{R}}{4}\\)
\nSolution:
\n\\(P=\\frac{V^{2}}{R}\\)
\nIndia, V = 220V, P = 60W, R = R
\nUSA, V = 110V, P’ = 60W, R’ = ?
\n\u2234 R = \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{P}}\\)
\nP = P’
\n\u2234 \\(\\mathrm{R}^{\\prime}=\\frac{\\mathrm{V}^{2}}{\\mathrm{~V}^{2}}=\\frac{110 \\times 110}{220 \\times 220}=\\frac{1}{4} \\quad {\\mathrm{R}^{\\prime}=\\frac{\\mathrm{R}}{4}}\\)<\/p>\n

Question 9.
\nIn a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W, and 1 heater of 1 kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be
\na) 14 A
\nb) 8 A
\nc) 10 A
\nd) 12 A
\nAnswer:
\nd) 12A
\nSolution:
\n\"Samacheer<\/p>\n

Question 10.
\nThere is a current of 1.0 A in the circuit shown below. What is the resistance of P?
\n\"Samacheer
\na) 1.5 \u03a9
\nb) 2.5 \u03a9
\nc) 3.5 \u03a9
\nd) 4.5 \u03a9
\nAnswer:
\nc) 3.5 \u03a9
\nSolution:
\nBy Kirchhoff\u2019s voltage law,
\n9 = (3 \u00d7 1) + (2.5 \u00d7 1) + (P \u00d7 1)
\n9 =3 + 2.5 + P
\n9 = 5.5 + P
\nP = 9 – 5.5
\nP = 3.5 \u03a9<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nWhat is the current drawn out from the battery?
\n\"Samacheer
\na) 1 A
\nb) 2 A
\nc) 3 A
\nd) 4 A
\nSolution:
\n\"Samacheer<\/p>\n

Question 12.
\nThe temperature deficient of resistance of a wire is 0.00125 per\u00b0C At 20\u00b0C, its resistance is 1\u03a9. The resistance of the wire will be 2\u03a9 at
\na) 800\u00b0C
\nb) 700\u00b0C
\nc) 850\u00b0C
\nd) 820\u00b0C
\nAnswer:
\nd) 820\u00b0C<\/p>\n

Question 13.
\nThe internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 \u03a9 is
\na) 0.2 \u03a9
\nb) 0.5 \u03a9
\nc) 0.8 \u03a9
\nd) 1.0 \u03a9
\nAnswer:
\nb) 0.5 \u03a9
\nSolution:
\n\\(r=\\left(\\frac{E-V}{V}\\right) R\\)
\nE = 2.1 V
\nV = IR = 0.2 \u00d7 10
\nV = 2 V
\n\\(\\begin{aligned}
\nr &=\\left(\\frac{2.1-2}{2}\\right) 10 \\\\
\n&=\\frac{0.1}{2} \\times 10
\n\\end{aligned}\\)
\nr = 0.5 \u03a9<\/p>\n

Question 14.
\nA piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
\na) each of them increases
\nb) each of them decreases
\nc) copper increases and germanium decreases
\nd) copper decreases and germanium increases
\nAnswer:
\nd) copper decreases and germanium increases
\nSolution:
\nCopper is a positive temperature coefficient of resistance, i.e their resistance decreases with a decrease in temperature. Germanium is a negative temperature coefficient of resistance, i.e their resistance increases with a decrease in temperature.<\/p>\n

Question 15.
\nIn Joule\u2019s heating law, when R and t are constant, if the H is taken along the y axis and I2<\/sup> along the x-axis, the graph is _______
\na) straight line
\nb) parabola
\nc) circle
\nd) ellipse
\nAnswer:
\na) straight line
\nSolution:
\nH = I2<\/sup>RT
\nH \u03b1 I2<\/sup>
\n\u2234 The graph is a straight line.<\/p>\n

\"Samacheer<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nWhy current is a scalar?
\nAnswer:
\nThe current I is defined as the scalar product of current density and area vector in which the charges cross.
\nI = \\(\\vec { j } \\) . \\(\\vec { A } \\)
\nThe dot product of two vector quantity is a scalar form. Hence, the current is called a scalar quantity.<\/p>\n

Question 2.
\nDefine Current density.
\nAnswer:
\nThe current density is defined as the current per unit area of the cross-section of the conductor.
\nJ = \\(\\mathrm{I} \/ \\mathrm{A}\\)
\nS.I unit is A\/m2<\/sup><\/p>\n

Question 3.
\nDistinguish between drift velocity and mobility.
\nAnswer:
\n\"Samacheer<\/p>\n

Question 4.
\nState microscopic form of Ohm\u2019s law.
\nAnswer:
\nThe macroscopic form of Ohm\u2019s Law relates voltage, current, and resistance. Ohm\u2019s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object\u2019s resistance.
\nV = IR.<\/p>\n

Question 5.
\nState macroscopic form of Ohm\u2019s law.
\nAnswer:
\nThe macroscopic form of Ohm\u2019s law is V = IR
\nWhere V \u2192 Potential difference
\nI \u2192 Current
\nR \u2192 Resistance.<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nWhat are the ohmic and non-ohmic devices?
\nAnswer:
\nMaterials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm\u2019s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm\u2019s law are said to be non-ohmic.<\/p>\n

Question 7.
\nDefine electrical resistivity.
\nAnswer:
\nThe electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having a unit area of cross-section.
\n\\(\\rho=\\frac{R A}{L}\\)
\nIts unit is ohm-meter (\u03a9m)<\/p>\n

Question 8.
\nDefine temperature coefficient of resistance.
\nAnswer:
\nThe ability of certain metals, their compounds, and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.<\/p>\n

Question 9.
\nWhat are superconductors?
\nAnswer:<\/p>\n

    \n
  1. The resistance of certain materials becomes zero below a certain temperature, known as critical or transition temperature.<\/li>\n
  2. The materials that exhibit this property are known as superconductors and the phenomenon is known as superconductivity.<\/li>\n<\/ol>\n

    Question 10.
    \nWhat are electric power and electric energy?
    \nAnswer:<\/p>\n\n\n\n\n\n\n
    Electric Energy<\/td>\nElectric Power<\/td>\n<\/tr>\n
    Electric potential energy gained by the charge carries as they move through potential difference v is dw = v dQ<\/td>\nElectric Power is the rate at which energy is transformed.<\/p>\n

    P = dw\/dt<\/td>\n<\/tr>\n

    dw = P dt<\/td>\nP = VI<\/td>\n<\/tr>\n
    Unit: 1 kWh<\/td>\nUnit: Watt<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    \"Samacheer<\/p>\n

    Question 11.
    \nDerive the expression for power P = VI in an electrical circuit.
    \nAnswer:
    \nPower is defined as the rate at which the electrical potential energy is delivered.
    \n\\(\\begin{array}{l}
    \n\"Samacheer<\/p>\n

    Question 12.
    \nWrite down the various forms of expression for power in electrical circuit.
    \nAnswer:
    \nThe electric power P is the rate at which the electrical potential energy is delivered,
    \nP = [latex]\\frac { dU }{ dt }\\) = \\(\\frac { 1 }{ dt }\\) (V.dQ) = V.\\(\\frac { dQ }{ dt }\\)
    \n[dU = V.dQ]
    \nThe electric power delivered by the battery to any electrical system.
    \nP = VI
    \nThe electric power delivered to the resistance R is expressed in other forms.
    \nP = VI = I(IR) = I2<\/sup>R
    \nP = IV = \\((\\frac { V }{ R })\\) V = \\(\\frac {{ V }^{2}}{ R }\\).<\/p>\n

    Question 13.
    \nState Kirchhoff\u2019s current rule.
    \nAnswer:<\/p>\n

      \n
    1. Kirchhoff’s current rule states that the algebraic sum of the currents at any junction of a circuit is zero.
      \ni.e \u03a3junction<\/sub> i = 0<\/li>\n
    2. It is a statement of conservation of electric charge.<\/li>\n<\/ol>\n

      Question 14.
      \nState Kirchhoff\u2019s voltage rule.
      \nAnswer:
      \nIt states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.<\/p>\n

      Question 15.
      \nState the principle of the potentiometer.
      \nAnswer:
      \nThe emf (E) of the cell is directly proportional to the balancing length (l)
      \n\u03be \u221d l
      \n\u03be = Irl
      \nWhere I \u2192current
      \nr \u2192 resistance per unit length of the wire.<\/p>\n

      \"Samacheer<\/p>\n

      Question 16.
      \nWhat do you mean by the internal resistance of a cell?
      \nAnswer:
      \nThe resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equal to its emf. But a real battery is made of electrodes and electrolytes, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with aging.<\/p>\n

      Question 17.
      \nState Joule\u2019s law of heating.
      \nAnswer:
      \nThe heat developed in an electrical circuit due to the flow of current varies directly as<\/p>\n

        \n
      • the square of the current<\/li>\n
      • the resistance of the circuit and<\/li>\n
      • the time of flow H = I2<\/sup>Rt<\/li>\n<\/ul>\n

        Question 18.
        \nWhat is the Seebeck effect?
        \nAnswer:
        \nSeebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.<\/p>\n

        Question 19.
        \nWhat is the Thomson effect?
        \nAnswer:<\/p>\n

          \n
        1. When two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result, me potential difference is created between these points.<\/li>\n
        2. This is known as the Thomson effect. Thomson effect is also reversible.<\/li>\n<\/ol>\n

          Question 20.
          \nWhat is the Peltier effect?
          \nAnswer:
          \nWhen an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as the Peltier effect.<\/p>\n

          Question 21.
          \nState the applications of the Seebeck effect.
          \nAnswer:<\/p>\n

            \n
          1. The Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.<\/li>\n
          2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.<\/li>\n
          3. The Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.<\/li>\n<\/ol>\n

            \"Samacheer<\/p>\n

            III. Long Answer Questions:<\/span><\/p>\n

            Question 1.
            \nDescribe the microscopic model of current and obtain the general form of Ohm’s law.
            \nAnswer:<\/p>\n

              \n
            1. XY is a conductor of an area of cross-section A<\/li>\n
            2. E is the applied electric field.<\/li>\n
            3. n is the number of electrons per unit volume with the same drift velocity (vd<\/sub>)<\/li>\n<\/ol>\n

              Let electrons move through a distance dx in time interval dt.
              \n\"Samacheer
              \nMicroscopic model of the current
              \n\u2234 \\(v d=\\frac{d x}{d t}\\)
              \ndx = vd<\/sub> . dt …………(1)<\/p>\n

              Electrons available in the given volume
              \n= volume \u00d7 number per unit volume
              \n= A. dx \u00d7 n
              \n= A . vd<\/sub> dt \u00d7 n [From (1)]<\/p>\n

              Total charge in volume dQ = charge \u00d7 number of electrons
              \ndQ = (e) (Avd<\/sub> dt) n<\/p>\n

              \"Samacheer<\/p>\n

              Question 2.
              \nObtain the macroscopic form of Ohm\u2019s law from its microscopic form and discuss its limitation.
              \nAnswer:<\/p>\n

                \n
              1. Consider a segment of wire of length l and area of cross section A<\/li>\n
              2. Electric field is created when a potential difference V is applied<\/li>\n
              3. If \u2018E\u2019 is uniform, then, V = El<\/li>\n
              4. Current density, J = \u03c3 E = \u03c3\\(\\frac{\\mathrm{V}}{l}\\)<\/li>\n<\/ol>\n

                Since J = \\(\\frac{\\mathrm{I}}{\\mathrm{A}}\\)
                \n\\(\\frac{\\mathrm{I}}{\\mathrm{A}}=\\sigma \\frac{\\mathrm{V}}{l}\\)
                \nRearranging the above equation, Current through the conductor<\/p>\n

                \"Samacheer
                \n\"Samacheer
                \nLimitations:
                \nThere are certain materials and devices where the proportionality of V and I does not hold good.<\/p>\n

                Question 3.
                \nExplain the equivalent resistance of a series and parallel resistor network.
                \nAnswer:
                \nResistors in series:
                \n\"Samacheer
                \n\"Samacheer<\/p>\n

                  \n
                1. Let V1<\/sub>, V2<\/sub>, V3<\/sub> be the potential difference across R1<\/sub>, R2<\/sub>, R3<\/sub><\/li>\n
                2. In the series network, the current is the same.<\/li>\n
                3. The net potential difference V = V1<\/sub> + V2<\/sub> + V3<\/sub><\/li>\n
                4. By Ohm’s law: V1<\/sub> = IR1<\/sub>, V2<\/sub> = IR2<\/sub>, V3<\/sub>= IR3<\/sub><\/li>\n
                5. The equivalent resistance is RS<\/sub> and V = IRS<\/sub> in series.<\/li>\n
                6. IRS<\/sub> = IR1<\/sub> + IR2<\/sub> + IR3<\/sub><\/li>\n
                7. RS<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub><\/li>\n
                8. The equivalent resistance is the sum of the individual resistances.<\/li>\n<\/ol>\n

                  Resistors in Parallel:
                  \n\"Samacheer
                  \n\"Samacheer<\/p>\n

                    \n
                  1. In a parallel network, the potential difference across each resistor is the same.<\/li>\n
                  2. current I is divided into I1<\/sub>, I2<\/sub>, I3<\/sub>, same that
                    \nI = I1<\/sub> + I2<\/sub> + I3<\/sub> ………..(1)
                    \nBy ohm’s law
                    \n\"Samacheer<\/li>\n
                  3. In parallel network, the reciprocal of the equivalent resistance is equal to the sum of the reciprocal of the resistance of the individual resistors.<\/li>\n<\/ol>\n

                    \"Samacheer<\/p>\n

                    Question 4.
                    \nExplain the determination of the internal resistance of a cell using voltmeter.
                    \nAnswer:<\/p>\n

                      \n
                    1. The emf of the cell \u03be is found by connecting a high resistance voltmeter across it. Without connecting the external resistance R.<\/li>\n
                    2. Since the voltmeter draws very little current for deflection, the circuit is called an open circuit.<\/li>\n
                    3. Voltmeter reading gives the emf of the cell.<\/li>\n
                    4. Resistance \u2018R is included in the circuit and current I is established in the circuit.<\/li>\n
                    5. The potential drop across the resistor.
                      \nR is V = IR …………(1)<\/li>\n
                    6. \u00a0Due to internal resistance (r) the voltmeter reads a value V, less than the emf of cell (E).<\/li>\n
                    7. \u00a0V = \u03be – Ir<\/li>\n
                    8. Ir = \u03be – V …………(2)<\/li>\n
                    9. Dividing (2) by (1)
                      \n\"Samacheer<\/li>\n
                    10. Internal resistance of the cell
                      \n\\(\\frac{\\mathrm{Ir}}{\\mathrm{IR}}=\\frac{\\xi-\\mathrm{V}}{\\mathrm{V}}\\)
                      \n\\(\\mathrm{r}=\\left(\\frac{\\xi-\\mathrm{V}}{\\mathrm{V}}\\right) \\mathrm{R}\\)<\/li>\n
                    11. Since \u03be, V and R are known, internal resistance r can be calculated<\/li>\n<\/ol>\n

                      Question 5.
                      \nState and explain Kirchhoff\u2019s rules.
                      \nAnswer:
                      \nKirchhoff\u2019s rules are used to find current and voltage for more complex circuits. There are 2 rules.<\/p>\n

                        \n
                      • Kirchhoff\u2019s current rule.<\/li>\n
                      • Kirchhoff\u2019s voltage rule.<\/li>\n<\/ul>\n

                        Kirchhoff\u2019s current rule:
                        \nThe algebraic sum of the currents at any function of a circuit is zero.<\/p>\n

                        \"Samacheer<\/p>\n

                        Sign convention:<\/p>\n

                          \n
                        1. Current flowing towards a junction is positive<\/li>\n
                        2. Current flowing away from a junction is negative.<\/li>\n
                        3. From the figure
                          \nI1<\/sub> + (-I2<\/sub>) + (-I3<\/sub>) + I4<\/sub> + I5<\/sub> = 0 (or)
                          \nI1<\/sub> + I4<\/sub> + I5<\/sub> = I2<\/sub> + I3<\/sub><\/li>\n<\/ol>\n

                          Kirchhoff\u2019s second [voltage rule or Loop rule]:
                          \nIn a closed circuit, the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit.
                          \n\"Samacheer<\/p>\n

                          Sign convention:<\/p>\n

                            \n
                          1. The product of current and resistance is positive if the current direction is followed (a)<\/li>\n
                          2. If current is opposite to the direction of the loop, then the product of current and resistance is negative (h)<\/li>\n
                          3. The emf is positive when moving from negative to the positive terminal (c) & (d).<\/li>\n<\/ol>\n

                            Question 6.
                            \nObtain the condition for bridge balance in Wheatstone\u2019s bridge.
                            \nAnswer:<\/p>\n

                              \n
                            1. Let P, Q, R, and S be the four resistances in the form of a bridge.<\/li>\n
                            2. A cell of emf E is connected between points A and C<\/li>\n
                            3. The current I is divided into I1<\/sub>, I2<\/sub>, I3,<\/sub> and I4<\/sub> across the four branches.
                              \nBy Kirchhoff\u2019s current rule<\/li>\n
                            4. At junction B
                              \nI1<\/sub> – Ig<\/sub> – I3<\/sub> = 0 ………..(1)
                              \n\"Samacheer<\/li>\n
                            5. At junction D
                              \nI2<\/sub> + Ig<\/sub> – I4<\/sub> = 0 …………..(2)
                              \nBy Kirchoff’s voltage rule<\/li>\n
                            6. At closed path ABDA
                              \nI1<\/sub>P + Ig<\/sub>G – I2<\/sub>R = 0 ……….(3)<\/li>\n
                            7. At closed path ABCDA
                              \nI1<\/sub>P + I3<\/sub>Q – I4<\/sub>S – I2<\/sub>R = 0 ………(4)<\/li>\n
                            8. When the galvanometer shows zero deflection, the points B and D are at the same potential, and Ig<\/sub> = 0<\/li>\n
                            9. Substitute Ig<\/sub> = 0 in (1), (2) and (3)
                              \nI1<\/sub> = I3<\/sub> …………….(5)
                              \nI2<\/sub> = I4<\/sub> …………….(6)
                              \nI1<\/sub>P = I2<\/sub>R …………….(7)<\/li>\n
                            10. Substitute (5) and (6) in (4)
                              \nI1<\/sub>P + I1<\/sub>Q – I2<\/sub>S – I2<\/sub>R = 0
                              \nI1<\/sub>(P + Q) = I2<\/sub>(R + S) …………….(8)
                              \nDividing (8) by (7),
                              \n\"Samacheer
                              \nThis is the condition of bridge balance.<\/li>\n<\/ol>\n

                              \"Samacheer<\/p>\n

                              Question 7.
                              \nExplain the determination of unknown resistance using a meter bridge.
                              \nAnswer:<\/p>\n

                                \n
                              1. It consists of a uniform manganin wire AB of 1-meter length.<\/li>\n
                              2. The wire is stretched along a meter scale on a wooden board between 2 copper strips C and D<\/li>\n
                              3. Between the 2 copper strips, another copper strip E is mounted to enclose gaps G1<\/sub> and G2<\/sub><\/li>\n
                              4. An unknown resistance P is connected in G1<\/sub><\/li>\n
                              5. A standard resistance Q is connected in G2<\/sub>.<\/li>\n
                              6. A jockey is connected to E through a galvanometer (G) and a high resistance (HR)<\/li>\n
                              7. The exact position of the jockey can be read on the scale.<\/li>\n
                              8. A Lechianche cell and a key (k) are connected across the ends of the wire.<\/li>\n
                              9. By measuring the radius r and length<\/li>\n
                              10. l of the wire P1<\/sub> then specific resistance of the wire is found.
                                \nP = Resistance \u00d7 \\(\\frac{A}{l}\\)<\/li>\n
                              11. If l is unknown resistance specific Resistance
                                \n\\(P=\\frac{P \\cdot \\pi r^{2}}{l}\\)
                                \n\"Samacheer<\/li>\n
                              12. The jockey is adjusted for zero deflection. Let the point be J.<\/li>\n
                              13. Lengths AJ and JB replace resistances R and S in Wheatstone’s bridge<\/li>\n
                              14. Then, \\(\\frac{P}{Q}=\\frac{R}{S}=\\frac{\\left.R^{\\prime} A\\right]}{R^{\\prime} J B}\\)
                                \nWhere R’ \u2192 resistance per unit length of wire
                                \n\\(\\frac{P}{Q}=\\frac{A J}{J B}=\\frac{l_{1}}{l_{2}}\\)
                                \n\\(P=Q \\cdot \\frac{l_{1}}{l_{2}}\\)<\/li>\n
                              15. Due to the soldering of copper strips, there occurs an error called end resistance.<\/li>\n
                              16. This can be eliminated by taking another set of readings by interchanging P and Q.<\/li>\n
                              17. By measuring the radius a and length l of the wire P, then specific resistance of the wire is found
                                \n\\(\\rho=\\text { Resistance } \\times \\frac{A}{l}\\)
                                \nIf \\(\\rho\\) is unknown resistance equation becomes
                                \n\\(\\rho=\\frac{P \\cdot \\pi a^{2}}{l}\\)<\/li>\n<\/ol>\n

                                Question 8.
                                \nHow the emf of two cells are compared using a potentiometer?
                                \nAnswer:
                                \nThe primary circuit consists of battery (Bt), Key (k), and rheostat (Rh)<\/p>\n

                                  \n
                                1. The end C is connected to M of a DPPT switch.<\/li>\n
                                2. The terminal N is connected to the jockey (J) through a galvanometer (G) and high resistance (HR)<\/li>\n
                                3. cell 1 is connected between M1<\/sub> and N1<\/sub><\/li>\n
                                4. cell 2 is connected between M2<\/sub> and N2<\/sub><\/li>\n
                                5. The jockey is adjusted for zero deflection.<\/li>\n
                                6. The DPDT has pressed towards \u03be1<\/sub> The potential difference across balancing length l1<\/sub> is Irl1<\/sub><\/li>\n
                                7. \u03be1<\/sub> = Irl1<\/sub> ………..(1)<\/li>\n
                                8. DPDT switch is pressed towards \u03be2<\/sub><\/li>\n
                                9. \u03be2<\/sub> = Irl2<\/sub> ………..(2)
                                  \n\"Samacheer
                                  \nComparison of EMF of two cells<\/li>\n
                                10. Dividing (1) by (2)
                                  \n\\(\\frac{\\xi 1}{\\xi_{2}}=\\frac{l_{1}}{l_{2}}\\)<\/li>\n
                                11. If \u03be1<\/sub> is known, then unknown emf
                                  \n\\(\\xi_{2}=\\xi_{1} \\cdot \\frac{l_{2}}{l_{1}}\\)<\/li>\n
                                12. The experiment can be repeated several times by changing the current emf of flowing through it.<\/li>\n<\/ol>\n

                                  IV. Numeric Problems:<\/span><\/p>\n

                                  Question 1.
                                  \nThe following graphs represent the current versus voltage and voltage versus current for the six conductors A, B, C, D, E, and F. Which conductor has the least resistance and which has maximum resistance?<\/p>\n

                                  \"Samacheer
                                  \nAnswer:
                                  \n\"Samacheer<\/p>\n

                                  Question 2.
                                  \nLightning is a very good example of a natural current. In typical lightning, there is 109<\/sup> J energy transfer across the potential difference of 5 \u00d7 107<\/sup> V during a time interval of 0.2 s. Using this information, estimate the following quantities:
                                  \n(a) the total amount of charge transferred between cloud and ground
                                  \n(b) the current in the lightning bolt
                                  \n(c) the power delivered in 0.2 s.
                                  \n\"Samacheer
                                  \nAnswer:
                                  \nGiven data: E = 109<\/sup>J, V = 5 \u00d7 107<\/sup> V, t = 0.2s
                                  \n\"Samacheer<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 3.
                                  \nA copper wire of 10-6<\/sup> m2<\/sup> area of cross-section, carries a current of 2 A. If the number of electrons per cubic meter is 8 \u00d7 1028<\/sup>, calculate the current density and average drift velocity.
                                  \nAnswer:
                                  \nGiven data:
                                  \nA = 10-6<\/sup> m2<\/sup>, I = 2A, n = 8 \u00d7 1028<\/sup>
                                  \nFormula:
                                  \nCurrent density, J = \\(J=\\frac{I}{A}\\)
                                  \n\"Samacheer<\/p>\n

                                  Question 4.
                                  \nThe resistance of a nichrome wire at 0\u00b0C is 10\u03a9. If its temperature coefficient of resistivity of nichrome is 0.004\/ \u00b0C, find its resistance of the wire at boiling point of water. Comment on the result.
                                  \nAnswer:
                                  \nGiven data:
                                  \nT1<\/sub> = 10\u00b0C, R0<\/sub> = 10\u03a9
                                  \nThe boiling point of water T = 100\u00b0C
                                  \n\u03b1 = 0.0041\u00b0C
                                  \nRT<\/sub>=?
                                  \nFormula:
                                  \nRT<\/sub> = R0 [l + \u03b1(T – T0<\/sub>)]
                                  \nRT<\/sub> = 10 [1+0.004(100 – 20)]
                                  \n= 10(1 + 0.32)
                                  \n= 10 (1.32)
                                  \nRT = 13.2\u03a9
                                  \nAs the temperature increase the resistance of the wire also increases.<\/p>\n

                                  Question 5.
                                  \nThe rod given in the figure is made up of two different materials.<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Both have square cross-sections of 3 mm side. The resistivity of the first material is 4 \u00d7 10-3<\/sup>\u03a9 and that of the second material has a resistivity of 5 \u00d7 10-3<\/sup>\u03a9m. What is the resistivity of the rod between its ends?
                                  \nAnswer:
                                  \n\"Samacheer<\/p>\n

                                  Question 6.
                                  \nThree identical lamps each having a resistance R are connected to the battery of emf E as shown in the figure. Suddenly the switch S is closed.
                                  \n(a) Calculate the current in the circuit when S is open and closed
                                  \n(b) What happens to the intensities of the bulbs A, B, and C.
                                  \n(c) Calculate the voltage across the three bulbs when S is open and closed
                                  \n(d) Calculate the power delivered to the circuit when S is opened and closed
                                  \n(e) Does the power delivered to the circuit decreases, increases or remain the same?
                                  \n\"Samacheer
                                  \nAnswer:
                                  \n\"Samacheer<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 7.
                                  \nAn electronics hobbyist is building a radio which requires 150\u03a9 in her circuit, but she has only 220\u03a9, 79\u03a9, and 92\u03a9 resistors available. How can she connect the available resistors to get the desired value of resistance?
                                  \nAnswer:
                                  \nAvailable resistances = 220\u03a9, 79\u03a9 92\u03a9
                                  \nCase I:
                                  \nIf 3 resistors are connected in series, then
                                  \nRS<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub> = 220 + 79 + 92 = 391
                                  \nThis value is greater than the required resistance so it is not possible.
                                  \nCase II:
                                  \nIf 3 resistors are connected in parallel, then
                                  \n\"Samacheer<\/p>\n

                                  Question 8.
                                  \nA cell supplies a current of 0.9 A through a 2\u03a9 resistor and a current of 0.3 A through a 7\u03a9 resistor. Calculate the internal resistance of the cell.
                                  \nAnswer:
                                  \n\"Samacheer
                                  \n\u03be = 0.3(7 + r)
                                  \nSince ‘\u03be’ is constant
                                  \n0.9(2+r) = 0.3(7+r)
                                  \n1.8 + 0.9r = 2.1 + 0.3r
                                  \nO.6r = 0.3
                                  \nr = \\(\\frac{0.3}{0.6}\\)
                                  \nr = \\(\\frac{1}{2}\\) = O.5\u03a9<\/p>\n

                                  Question 9.
                                  \nCalculate the currents in the following circuit.
                                  \n\"Samacheer
                                  \nAnswer:
                                  \nAt junction B, applying current law,
                                  \nI1<\/sub> – I2<\/sub> – I3<\/sub> = 0
                                  \nI1<\/sub> = I2<\/sub> + I3<\/sub> ………..(1)
                                  \nKirchhoff\u2019s voltage law in loop ABEFA
                                  \n100I3<\/sub> + 100I1<\/sub> = 15
                                  \nUsing (1), we get,
                                  \n100I3<\/sub> + 100 (I2<\/sub> + I3<\/sub>) = 15
                                  \n100I3<\/sub> + 100I2<\/sub> + 1003<\/sub> = 15
                                  \n100I2<\/sub> + 2003<\/sub> = 15 ……………(2)
                                  \nVoltage law in loop BCDEB
                                  \n100I2<\/sub> – 100I3<\/sub> = -9
                                  \n100I3<\/sub> – 100I2<\/sub> = 9 ……………(3)
                                  \nAdding (1) & (2), we get
                                  \n\\(\\begin{array}{l}
                                  \n200 \\mathrm{I}_{3}+100 \\mathrm{I}_{2}=15 \\\\
                                  \n100 \\mathrm{I}_{3}-100 \\mathrm{I}_{2}=9 \\\\
                                  \n300 \\mathrm{I}_{3} \\quad=24 \\quad \\mathrm{I}_{3}=\\frac{24}{300}=0.08
                                  \n\\end{array}\\)
                                  \nI3<\/sub> = 0.08A
                                  \n(3) \u2192 100I3<\/sub> – 100I2<\/sub> = 9
                                  \nSubstituting the value of I3<\/sub>, we get
                                  \n100 \u00d7 0.08 – 100I2<\/sub> = 9
                                  \n8 – 100I2<\/sub> = 9
                                  \n-100I2<\/sub> = 1
                                  \n\\(\\Rightarrow \\mathrm{I}_{2}=\\frac{-1}{100}=-0.01 \\mathrm{~A}\\)
                                  \nI2<\/sub> = -0.01A
                                  \nFrom (1),
                                  \nI1<\/sub> = I2<\/sub> + I3<\/sub>
                                  \n= -0.01 + 0.08
                                  \nI1<\/sub> = 0.07A<\/p>\n

                                  Question 10.
                                  \nA potentiometer wire has a length of 4 m and a resistance of 20\u03a9. It is connected in series with a resistance of 2980\u03a9 and a cell of emf 4 V. Calculate the potential along the wire.
                                  \nAnswer:
                                  \nGiven data:
                                  \nl = 4m of R = 20\u03a9
                                  \nIn series with R’ = 2980\u03a9
                                  \nE = 4 V
                                  \n\"Samacheer<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 11.
                                  \nDetermine the current flowing through the galvanometer (G) as shown in the figure.
                                  \n\"Samacheer
                                  \nAnswer:
                                  \nI2<\/sub> = I – I1<\/sub>
                                  \nCurrent flowing through the circuit I = 2A
                                  \nApplying Kirchhoffs II law to PQSP
                                  \n5I1<\/sub> + 10Ig<\/sub> – 15I2<\/sub> = 0
                                  \n5I1<\/sub> + 10Ig<\/sub> – 15(I – I1<\/sub>) = 0
                                  \n5I1<\/sub> + 10Ig<\/sub> – 15I + 15I1<\/sub> = 0
                                  \n20I1<\/sub> + 10Ig<\/sub> = 15 \u00d7 I
                                  \n20I1<\/sub> + 10Ig<\/sub> = 30
                                  \n2I1<\/sub> + Ig<\/sub> = 3 ………..(1)
                                  \nApplying Kirchhoffs II law to QRSQ
                                  \n10(I1<\/sub> – Ig<\/sub>) – 20 (I2<\/sub> + Ig<\/sub>) – 10 Ig<\/sub> = 0
                                  \n10I1<\/sub> – 10 Ig<\/sub> – 20 (I – I1<\/sub> + Ig<\/sub>) – 10 Ig<\/sub> = 0
                                  \n10I1<\/sub> – 10 Ig<\/sub> – 20I + 20I1<\/sub> – 20Ig<\/sub> – 10Ig<\/sub> = 0
                                  \n30I1<\/sub> – 40Ig<\/sub> = 20I
                                  \n30I1<\/sub> – 40Ig<\/sub> = 20 \u00d7 2
                                  \n3I1<\/sub> – 4Ig<\/sub> = 4 ………..(2)
                                  \n(1) \u00d7 3 \u21d2 6I1<\/sub> + 3Ig<\/sub> = 9 …………..(3)
                                  \n(2) \u00d7 2 \u21d2 6I1<\/sub> – 8Ig<\/sub> = 8 ……………(4)
                                  \nSolving (3) and (4)
                                  \n+ 11 Ig<\/sub> = 1
                                  \n\\(\\mathrm{I}_{\\mathrm{g}}=\\frac{1}{11} \\mathrm{~A}\\)<\/p>\n

                                  Question 12.
                                  \nTwo cells each of 5V are connected in series across an 8\u03a9 resistor and three parallel resistors of 4\u03a9, 6\u03a9, and 12\u03a9. Draw a circuit diagram for the above arrangement. Calculate
                                  \n(i) the current drawn from the cell
                                  \n(ii) current through each resistor
                                  \nAnswer:
                                  \nCircuit Diagram:
                                  \n\"Samacheer
                                  \ni) Current drawn from the cell
                                  \nEeq<\/sub> = 5+5 = 10V
                                  \nReff<\/sub> = Rs<\/sub>+ Rp<\/sub>
                                  \nRs<\/sub> = 8\u03a9
                                  \n\\(\\frac{1}{R_{P}}=\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}=\\frac{18+12+6}{72}=\\frac{36}{72}\\)
                                  \n\u2234Rp<\/sub> = \\(\\frac{72}{36}\\) = 2\u03a9
                                  \nReff<\/sub> = 8 + 2 = 10\u03a9
                                  \n\\(I=\\frac{E_{e g}}{R_{e f f}}=\\frac{10}{10}=1 A\\)
                                  \n\u2234 I = 1A
                                  \nVoltage drop V= IR =1 \u00d7 2 = 2V
                                  \nV = 2V<\/p>\n

                                  ii) Current through each resistor.
                                  \n\u2234 current in 4\u03a9 resistor
                                  \nI = \\(\\frac{2}{4}\\) = O.5A
                                  \nI = O.5A
                                  \ncurrent in 6\u03a9 resistor, I = \\(\\frac{2}{6}\\) = 0.33A
                                  \nI = 0.33A
                                  \ncurrent in 12\u03a9 resistor, I = \\(\\frac{2}{12}\\) = \\(\\frac{1}{6}\\) = 0.17A
                                  \nI = 0.17A<\/p>\n

                                  Question 13.
                                  \nFour light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.<\/p>\n\n\n\n\n\n\n\n
                                  <\/td>\nP<\/td>\nQ<\/td>\nR<\/td>\nS<\/td>\n<\/tr>\n
                                  P removed<\/td>\n*<\/td>\non<\/td>\non<\/td>\non<\/td>\n<\/tr>\n
                                  Q removed<\/td>\non<\/td>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 *<\/td>\non<\/td>\noff<\/td>\n<\/tr>\n
                                  R removed<\/td>\noff<\/td>\noff<\/td>\n*<\/td>\noff<\/td>\n<\/tr>\n
                                  S removed<\/td>\non<\/td>\noff<\/td>\non<\/td>\n*<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                  Draw the circuit diagram for these bulbs.
                                  \nAnswer:
                                  \n\"Samacheer<\/p>\n

                                  Question 14.
                                  \nIn a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
                                  \nAnswer:
                                  \nE1<\/sub> = 1.25 V
                                  \nl1<\/sub> = 35 cm
                                  \nE2<\/sub> =?
                                  \nl2<\/sub> = 63cm
                                  \nimg 81
                                  \nEmf of the second cell is 2.25 V
                                  \n\"Samacheer<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Part II:<\/span><\/p>\n

                                  12th Physics Guide Current Electricity Additional Questions and Answers<\/h3>\n

                                  I(a). Match the following:<\/span><\/p>\n\n\n\n\n\n\n
                                  a. current<\/td>\ni. coulomb<\/td>\n<\/tr>\n
                                  b. Drift velocity<\/td>\nii. m2<\/sup>\/Vs<\/td>\n<\/tr>\n
                                  c. charge<\/td>\niii. m\/s<\/td>\n<\/tr>\n
                                  d. mobility<\/td>\niv. Ampere<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                  Answer:
                                  \na. iv
                                  \nb. iii
                                  \nc. i
                                  \nd. ii<\/p>\n

                                  I(b). Match the following:<\/span><\/p>\n\n\n\n\n\n\n
                                  a. microscopic form of Ohm’s law<\/td>\ni. V = IR<\/td>\n<\/tr>\n
                                  b. microscopic form of Ohm’s law<\/td>\nii. a \\(\\tau\\)<\/td>\n<\/tr>\n
                                  c. Drift velocity (vd<\/sub>)<\/td>\niii.\\(\\overrightarrow{\\mathrm{J}}=\\frac{\\mathrm{ne}^{2} \\tau}{\\mathrm{m}} \\overrightarrow{\\mathrm{E}}\\)<\/td>\n<\/tr>\n
                                  d. Current density (J)<\/td>\niv. nevd<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                  Answer:
                                  \na. iii
                                  \nb. i
                                  \nc. ii
                                  \nd. iv<\/p>\n

                                  I(c). Match the following:<\/span><\/p>\n\n\n\n\n\n\n
                                  a. Heating element<\/td>\ni. 3000\u00b0<\/td>\n<\/tr>\n
                                  b. Fuse wire<\/td>\nii. Copper<\/td>\n<\/tr>\n
                                  c. carbon are furnace<\/td>\niii. Copper<\/td>\n<\/tr>\n
                                  d. Electrical lamp<\/td>\niv. Nichrome<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                  Answer:
                                  \na. iv
                                  \nb. iii
                                  \nc. i
                                  \nd. ii<\/p>\n

                                  I(d). Match the following:<\/span><\/p>\n\n\n\n\n\n\n
                                  a. Power<\/td>\ni. Ohm-metre<\/td>\n<\/tr>\n
                                  b. Energy<\/td>\nii. Watt<\/td>\n<\/tr>\n
                                  c. electrical resistivity<\/td>\niii. \u03a91<\/sub> m \u03a91<\/sub><\/td>\n<\/tr>\n
                                  d. electrical conductivity<\/td>\niv. Joule<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                  Answer:
                                  \na. ii
                                  \nb. iv
                                  \nc. i
                                  \nd. iii<\/p>\n

                                  \"Samacheer<\/p>\n

                                  II. Fill in the Blanks.<\/span>
                                  \nQuestion 1.
                                  \nThe graph plotted with current against the voltage that obeys Ohm\u2019s law is _______.
                                  \nAnswer:
                                  \nStraight line<\/p>\n

                                  Question 2.
                                  \nCurrent is a _______ quantity.
                                  \nAnswer:
                                  \nScalar<\/p>\n

                                  Question 3.
                                  \n________ is used in complicated circuits.
                                  \nAnswer:
                                  \nKirchhoff’s laws<\/p>\n

                                  Question 4.
                                  \nMercury exhibits superconductor behaviour at _______.
                                  \nAnswer:
                                  \n4.2K<\/p>\n

                                  III. Choose the Odd One Out:<\/span><\/p>\n

                                  Question 1.
                                  \na) Current
                                  \nb) current density
                                  \nc) drift velocity
                                  \nd) Electric field
                                  \nAnswer:
                                  \na) Current – it is scalar and others are vector.<\/p>\n

                                  Question 2.
                                  \na) Current
                                  \nb) resistance
                                  \nc) time
                                  \nd) electromotive force
                                  \nAnswer:
                                  \nd) electromotive force – because heat dissipated depends only on the first 3 factors.<\/p>\n

                                  Question 3.
                                  \na) Current rule
                                  \nb) Voltage rule
                                  \nc) Wheatstone\u2019s bridge
                                  \nd) Joule\u2019s law
                                  \nAnswer:
                                  \nd) Joule\u2019s law – Because the first 3 comes under Kirchhoff\u2019s laws.<\/p>\n

                                  Question 4.
                                  \na) 1kWh
                                  \nb) 1Js-1
                                  \n<\/sup>c) 1000Wh
                                  \nd) 3.6 \u00d7 106<\/sup>J
                                  \nAnswer:
                                  \nb) 1Js-1<\/sup> – Others are equivalent of 1 unit of electrical energy<\/p>\n

                                  \"Samacheer<\/p>\n

                                  IV. Choose the Incorrect Pair:<\/span><\/p>\n

                                  Question 1.
                                  \na) Ohmic material – Straight line
                                  \nb) non – Ohmic material – non-linear
                                  \nc) macroscope form – V = IR
                                  \nd) macroscope form – V = El
                                  \nAnswer:
                                  \nd) macroscope form – V = El<\/p>\n

                                  Question 2.
                                  \na) Electrical resistivity – \u03a9m
                                  \nb) Electrical conductivity – \u03a9-1<\/sup> m-1<\/sup>
                                  \nc) Internal resistance – \u03a9m-1<\/sup>
                                  \nd) Electrical resistance – \u03a9
                                  \nAnswer:
                                  \nc) Internal resistance – \u03a9m-1<\/sup><\/p>\n

                                  Question 3.
                                  \na) Resistors in series – Rs<\/sub> = R1<\/sub> + R2<\/sub>
                                  \nb) Resistors in parallel – Rp<\/sub> = \\(\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\)
                                  \nc) Internal resistance – r = \\(\\left(\\frac{\\mathrm{E}-\\mathrm{V}}{\\mathrm{V}}\\right) \\mathrm{R}\\)
                                  \nd) Energy – E = VI
                                  \nAnswer:
                                  \nb) Resistors in parallel – R = Rp<\/sub> = \\(\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\)<\/p>\n

                                  Question 4.
                                  \na) Electromotive force – force acting
                                  \nb) Electric energy – work done
                                  \nc) Electric power – rate of change in electric energy
                                  \nd) Internal resistance – resistance offered
                                  \nAnswer:
                                  \na) Electromotive force – force acting because emf is the potential difference<\/p>\n

                                  V. Choose the Correct Pair:<\/span><\/p>\n

                                  Question 1.
                                  \na) Household appliances – Parallel network
                                  \nb) HousehoLd appliances – Series network
                                  \nc) \\(\\frac{1}{\\mathrm{R}_{\\mathrm{P}}}\\) – R1<\/sub> + R2<\/sub>
                                  \nd) RRp<\/sub> – \\(\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\)
                                  \nAnswer:
                                  \na) Household appliances \u2014 Parallel network<\/p>\n

                                  Question 2.
                                  \nColour code in carbon resistor
                                  \na) Brown ring – 0
                                  \nb) Yellow ring – 1
                                  \nc) Silver ring – 10%
                                  \nd) Colourless – 5%
                                  \nAnswer:
                                  \na) Silver – 10%<\/p>\n

                                  Question 3.
                                  \na) Voltmeter – used to measure current
                                  \nb) Ammeter – used to measure voltage
                                  \nc) Multimeter – used to measure voltage current and resistance
                                  \nAnswer:
                                  \nc) Multimeter – used to measure voltage current and resistance<\/p>\n

                                  Question 4.
                                  \na) Electric heater – tungsten
                                  \nb) Electric fuse – trippers
                                  \nc) Electric furnace – Nichrorne
                                  \nd) Electric lamp – Carbon arc
                                  \nAnswer:
                                  \nb) Electric fuse – trippers<\/p>\n

                                  \"Samacheer<\/p>\n

                                  VI. Assertion and Reason:<\/span><\/p>\n

                                  Question 1.
                                  \nA: Fuses are connected in series to protect electric devices.
                                  \nR: It melts and breaks the circuit if the current exceeds a certain value.
                                  \na) R does not explain A
                                  \nb) R explains A
                                  \nc) A is correct, R is wrong
                                  \nd) Both A and R are wrong
                                  \nAnswer:
                                  \nb) R explains A<\/p>\n

                                  Question 2.
                                  \nA: Positive charge flows from higher electric potential to lower electric potential.
                                  \nR: These positive charges constitute the flow of current.
                                  \na) R does not explain A
                                  \nb) R explains A
                                  \nc) A is correct, R is wrong
                                  \nd) Both A and R are wrong
                                  \nAnswer:
                                  \nc) A is correct, R is wrong<\/p>\n

                                  Question 3.
                                  \nA: Current is a scalar quantity.
                                  \nR: Current does not obey the law of vector addition.
                                  \na) R does not explain A
                                  \nb) R explains A
                                  \nc) A is correct, R is wrong
                                  \nd) Both A and R are wrong
                                  \nAnswer:
                                  \nb) R explains A<\/p>\n

                                  Question 4.
                                  \nA: A semiconductor with a negative temperature coefficient of resistance is called a thermopile.
                                  \nR: The value of a remains the same for all materials
                                  \na) R does not explain A
                                  \nb) R explains A
                                  \nc) A is correct, R is wrong
                                  \nd) Both A and R are wrong
                                  \nAnswer:
                                  \nd) Both A and R are wrong<\/p>\n

                                  VII. Choose the Correct Statement:<\/span><\/p>\n

                                  Question 1.
                                  \na) All carbon resistors have a tolerance value.
                                  \nb) The positive charges constitute the electric current
                                  \nc) Allessandro Volta invented the telephone
                                  \nd) Ohm’s law is used for complicated circuits.
                                  \nAnswer:
                                  \na) All carbon resistors have a tolerance value<\/p>\n

                                  Question 2.
                                  \na) The innermost electrons in the atom is called the free electrons.
                                  \nb) All atoms are neutral with an equal number of protons and electrons.
                                  \nc) Conventional current flows from negative to positive charges.
                                  \nd) The outermost electrons are core electrons.
                                  \nAnswer:
                                  \nb) All atoms are neutral with an equal number of protons and electrons<\/p>\n

                                  Question 3.
                                  \na) Charging the battery on my mobile.
                                  \nb) My mobile phone battery has no charge
                                  \nc) Battery has no charge.
                                  \nd) My mobile is charging.
                                  \nAnswer:
                                  \nd) My mobile is charging.<\/p>\n

                                  Question 4.
                                  \na) Ohmic materials do not have constant resistance.
                                  \nb) The V – I graph for ohmic materials is non-linear.
                                  \nc) Non-ohmic materials do not have constant resistance.
                                  \nd) The V – I graph for non-ohmic materials is linear. constant resistance.
                                  \nAnswer:
                                  \nc) Non-ohmic materials do not have constant resistance.<\/p>\n

                                  \"Samacheer<\/p>\n

                                  VIII. Choose the Incorrect Statement:<\/span><\/p>\n

                                  Question 1.
                                  \na) The value of the equivalent resistance in series connection is lesser than each individual resistance.
                                  \nb) The value of equivalent resistance in the parallel network is lesser than individual resistance.
                                  \nc) The value of equivalent resistance in the series network is Rs<\/sub> = R1<\/sub> + R2<\/sub>
                                  \nd) The value of equivalent resistance in parallel network is \\(\\frac{1}{\\mathrm{R}_{\\mathrm{P}}}=\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\)
                                  \nAnswer:
                                  \na) The value of the equivalent resistance in series connection is lesser than each individual resistance.<\/p>\n

                                  Question 2.
                                  \na) Superconductivity was discovered by H.K. Onnes.
                                  \nb) The resistance of certain materials becomes zero.
                                  \nc) That temperature is called critical temperature.
                                  \nd) Experiments showed Ag exhibits superconductivity at 4.2K
                                  \nAnswer:
                                  \nd) Experiments showed Ag exhibits superconductivity at 4.2K<\/p>\n

                                  Question 3.
                                  \na) Electric power produced by a resistor is I\u201cR.
                                  \nb) It depends on the square of the current.
                                  \nc) If the current is doubled, the power will increase by 8 times.
                                  \nd) The SI unit of power is a watt.
                                  \nAnswer:
                                  \nc) If the current is doubled, the power will increase by 8 times.<\/p>\n

                                  Question 4.
                                  \na) Kirchhoff’s current law obeys the laws of conservation of charges
                                  \nb) Kirchhoff’s voltage law obeys the laws of conservation of energy
                                  \nc) Important application of Kirchhoff s rules is Wheatstone’s bridge
                                  \nd) This rule is used for simple electrical circuits.
                                  \nAnswer:
                                  \nd) This rule is used for simple electrical circuits.<\/p>\n

                                  \"Samacheer<\/p>\n

                                  IX. Choose the best answer:<\/span><\/p>\n

                                  Question 1.
                                  \nWhen current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
                                  \n(a) \\(\\frac { v }{ 4 }\\)
                                  \n(b) \\(\\frac { v }{ 2 }\\)
                                  \n(c) v
                                  \n(d) 2 v
                                  \nAnswer:
                                  \n(c) v
                                  \nHint:
                                  \nVd<\/sub> = \\(\\frac { 1 }{ nAe }\\); v’d<\/sub> = \\(\\frac { 2I }{ (2A)ne }\\) = vd<\/sub><\/p>\n

                                  Question 2.
                                  \nThe resistance of the wire varies inversely as _______.
                                  \na) Area of the cross-section
                                  \nb) Resistivity
                                  \nc) Length
                                  \nd) Temperature
                                  \nAnswer:
                                  \na) Area of the cross-section<\/p>\n

                                  Question 3.
                                  \nThe curve representing Ohms law is a _______.
                                  \na) linear
                                  \nb) cosme function
                                  \nc) parabola
                                  \nd) Hyperbola
                                  \nAnswer:
                                  \na) linear<\/p>\n

                                  Question 4.
                                  \nA 10 m long wire of resistance 20\u2126 is connected in series with a battery of emf 3V and a resistance of 10 \u2126. The potential gradient along the wire in volt per meter is
                                  \n(a) 6.02
                                  \n(b) 0.1
                                  \n(c) 0.2
                                  \n(d) 1.2
                                  \nAnswer:
                                  \n(c) 0.2
                                  \nHint:
                                  \nPotential difference across the wire = \\(\\frac { 20 }{ 3 }\\) x 3 = 2 V
                                  \nPotential gradient = \\(\\frac { v }{ l }\\) = \\(\\frac { 2 }{ 10 }\\) = 0.2 V\/m<\/p>\n

                                  Question 5.
                                  \nTo produce an electric current what is the requirement?
                                  \na) A voltage source
                                  \nb) a source of energy that moves charges
                                  \nc) an electric field
                                  \nd) Any of the above
                                  \nAnswer:
                                  \nd) Any of the above<\/p>\n

                                  Question 6.
                                  \nIn Joule\u2019s heating law, when R and t are constant, if the H is taken along the y-axis and I along the x-axis, the graph is:
                                  \na) straight line
                                  \nb) parabola
                                  \nc) circle
                                  \nd) ellipse
                                  \nAnswer:
                                  \na) straight line<\/p>\n

                                  Question 7.
                                  \nA series circuit consists of 3 resistors with 140\u03a9, 250\u03a9, and 220\u03a9. The total resistance is _______.
                                  \na) 330\u03a9
                                  \nb) 610\u03a9
                                  \nc) 720\u03a9
                                  \nd) None of the above
                                  \nAnswer:
                                  \nb) 610\u03a9<\/p>\n

                                  Question 8.
                                  \nA cell has an emf of 1.5 V. When short-circuited, it gives a current of 3A. The internal resistance of the cell is
                                  \n(a) 0.5 \u2126
                                  \n(b) 2.0 \u2126
                                  \n(c) 4.5 \u2126
                                  \n(d) \\(\\frac { 1 }{ 4.5 }\\) \u2126
                                  \nAnswer:
                                  \n(a) 0.5 \u2126
                                  \nHint:
                                  \nr = \\(\\frac { \u03be }{ I }\\) = \\(\\frac { 1.5 }{ 3 }\\) = 0.5 \u2126.<\/p>\n

                                  Question 9.
                                  \nThe instrument used for measuring electric current is ________.
                                  \na) galvanometer
                                  \nb) ammeter
                                  \nc) voltmeter
                                  \nd) potentiometer
                                  \nAnswer:
                                  \nb) ammeter<\/p>\n

                                  Question 10.
                                  \nWhat is the most commonly used conductor in electronics?
                                  \na) copper
                                  \nb) aluminum
                                  \nc) gold
                                  \nd) silver
                                  \nAnswer:
                                  \na) copper<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 11.
                                  \nKirchhoff\u2019s two laws for electrical circuits are manifestations of the conservation of
                                  \n(a) charge only
                                  \n(b) both energy and momentum
                                  \n(c) energy only
                                  \n(d) both charge and energy
                                  \nAnswer:
                                  \n(d) both charge and energy<\/p>\n

                                  Question 12.
                                  \nWhy does a circuit require a battery?
                                  \na) measure current
                                  \nb) maintain a potential difference
                                  \nc) oppose the current
                                  \nd) measure potential
                                  \nAnswer:
                                  \nb) maintain a potential difference<\/p>\n

                                  Question 13.
                                  \nIn the above circuit, the equivalent resistance between A and B is
                                  \n\"Samacheer
                                  \na) \\(\\frac{20}{3}\\) \u03a9
                                  \nb)10\u03a9
                                  \nc)16\u03a9
                                  \nd)20\u03a9
                                  \nAnswer:
                                  \nc) 16\u03a9<\/p>\n

                                  Question 14.
                                  \nA flow of 107<\/sup> electrons per second in a conduction wire constitutes a current of
                                  \n(a) 1.6 x 10-26<\/sup> A
                                  \n(b) 1.6 x 1012<\/sup> A
                                  \n(c) 1.6 x 10-12<\/sup> A
                                  \n(d) 1.6 x 1026<\/sup> A
                                  \nAnswer:
                                  \n(c) 1.6 x 10-12<\/sup> A
                                  \nHint:
                                  \nI = \\(\\frac { Q }{ t }\\) = \\(\\frac{10^{7} \\times 1.6 \\times 10^{-19}}{1}\\) = 1.6 x 1012<\/sup> A.<\/p>\n

                                  Question 15.
                                  \nA short circuit has _______.
                                  \na) no resistance
                                  \nb) no conductance
                                  \nc) low current
                                  \nd) None of the above
                                  \nAnswer:
                                  \nb) no conductance<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 16.
                                  \nA current of 2A flows through a 12V bulb then calculates the resistance?
                                  \na) 6\u03a9
                                  \nb) 0.16\u03a9
                                  \nc) 32\u03a9
                                  \nd) 0.32\u03a9
                                  \nAnswer:
                                  \nc) 32\u03a9<\/p>\n

                                  Question 17.
                                  \nA galvanometer is converted into an ammeter when we connect a
                                  \n(a) high resistance in series
                                  \n(b) high resistance in parallel
                                  \n(c) low resistance in series
                                  \n(d) low resistance in parallel
                                  \nAnswer:
                                  \n(d) low resistance in parallel<\/p>\n

                                  Question 18.
                                  \nThe resistors each on resistance 1\u03a9 is connected as shown in the figure. The resultant resistance between A and B is
                                  \n\"Samacheer
                                  \na) \\(\\frac{11}{5} \\Omega\\)
                                  \nb) \\(\\frac{5}{11} \\Omega\\)
                                  \nc) \\(\\frac{6}{5} \\Omega\\)
                                  \nd) \\(\\frac{5}{6} \\Omega\\)
                                  \nAnswer:
                                  \nb) \\(\\frac{5}{11} \\Omega\\)<\/p>\n

                                  Question 19.
                                  \nThe rings in the carbon resistor are red red red silver. What is its value?
                                  \na) 22 \u00d7 102<\/sup> \u00b1 10%
                                  \nb) 22 \u00d7 103<\/sup> \u00b1 10%
                                  \nc) 22 \u00d7 102<\/sup> \u00b1 20%
                                  \nd) 22 \u00d7 103<\/sup> \u00b1 20%
                                  \nAnswer:
                                  \na) 22 \u00d7 102<\/sup> \u00b1 10%<\/p>\n

                                  Question 20.
                                  \nThe electrical resistivity of a thin copper wire and a thick copper rod are respectively \u03c11<\/sub>m and \u03c12<\/sub>m then _______
                                  \na) \\(\\rho_{1}>\\rho_{2}\\)
                                  \nb) \\(\\rho_{2}^{>} \\rho_{1}\\)
                                  \nc) \\(\\rho_{1}=\\rho_{2}\\)
                                  \nd) \\(\\frac{\\rho_{2}}{\\rho_{1}}=\\infty\\)
                                  \nAnswer:
                                  \nc) \\(\\rho_{1}=\\rho_{2}\\)<\/p>\n

                                  Question 21.
                                  \nA material with a negative temperature coefficient of resistance is called _______.
                                  \na) metal
                                  \nb) alloy
                                  \nc) thermistor
                                  \nd) thermometer
                                  \nAnswer:
                                  \nc) thermistor<\/p>\n

                                  \"Samacheer<\/p>\n

                                  Question 22.
                                  \nKirchoff\u2019s first law is a consequence of conservation of ________
                                  \na) current
                                  \nb) charges
                                  \nc) energy
                                  \nd) power
                                  \nAnswer:
                                  \nb) charges<\/p>\n

                                  Question 23.
                                  \nCondition for bridge balance of Wheatstone\u2019s bridge is _______.
                                  \na) \\(\\frac{P}{Q}=\\frac{S}{R}\\)
                                  \nb) \\(\\frac{P}{Q}=R S\\)
                                  \nc) \\(\\frac{P}{Q}=\\frac{R}{S}\\)
                                  \nd) \\(\\frac{Q}{P}=\\frac{R}{S}\\)
                                  \nAnswer:
                                  \nc) \\(\\frac{P}{Q}=\\frac{R}{S}\\)<\/p>\n

                                  Question 24.
                                  \nFive cells, each of emf E, are joined in parallel. The total emf of the combination is
                                  \n(a) 5E
                                  \n(b) \\(\\frac { E }{ 5 }\\)
                                  \n(c) E
                                  \n(d) \\(\\frac { 5E }{ 2 }\\)
                                  \nAnswer:
                                  \n(c) E<\/p>\n

                                  Question 25.
                                  \nNichrome wire is used as the heating element because it has ________
                                  \na) low specific resistance
                                  \nb) low melting point
                                  \nc) high specific resistance
                                  \nd) high conductivity
                                  \nAnswer:
                                  \nc) high specific resistance<\/p>\n

                                  Question 26.
                                  \nPeltier effect is the converse of ________
                                  \na) Joule effect
                                  \nb) Raman effect
                                  \nc) Thomson effect
                                  \nd) Seebeck effect
                                  \nAnswer:
                                  \nd) Seebeck effect<\/p>\n

                                  Question 27.
                                  \nThe potential difference between the points A and B in the given figure is ________.
                                  \n\"Samacheer
                                  \na) -3V
                                  \nb) +3V
                                  \nc) +6V
                                  \nd) +9V
                                  \nAnswer:
                                  \nd) +9V<\/p>\n

                                  Question 28.
                                  \nPotentiometer measures potential more accurately because
                                  \n(a) It measures potential in the open circuit.
                                  \n(b) It uses a sensitive galvanometer for null detection.
                                  \n(c) It uses high resistance potentiometer wire.
                                  \n(d) It measures potential in the closed circuit.
                                  \nAnswer:
                                  \n(a) It measures potential in the open circuit.<\/p>\n

                                  \"Samacheer<\/p>\n

                                  X. Two Mark Questions:<\/span><\/p>\n

                                  Question 1.
                                  \nDefine current?
                                  \nAnswer:
                                  \nCurrent is defined as a net charge Q passes through any cross-section of a conductor in time t
                                  \nthen, I = \\(\\frac { Q }{ t }\\).<\/p>\n

                                  Question 2.
                                  \nDefine 1 ampere.
                                  \nAnswer:
                                  \n1A of current is equivalent to 1 Coulomb of charge passing through a perpendicular cross-section in 1 second.<\/p>\n

                                  Question 3.
                                  \nDoes lightning produce an electric current? How?
                                  \nAnswer:<\/p>\n

                                    \n
                                  1. The lightning bolt produces an enormous electric current in a short time.<\/li>\n
                                  2. During lightning, a very high potential difference is created between the clouds and ground so charges flow between the clouds and ground.<\/li>\n<\/ol>\n

                                    Question 4.
                                    \nWhat is meant by transition temperature?
                                    \nAnswer:
                                    \nThe resistance of certain materials becomes zero below a certain temperature Tc<\/sub>. This temperature is known as critical temperature or transition temperature.<\/p>\n

                                    Question 5.
                                    \nDefine mobility.
                                    \nAnswer:<\/p>\n

                                      \n
                                    1. The mobility of the electron is defined as the magnitude of the drift velocity per unit electric field.
                                      \n\\(\\mu=\\frac{\\left|\\vec{V}_{d}\\right|}{|E|}\\)<\/li>\n
                                    2. The SI unit of mobility is \\(\\frac{\\mathrm{m}^{2}}{\\mathrm{~V}_{\\mathrm{S}}}\\)<\/li>\n<\/ol>\n

                                      \"Samacheer<\/p>\n

                                      Question 6.
                                      \nDefine resistance.
                                      \nAnswer:
                                      \nThe resistance is the ratio of potential difference across the given conductor to the current passing through the conductor.<\/p>\n

                                      Question 7.
                                      \nThree bulbs 40w, 60w, and 100w are connected to 220 V mains which bulb will glow brightly. If they are joined in series?
                                      \nAnswer:
                                      \nFrom the relation = \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{p}}\\)
                                      \nIt follows that resistance of 40w bulb will be maximum.<\/p>\n

                                      Question 8.
                                      \nWhat is a thermistor?
                                      \nAnswer:
                                      \nA material with a negative temperature coefficient is called a thermistor.
                                      \nEg:<\/p>\n

                                        \n
                                      1. Insulator<\/li>\n
                                      2. Semiconductor.<\/li>\n<\/ol>\n

                                        Question 9.
                                        \nDefine equivalent resistance when resistors are connected in series.
                                        \nAnswer:
                                        \nWhen several resistances are connected in series, the total or equivalent resistance is the sum of the individual) resistances.
                                        \nRS<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub><\/p>\n

                                        Question 10.
                                        \nDefine equivalent resistance in a parallel resistance network.
                                        \nAnswer:
                                        \nWhen a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination
                                        \n\\(\\frac{1}{\\mathrm{R}_{\\mathrm{P}}}=\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}+\\frac{1}{\\mathrm{R}_{3}}\\)<\/p>\n

                                        Question 11.
                                        \nHousehold appliances are always connected in parallel, why?
                                        \nAnswer:
                                        \nHousehold appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.<\/p>\n

                                        Question 12.
                                        \nWhat is a carbon resistor?
                                        \nAnswer:<\/p>\n

                                          \n
                                        1. Carbon resistors consist of a ceramic core, on which a thin layer of crystalline Carbon is deposited.<\/li>\n
                                        2. These resistors are inexpensive, stable, and compact in size.<\/li>\n<\/ol>\n

                                          \"Samacheer<\/p>\n

                                          Question 13.
                                          \nWhat is a multimeter?
                                          \nAnswer:
                                          \nA multimeter is a very useful electronic instrument used to measure voltage, current, resistance, and capacitance. In fact, it can also measure AC voltage and AC current.<\/p>\n

                                          Question 14.
                                          \nWhat is a thermistor?
                                          \nAnswer:
                                          \nA semiconductor with a negative temperature coefficient of resistance is called a thermistor.<\/p>\n

                                          Question 15.
                                          \nDefine the internal resistance of a battery.
                                          \nAnswer:
                                          \nA real battery is made of electrodes and electrolytes, there is resistance to the flow of charges within the battery.<\/p>\n

                                          Question 16.
                                          \nThe headlights of a car turn dim on starting the car, why?
                                          \nAnswer:
                                          \nWhen the car engine is started with headlights turned on, they sometimes become dim. This is due to the internal resistance of the car battery.<\/p>\n

                                          Question 17.
                                          \nWhy is a potentiometer preferred over a voltmeter for comparison of emf of cells?
                                          \nAnswer:
                                          \nThe potentiometer reads actual emf as it does not draw any current from the cell whereas the voltmeter measures terminal potential difference as it draws current from the cell.<\/p>\n

                                          Question 18.
                                          \nWhat are the factors that affect the heat produced in an electrical circuit?
                                          \nAnswer:<\/p>\n

                                            \n
                                          1. Current I flows through a conductor kept across a potential difference V for a time t, the work done or the electric potential energy spent is W = Vlt<\/li>\n
                                          2. In the absence of any other external effect, this energy is spent heating the conductor.<\/li>\n<\/ol>\n

                                            The amount of heat(H) produced is H = VIt
                                            \nFor a resistance R,
                                            \nH = I2<\/sup>Rt
                                            \nThis is known as Joule’s law of heating.
                                            \nThe heat developed in an electrical circuit due to the flow of current varies directly as<\/p>\n

                                              \n
                                            • \n
                                                \n
                                              • the square of the current<\/li>\n
                                              • the resistance of the circuit and<\/li>\n
                                              • the time of flow.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n

                                                Question 19.
                                                \nWhy nichrome is used as a heating element?
                                                \nAnswer:
                                                \nNichrome has a high specific resistance arid can be heated to very high temperatures without oxidation.<\/p>\n

                                                Question 20.
                                                \nWhat is the use of trippers in houses?
                                                \nAnswer:
                                                \nWhenever there is an excessive current produced due to a faulty wire connection, the circuit breaker switch opens. After repairing the faulty connection, we can close the circuit breaker switch.<\/p>\n

                                                \"Samacheer<\/p>\n

                                                Question 21.
                                                \nWhat are the applications of the heating effect of current?
                                                \nAnswer:
                                                \nElectric discharge lamps, electric welding, and electric arc also utilize the heating effect of current<\/p>\n

                                                Question 22.
                                                \nWhat is the thermoelectric effect?
                                                \nAnswer:
                                                \nConversion of temperature differences into electrical voltage and vice versa is known as the thermoelectric effect.<\/p>\n

                                                Question 23.
                                                \nA wire of resistance 10\u03a9 is stretched uniformly, to thrice its original length calculate the resistance of the stretched wire.
                                                \nAnswer:
                                                \nR = 10\u03a9 \u21d2 R = \\(\\frac{\\rho l}{A}\\)
                                                \nOriginal length = l
                                                \nNew length = l’ =3l
                                                \nOriginal area = A
                                                \nIf length increases, area decreases
                                                \n\"Samacheer<\/p>\n

                                                Question 24.
                                                \nCalculate the internal resistance of a 2.1V cell which gives a current of 0.2A through a resistance of 10\u03a9
                                                \nAnswer:
                                                \n\\(\\mathrm{r}=\\left(\\frac{\\xi-\\mathrm{V}}{\\mathrm{V}}\\right) \\mathrm{R}\\)
                                                \nV = IR = 0.2 \u00d7 10 = 2V
                                                \n\\(r=\\left(\\frac{2.1-2}{2}\\right) 10\\) = 0.5\u03a9
                                                \nr = 0.5\u03a9<\/p>\n

                                                Question 25.
                                                \nDifferentiate between the Joule heating effect and the Peltier effect.
                                                \nAnswer:<\/p>\n\n\n\n\n
                                                Joule heating effect<\/td>\nPeltier effect<\/td>\n<\/tr>\n
                                                1. When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated.<\/td>\n1. An electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                \"Samacheer<\/p>\n

                                                Question 26.
                                                \nWhat is the transition or critical temperature in superconductivity?
                                                \nAnswer:
                                                \nThe temperature at which the resistance of the material becomes zero and the material changes from normal to superconductor is known as critical or transition temperature.<\/p>\n

                                                Question 27.
                                                \nWrite the formula for electrical conductivity and write its unit?
                                                \nAnswer:<\/p>\n

                                                  \n
                                                1. Electrical conductivity,
                                                  \n\\(\\sigma=\\frac{\\mathrm{ne}^{2} \\tau}{\\mathrm{m}}\\)<\/li>\n
                                                2. It’s unit is \u03a9-1<\/sup> m-1<\/sup><\/li>\n<\/ol>\n

                                                  Question 28.
                                                  \nCalculate the current in the wire, if a charge of 180C flows through a wire for 1 minute.
                                                  \nAnswer:
                                                  \nCurrent I = \\(\\frac{\\mathrm{dQ}}{\\mathrm{dt}}=\\frac{180}{60}\\) = 3A
                                                  \nI = 3A<\/p>\n

                                                  Question 29.
                                                  \nA parallel combination of two cells of emf’s E1<\/sub> and E2<\/sub>, internal resistances r1<\/sub> and r2<\/sub> are used to supply current to load of resistance R – Write the expression for the current through the load in terms E1<\/sub>, E2<\/sub>, r1,<\/sub> and r2<\/sub>.
                                                  \nAnswer:
                                                  \n\"Samacheer<\/p>\n

                                                  Question 30.
                                                  \nA wire with uniform cross-section A, length l, and resistance R is bent into a complete circle. Calculate the resistance between two diametrically opposite points.
                                                  \nAnswer:
                                                  \n\"Samacheer
                                                  \nThe resistance between two diametrically opposite points is R\/4\u03a9.<\/p>\n

                                                  \"Samacheer<\/p>\n

                                                  XI. Three Mark Questions.<\/span><\/p>\n

                                                  Question 1.
                                                  \nDerive a relation between drift velocity and mobility.
                                                  \nAnswer:<\/p>\n

                                                    \n
                                                  1. The drift velocity is the average velocity acquired by the electrons inside the conductor<\/li>\n
                                                  2. The average time between successive collisions is called the mean free time denoted by \u03c4.<\/li>\n
                                                  3. The acceleration \\(\\overrightarrow{\\mathrm{a}\\) experienced by the electron in an electric field \\(\\overrightarrow{\\mathrm{E}\\) is given by<\/li>\n<\/ol>\n

                                                    \"Samacheer<\/p>\n

                                                    Question 2.
                                                    \nExplain the concept of colour code for carbon resistors.
                                                    \nAnswer:
                                                    \nColor code for Carbon resistors:
                                                    \nCarbon resistors consist of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable, and compact in size. Color rings are used to indicate the value of the resistance according to the rules.<\/p>\n

                                                    Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 103<\/sup> (orange) and tolerance = 5% (gold). The value of resistance = 56 x 103<\/sup> Q or 56 k\u03a9 with a tolerance value of 5%.<\/p>\n

                                                    Question 3.
                                                    \nDraw the current versus voltage graph for the ohmic and non-ohmic devices. Give one example for each.
                                                    \nAnswer:<\/p>\n

                                                      \n
                                                    • Ohmic devices – Conductors.<\/li>\n
                                                    • Non-ohmic devices – Diode.<\/li>\n<\/ul>\n

                                                      \"Samacheer<\/p>\n

                                                      \"Samacheer<\/p>\n

                                                      Question 4.
                                                      \nRepairing the electrical connection with the wet skin is always dangerous why?
                                                      \nAnswer:<\/p>\n

                                                        \n
                                                      1. The human body contains a large amount of water which has a low resistance of around 200\u03a9 and the dry skin has high resistance of around 500 k\u03a9.<\/li>\n
                                                      2. But when the skin is wet, the resistance is reduced to around 1000\u03a9. This is the reason, repairing the electrical connection with the wet skin is always dangerous.<\/li>\n<\/ol>\n

                                                        Question 5.
                                                        \nUnder what condition, is the heat produced in an electric circuit
                                                        \n(i) directly proportional and
                                                        \n(ii) inversely proportional to the resistance of the circuit?
                                                        \nAnswer:
                                                        \nH = I2<\/sup>Rt
                                                        \nIts constant current is flowing through a circuit then heat produced is directly proportional to resistance R
                                                        \nBut H = I2<\/sup>Rt
                                                        \n\\(\\begin{array}{l}
                                                        \nI=\\frac{V}{R} \\\\
                                                        \n\\text { then } H=\\frac{V^{2}}{R^{2}} \\times k t=\\frac{V^{2}}{R} t
                                                        \n\\end{array}\\)
                                                        \nIf a constant voltage is flowing through a circuit then heat produced is inversely proportional to resistance R.<\/p>\n

                                                        Question 6.
                                                        \nHow are batteries formed?
                                                        \nAnswer:<\/p>\n

                                                          \n
                                                        1. An electric cell converts chemical energy into electrical energy to produce electricity. It contains two electrodes immersed in an electrolyte.<\/li>\n
                                                        2. Several electric cells connected together form a battery.<\/li>\n
                                                        3. When a cell or battery is connected to a circuit, electrons flow from the negative terminal to the positive terminal through the circuit.<\/li>\n
                                                        4. By using chemical reactions, a battery produces a potential difference across its terminals.<\/li>\n
                                                        5. This potential difference provides the energy to move the electrons through the circuit.<\/li>\n<\/ol>\n

                                                          Question 7.
                                                          \nHow will you determine the current in a circuit when 2 or more cells are connected in series.
                                                          \nAnswer:<\/p>\n

                                                          \"Samacheer<\/p>\n

                                                            \n
                                                          1. In a series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of the second cell is connected to the positive terminal of the third cell, and so on.<\/li>\n
                                                          2. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.<\/li>\n
                                                          3. Suppose n cells, each of emf \u03be volts and internal resistance r ohms is connected in series with an external resistance R<\/li>\n
                                                          4. The total emf of the battery = n\u03be<\/li>\n
                                                          5. The total resistance in the circuit = nr + R. By Ohm’s law, the current in the circuit is
                                                            \n\"Samacheer<\/li>\n
                                                          6. It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.<\/li>\n<\/ol>\n

                                                            Question 8.
                                                            \nWhat is end resistance and how will you eliminate it.
                                                            \nAnswer:<\/p>\n

                                                              \n
                                                            1. The bridge wire is soldered at the ends of the copper strips.<\/li>\n
                                                            2. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances.<\/li>\n
                                                            3. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.<\/li>\n<\/ol>\n

                                                              \"Samacheer<\/p>\n

                                                              Question 9.
                                                              \nWrite a note on the electric furnace.
                                                              \nAnswer:<\/p>\n

                                                                \n
                                                              1. Furnaces are used to manufacture a large number of technologically important materials such as steel, silicon carbide, quartz, gallium arsenide, etc.<\/li>\n
                                                              2. To produce temperatures up to 1500\u00b0C, a molybdenum-nichrome wire wound on a silica tube is used.<\/li>\n
                                                              3. Carbon arc furnaces produce temperatures up to 3000 \u00b0C.<\/li>\n<\/ol>\n

                                                                Question 10.
                                                                \nCompare the emf and potential difference
                                                                \nAnswer:<\/p>\n\n\n\n\n\n
                                                                Emf<\/td>\nPotential difference<\/td>\n<\/tr>\n
                                                                1. The difference of potentials between the 2 terminals of a cell in an open circuit is called emf.<\/td>\nThe difference in potentials between any 2 points in a closed circuit is called the potential difference.<\/td>\n<\/tr>\n
                                                                2. The emf is independent of the external resistance<\/td>\nIt is proportional to the resistance between any 2 points.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                Question 11.
                                                                \nThe colours of the carbon resistors are orange, orange, orange. What is the value of the resistor?
                                                                \nAnswer:<\/p>\n

                                                                  \n
                                                                1. The first orange ring corresponds to 3. The next orange ring corresponds to 3.<\/li>\n
                                                                2. The next orange ring corresponds to 3, that is the 3rd orange ring corresponds to 103<\/sup>.<\/li>\n
                                                                3. There is no coloured ring at the end. So tolerance is 20%.<\/li>\n
                                                                4. Total resistance is 33 \u00d7 103<\/sup>\u03a9 \u00b1 20%<\/li>\n<\/ol>\n

                                                                  Question 12.
                                                                  \nIn the given circuit, the heat produced per second in 6\u03a9 resistor is 32 J. Then calculate the heat produced per second in 3\u03a9 resistor.
                                                                  \n\"Samacheer
                                                                  \nAnswer:
                                                                  \n\"Samacheer<\/p>\n

                                                                  Question 13.
                                                                  \nFind the value of unknown resistance x in the given circuit.
                                                                  \n\"Samacheer
                                                                  \nAnswer:
                                                                  \nP.d across 1\u03a9 = P.d across x-axis
                                                                  \n6 \u00d7 1 = (11 – 9)x-axis
                                                                  \n6 = 2x
                                                                  \n[\u223411 = 3 + 6 + x, x = (11 – 9)]
                                                                  \nx = 3\u03a9<\/p>\n

                                                                  \"Samacheer<\/p>\n

                                                                  Question 14.
                                                                  \nAluminum and copper of equal length are found to have some resistance. If the ratio of their radii is 1:3, calculate the ratio of their resistivity.
                                                                  \nAnswer:
                                                                  \n\"Samacheer<\/p>\n

                                                                  Question 15.
                                                                  \nThe resistance of a metal wire of length AB to 2\u03a9. Another wire of length PQ of the same material with twice the diameter of AB is found to have the same resistance of 2\u03a9. what is the length of PQ?
                                                                  \nAnswer:
                                                                  \n\"Samacheer<\/p>\n

                                                                  Question 16.
                                                                  \nA filament bulb (500W, 100V) is to be used in a 230V main supply, when a resistance R is connected in series, it works perfectly and the bulb consumes 500W. What is the value of R?
                                                                  \n\"Samacheer
                                                                  \nAnswer:
                                                                  \n\"Samacheer<\/p>\n

                                                                  Question 17.
                                                                  \nDerive an expression for Joule’s law.
                                                                  \nAnswer:
                                                                  \nCurrent I flows through a conductor kept across a potential difference V for a time t, the work done or the electric potential energy spent is
                                                                  \nW = Vit
                                                                  \nIn the absence of any other external effect, this energy is spent heating the conductor.
                                                                  \nThe amount of heat(H) produced is
                                                                  \nH = Vit
                                                                  \nFor a resistance R,
                                                                  \nH = I2<\/sup>Rt
                                                                  \nThis is known as Joule\u2019s law of heating. The heat developed in an electrical circuit due to the flow of current varies directly as<\/p>\n

                                                                    \n
                                                                  • the square of the current<\/li>\n
                                                                  • the resistance of the circuit and<\/li>\n
                                                                  • the time of flow.<\/li>\n<\/ul>\n

                                                                    Question 18.
                                                                    \nWrite a note on electric fuses.
                                                                    \nAnswer:<\/p>\n

                                                                      \n
                                                                    1. Fuses are connected in series in a circuit to protect the electric devices from the heat developed by the passage of excessive current.<\/li>\n
                                                                    2. It is a short length of a wire made of a low melting point material.<\/li>\n
                                                                    3. It melts and breaks the circuit if the current exceeds a certain value.<\/li>\n
                                                                    4. The only disadvantage with the above fuses is that once fuse wire is burnt due to excessive current, they need to be replaced. Nowadays in houses, circuit breakers (trippers) are also used instead of fuses.<\/li>\n<\/ol>\n

                                                                      \"Samacheer<\/p>\n

                                                                      Question 19.
                                                                      \nDistinguish between Positive and Negative Thomson effect.
                                                                      \nAnswer:<\/p>\n\n\n\n\n\n\n
                                                                      Positive Thomson effect<\/td>\nNegative Thomson Effect<\/td>\n<\/tr>\n
                                                                      1. Heat is absorbed first and then evolved<\/td>\nHeat is evolved first and then absorbed<\/td>\n<\/tr>\n
                                                                      2. Heat is transferred in the direction of current<\/td>\nHeat is transferred in the direction opposite to the direction of the current.<\/td>\n<\/tr>\n
                                                                      3. Eg. Silver, Zinc, cadmium<\/td>\nEg. Platinum, nickel, cobalt, mercury<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                                      Question 20.
                                                                      \nV = I graph for a metallic wire at two different temperatures T1<\/sub> and T2<\/sub> is shown in the Fig which of the two temperatures T1<\/sub> and T2<\/sub> is higher and why?
                                                                      \n\"Samacheer
                                                                      \nAnswer:
                                                                      \nIf the same potential difference V is applied current equal to I1<\/sub> and I2<\/sub> will flow through the wire, when its temperature is kept at T1<\/sub> and T2<\/sub> respectively since R = V\/I it follows that resistance of the wire is more at temperature T2<\/sub> further, as in case of the metals the resistance increases with rising of temperature T2<\/sub> > T1<\/sub>.<\/p>\n

                                                                      Question 21.
                                                                      \nWhich material is used for the meter bridge wire and why?
                                                                      \nAnswer:
                                                                      \nThe wire used in the meter bridge is made up of all manganin due to its high specific resistance copper strips are used at the ends because it is a good conductor of electricity.<\/p>\n

                                                                      \"Samacheer<\/p>\n

                                                                      XII. Five Mark Questions:<\/span><\/p>\n

                                                                      Question 1.
                                                                      \nEstablish a relation between drift velocity and current density.
                                                                      \nAnswer:<\/p>\n

                                                                        \n
                                                                      1. XY is a conductor of the area of cross-section A<\/li>\n
                                                                      2. E is the applied electric field.<\/li>\n
                                                                      3. n is the number of electrons per unit volume with the same drift velocity (vd)<\/li>\n
                                                                      4. Let electrons move through a distance dx in time interval dt.
                                                                        \n\\(\\mathrm{v}_{\\mathrm{d}}=\\frac{\\mathrm{d} \\mathrm{x}}{\\mathrm{dt}} ; \\mathrm{d} \\mathrm{x}=\\mathrm{v}_{\\mathrm{d}} \\mathrm{dt} \\ldots \\ldots \\ldots \\ldots \\ldots(1)\\)
                                                                        \n\"Samacheer<\/li>\n
                                                                      5. Electrons available in the given volume = volume \u00d7 number per unit volume
                                                                        \n= Adx \u00d7 n
                                                                        \n= A vd<\/sub> \u00d7 n[from (1)]
                                                                        \nTotal charge in volume dQ = charge \u00d7 number of electrons
                                                                        \ndQ = (e)(A vd<\/sub>)n
                                                                        \nHence the current,
                                                                        \n\\(\\begin{array}{l}
                                                                        \nI=\\frac{d Q}{d t} \\\\
                                                                        \nI=\\frac{n e A v_{d} d t}{d t}
                                                                        \n\\end{array}\\)
                                                                        \nI = neAvd<\/sub>dt.<\/li>\n<\/ol>\n

                                                                        Question 2.
                                                                        \nDeduce ohm’s law using the concept of current density?
                                                                        \n\"Samacheer
                                                                        \nAnswer:<\/p>\n

                                                                          \n
                                                                        1. Consider a segment of wire of length l and area of cross section A.<\/li>\n
                                                                        2. Electric field is created when a potential difference V is applied.<\/li>\n
                                                                        3. If \u2018E\u2019 is uniform, then, V = El<\/li>\n
                                                                        4. Current density, J = \u03c3E = \u03c3\\(\\frac{V}{l}\\)<\/li>\n
                                                                        5. Since J = \\(\\frac{I}{A}\\)
                                                                          \n\\(\\frac{I}{A}=\\sigma \\frac{V}{l}\\)
                                                                          \nRearranging the above equation,
                                                                          \n\\(\\mathrm{V}=\\mathrm{I}\\left(\\frac{I}{\\sigma A}\\right)\\)
                                                                          \nWhere \\(\\left(\\frac{I}{\\sigma A}\\right)\\) – resistance of conductor (R)
                                                                          \n:. R \u03b1 \\(\\left(\\frac{I}{\\sigma A}\\right)\\)
                                                                          \n:. The macroscopic form of ohm\u2019s law is
                                                                          \nV = IR
                                                                          \nV \u03b1 I.<\/li>\n<\/ol>\n

                                                                          \"Samacheer<\/p>\n

                                                                          Question 3.
                                                                          \nExplain how the internal resistance of a cell can be determined using a potentiometer.
                                                                          \nAnswer:<\/p>\n

                                                                          \"Samacheer
                                                                          \nMeasurement of internal resistance<\/p>\n

                                                                            \n
                                                                          1. To measure the internal resistance of a cell, the circuit connections are made as shown in Figure.<\/li>\n
                                                                          2. The end C of the potentiometer wire is connected to the positive terminal of battery B and the negative terminal of the battery is connected to the end D through a key K1<\/sub> which forms the primary circuit.<\/li>\n<\/ol>\n

                                                                            Measurement of internal resistance:<\/p>\n

                                                                              \n
                                                                            1. The positive terminal of the cell \u03be whose internal resistance is to be determined is also connected to the end C of the wire.<\/li>\n
                                                                            2. The negative terminal of the cell \u00a3 is connected to a jockey through a galvanometer and a high resistance.<\/li>\n
                                                                            3. A resistance box R and key K2<\/sub> are connected across the cell \u03be<\/li>\n
                                                                            4. With K2<\/sub> open, the balancing point J is obtained and the balancing length CJ= l is measured since the cell is in an open circuit, its emf
                                                                              \n\u03be \u03b1 l1<\/sub> ………..(1)<\/li>\n
                                                                            5. With the help of resistance box R, introduce resistance, say R and then key K2<\/sub> is closed.<\/li>\n
                                                                            6. The current passing through the cell and the resistance R is given by
                                                                              \n\\(I=\\frac{\\xi}{R+r}\\)<\/li>\n
                                                                            7. The potential difference across R is
                                                                              \n\\(V=\\frac{\\xi R}{R+r}\\)<\/li>\n
                                                                            8. When this potential difference is balanced on the potentiometer wire, let l2<\/sub> be the balancing length, then
                                                                              \n\"Samacheer
                                                                              \n\\(\\mathrm{r}=\\mathrm{R}\\left(\\frac{l_{1}-l_{2}}{l_{2}}\\right)\\)<\/li>\n
                                                                            9. The internal resistance of the cell is not constant but increases with the increase of external resistance connected across its terminals.<\/li>\n<\/ol>\n

                                                                              Question 4.
                                                                              \nDefine electric power and electric potential energy. A potential difference V is applied across a resistance R. Find a relation for the power consumed.
                                                                              \nAnswer:
                                                                              \n1. The electrical power P, is the rate at which the electrical potential energy is delivered.
                                                                              \n2. If the charge q, flows between two points having a potential difference V, then the work done in moving the charge is dW=VdQ.
                                                                              \ni.e dW=V dQ<\/p>\n

                                                                              \"Samacheer
                                                                              \n3. Consider a circuit in which a battery of voltage V is connected to the resistor.
                                                                              \n4. Assume that a positive charge of dQ moves from point a to b through the battery and moves from point c to d through the resistor and back to point a.
                                                                              \n5. When the charge moves from point a to b, it gains potential energy dU =V dQ, and the chemical potential energy of the battery decreases by the same amount.
                                                                              \n6. When this charge dQ passes through the resistor it loses the potential energy dU=V dQ due to collision with atoms in the resistor and again reaches point a.
                                                                              \n7. This process occurs continuously till the battery is connected in the circuit.
                                                                              \n8. The electrical power P, is the rate at which the electrical potential energy is delivered,
                                                                              \n\\(P=\\frac{d W}{d t}\\)
                                                                              \nW.K.T dW = V dQ
                                                                              \n\\(P=\\frac{d}{d t}(V d Q)=V\\left(\\frac{d Q}{d t}\\right)\\)
                                                                              \nP = VI (\u2234\\(I=\\frac{d Q}{d t}\\))
                                                                              \nSI unit = Watt
                                                                              \n9. The bigger units of electric power are kilowatt = 1KW = 103<\/sup> W
                                                                              \nMegawatt = IMW = 106<\/sup> W
                                                                              \n10.Expression for electric power in terms of I and R. P = I2<\/sup>R (\u2234 V=IR)
                                                                              \n11. Expression for electric power interms of V and R.
                                                                              \n\\(\\mathrm{P}=\\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\\left(\\mathrm{I}=\\frac{\\mathrm{V}}{\\mathrm{R}}\\right)\\)<\/p>\n

                                                                              Question 5.
                                                                              \nExplain the principle of potentiometer.
                                                                              \nAnswer:
                                                                              \n\"Samacheer<\/p>\n

                                                                              Potentiometer:<\/p>\n

                                                                                \n
                                                                              1. A steady current is maintained across the wire CD by a battery.<\/li>\n
                                                                              2. The battery, key, and potentiometer wire are connected in series from the primary circuit.<\/li>\n
                                                                              3. The positive terminal of a primary cell of emf \u03be is connected to point c and negative terminal is connected to the Jockey through a galvanometer G and a high resistance HR. This forms the secondary circuit.<\/li>\n
                                                                              4. Let contact be made at any point J on the wire by a jockey.<\/li>\n
                                                                              5. If the potential difference across CJ is equal to the emf of the cell then no current will flow through the galvanometer and it will show zero deflection.<\/li>\n
                                                                              6. CJ is the balancing length L.<\/li>\n
                                                                              7. The potential difference across CJ is equal to IrL<\/li>\n
                                                                              8. \u03be = IrL<\/li>\n
                                                                              9. Where I – current, r – resistance per unit length of the wire.<\/li>\n
                                                                              10. Since I and r are constants, \u03be \u03b1 l<\/li>\n
                                                                              11. The emf of the cell is directly proportional to the balancing length.<\/li>\n<\/ol>\n

                                                                                \"Samacheer<\/p>\n

                                                                                Question 6.
                                                                                \nExplain the Thomson effect.
                                                                                \nAnswer:
                                                                                \nThomson effect:
                                                                                \nIf two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result, the potential difference is created between these points. Thomson effect is also reversible.<\/p>\n

                                                                                Positive Thomson effect:
                                                                                \n\"Samacheer<\/p>\n

                                                                                  \n
                                                                                1. If the current is passed through a copper bar AB which is heated at the middle point C, the point C will be at a higher potential.<\/li>\n
                                                                                2. This indicates that the heat is absorbed along with AC and evolved along with the CB of the conductor. Thus heat is transferred due to the current flow in the direction of the current. It is called the positive Thomson effect.<\/li>\n
                                                                                3. A similar effect is observed in metals like silver, zinc, and cadmium.<\/li>\n<\/ol>\n

                                                                                  Negative Thomson effect:
                                                                                  \n\"Samacheer<\/p>\n

                                                                                    \n
                                                                                  1. When the copper bar is replaced by an iron bar, heat is evolved along CA and absorbed along BC.<\/li>\n
                                                                                  2. Thus heat is transferred due to the current flow in the direction opposite to the direction of the current. It is called the negative Thomson effect.<\/li>\n
                                                                                  3. A similar effect is observed in metals like platinum, nickel, cobalt, and mercury.<\/li>\n<\/ol>\n

                                                                                    Question 7.
                                                                                    \nGive a short account on carbon resistor
                                                                                    \nAnswer:<\/p>\n

                                                                                      \n
                                                                                    1. Carbon resistors consist of a ceramic core, on which a thin layer of crystalline Carbon is deposited.<\/li>\n
                                                                                    2. These resistors are inexpensive, stable, and compact in size.<\/li>\n
                                                                                    3. Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth colour, silver or gold shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%.<\/li>\n
                                                                                    4. While reading the colour code, hold the resistor with colour bands to your left. Resistors never start with a metallic band on the left.<\/li>\n<\/ol>\n

                                                                                      Question 8a.
                                                                                      \nDefine electric power
                                                                                      \nAnswer:
                                                                                      \nElectric power is the rate at which electrical potential energy is delivered.
                                                                                      \n(i.e) \\(\\frac{\\mathrm{dW}}{\\mathrm{dt}}=\\mathrm{P}=\\mathrm{Vi}\\)
                                                                                      \nunit \u2192 watt<\/p>\n

                                                                                      Question 8b.
                                                                                      \nAn electric iron of resistance of 80\u03a9 is operated at 200 V for 2 hours, find the electrical energy consumed.
                                                                                      \nAnswer:
                                                                                      \nGiven data: R = 80\u03a9 t = 2hours v = 200V
                                                                                      \nP = VI
                                                                                      \nP = \\(\\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\\)
                                                                                      \n= \\(\\frac{200 \\times 200}{80}=500 \\mathrm{~W}\\)
                                                                                      \nEnergy consumed in 2 hour = Power in watt \u00d7 time in hour
                                                                                      \nE = 500 \u00d7 2 = 1000 Wh
                                                                                      \nE = 1 kWh<\/p>\n

                                                                                      Question 9.
                                                                                      \nTo balance the Wheatstone’s bridge shown in the figure, determine an additional resistance that has to be connected with 15\u2126
                                                                                      \nAnswer:
                                                                                      \n\"Samacheer
                                                                                      \nGiven data
                                                                                      \nP = 4.2\u2126, Q = 3\u2126, R = 2\u2126, S = 15\u2126
                                                                                      \n\"Samacheer
                                                                                      \n\"Samacheer<\/p>\n

                                                                                      Question 10.
                                                                                      \nDetermine the effective resistance of the given circuit between points A and B.
                                                                                      \n\"Samacheer
                                                                                      \nAnswer:
                                                                                      \n\"Samacheer
                                                                                      \n\"Samacheer
                                                                                      \nThe combined resistance between the points A and B is
                                                                                      \nR = R’ + R’ + R’
                                                                                      \n= 7.5 + 7.5 + 7.5
                                                                                      \nR = 22.5\u2126<\/p>\n

                                                                                      \"Samacheer<\/p>\n

                                                                                      Question 11.
                                                                                      \nFind the heat energy produced in a resistance of 10\u2126 when 5A current flows through it for 5 minutes.
                                                                                      \nAnswer:
                                                                                      \nR = 10\u2126, I = 5A, t = 5 minutes = 5 x 60s
                                                                                      \nH = I2<\/sup>Rt
                                                                                      \n= 52<\/sup> \u00d7 10 \u00d7 5 \u00d7 60
                                                                                      \n= 25 \u00d7 10 \u00d7 300
                                                                                      \n= 25 x 3000
                                                                                      \nH = 75000J or 75 KJ.<\/p>\n

                                                                                      Question 12.
                                                                                      \nThe resistance of a nichrome wire ar 0\u00b0C is 10\u2126. If its temperature coefficient of resistance is 0.004\/\u00b0C find its resistance at the boiling point of water. Comment on the result.
                                                                                      \nAnswer:
                                                                                      \nGiven data:
                                                                                      \nR0<\/sub> = 10\u2126, \u03b1 = 0.004\/\u00b0C, t = 100\u00b0C
                                                                                      \nRt<\/sub> = R0<\/sub> (1 + \u03b1 t)
                                                                                      \n=10 (1 + 0.004 \u00d7 100)
                                                                                      \n= 10 (1 + 0.4)
                                                                                      \n= 10 (1.4)
                                                                                      \nRt<\/sub> = 14\u2126
                                                                                      \nAs the temperature increases, the resistance of the wire increases.<\/p>\n

                                                                                      Question 13.
                                                                                      \nWhen two resistors connected in series and parallel their equivalent resistances are 15\u2126, 56\/15 \u2126 respectively. Find the two resistances.
                                                                                      \nAnswer:
                                                                                      \nRS<\/sub> = R1<\/sub> + R2<\/sub> = 15\u2126 ………….(1)
                                                                                      \n\\(R_{p}=\\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\\frac{56}{15} \\Omega\\) ………..(2)
                                                                                      \nFrom equation (1) substituting R1<\/sub> + R2<\/sub> in equation (2)
                                                                                      \n\\(\\frac{R_{1} R_{2}}{15}=\\frac{56}{15} \\Omega\\)
                                                                                      \n\u2234R1<\/sub>R2<\/sub> = 56
                                                                                      \nR2 <\/sub> = \\(\\frac{56}{15}\\) = 15 …………..(3)
                                                                                      \nSubstituting for R2<\/sub> in equation (1) from equation (3)
                                                                                      \nR1<\/sub> = \\(\\frac{56}{\\mathrm{R}_{1}}\\) = 15
                                                                                      \nThen, \\(\\frac{R_{1}^{2}+56}{R_{1}}=15\\)
                                                                                      \nR1<\/sub>2<\/sup> + 56 = 15R1<\/sub>
                                                                                      \nR1<\/sub>2<\/sup> – 15 R1<\/sub> + 56 = 0
                                                                                      \nThe above equation can be solved using the simple mathematics
                                                                                      \nR1<\/sub>2<\/sup> – 8R1<\/sub> – 7R1<\/sub> + 56=0
                                                                                      \nR1<\/sub> (R1<\/sub> – 8) – 7 (R1<\/sub> – 8) = 0
                                                                                      \n(R1<\/sub> – 8) (R1<\/sub> – 7) = 0
                                                                                      \n(R1<\/sub> = 8\u2126) or R1<\/sub> = 7\u2126
                                                                                      \nIf R1<\/sub> = 8\u2126
                                                                                      \nusing in equation (1)
                                                                                      \n8 + R2<\/sub> = 15
                                                                                      \nR2<\/sub> = 15 – 8 = 7\u2126
                                                                                      \nR2<\/sub> = 7\u2126 i.e, (when R1<\/sub> = 8\u2126; R2<\/sub> = 7\u2126)
                                                                                      \nIf R1<\/sub> = 7\u2126,
                                                                                      \nSubstituting the equation (1)
                                                                                      \n7 + R2<\/sub> = 15
                                                                                      \nR2<\/sub> = 8\u2126 i.e, (when R1<\/sub> = 7\u2126, R2<\/sub> = 8\u2126)<\/p>\n

                                                                                      Question 14.
                                                                                      \nIn the circuit, find the current through each branch of the circuit and the potential drop across the 10\u2126 resistor.
                                                                                      \nAnswer:
                                                                                      \n\"Samacheer
                                                                                      \nApIying kirchhof’s second rule for loop AFDCA
                                                                                      \n10 (I1<\/sub> + I2<\/sub>)+ 2 I1<\/sub> = 4
                                                                                      \n12 I1<\/sub> + 10 I2<\/sub> = 4 ……….(1)
                                                                                      \nFor loop AFEBA
                                                                                      \n10 (I1<\/sub>+ I2<\/sub>) + 4 I2<\/sub> = 5
                                                                                      \nFE I2<\/sub>4V 5V A B
                                                                                      \n10 I1<\/sub> + 14I2<\/sub> = 5 …………(2)
                                                                                      \n(1) \u00d7 1o \u21d2 120 I1<\/sub> + 100 I2<\/sub> = 40 ………….(3)
                                                                                      \n(2) \u00d7 12 \u21d2 120 I1<\/sub> + 168 I2<\/sub> = 60 …………..(4)
                                                                                      \n(4) – (3) 68 I12<\/sub> = 20
                                                                                      \n\\(I_{2}=\\frac{20}{68}=\\frac{10}{34}=0.2941\\)
                                                                                      \nI2<\/sub> = 0.2941A
                                                                                      \nSubstituting I2<\/sub> in eqn (1),
                                                                                      \n12 I1<\/sub> + 10(0.2941) = 4
                                                                                      \n12I1<\/sub> = 4 – 2.941 = 1.059
                                                                                      \nI1<\/sub> = \\(\\frac{1.059}{12}\\) = 0.08825A
                                                                                      \nI1<\/sub> = 0.08825A
                                                                                      \nThe voltage drop across 10\u03a9 resistor is
                                                                                      \nV = IR
                                                                                      \nV=(I1<\/sub> + I2<\/sub>)R
                                                                                      \n= (0.008825 + 0.2941) \u00d7 10
                                                                                      \nV = 3.824 V<\/p>\n

                                                                                      \"Samacheer<\/p>\n

                                                                                      Question 15.
                                                                                      \nObtain the condition for maximum current through a resistor, when a number of cells are connected cells in series
                                                                                      \n(i) in series and
                                                                                      \n(ii) in parallel suppose n cells, each of emf \u03be volts and internal resistance r ohms is connected in series with on external resistance R as shown in the figure.
                                                                                      \nAnswer:
                                                                                      \nCells in series:
                                                                                      \nThe total emf of the cells = n\u03be
                                                                                      \nThe total internal resistance of the cells = r + r + r +……..n times = nr
                                                                                      \n\"Samacheer
                                                                                      \nTherefore, total resistance of the circuit = R+nr
                                                                                      \nIf I is current flowing through the circuit then,
                                                                                      \n\\(I=\\frac{\\text { total emf }}{\\text { total resistance }}\\)
                                                                                      \n\\(I=\\frac{n \\xi}{n r+R}\\)
                                                                                      \ntotal resistance nr + R
                                                                                      \n1) When r << R, I = \\(\\frac{n \\xi}{R}\\) = nI,
                                                                                      \nI1<\/sub>, 5 the current due to a single cell<\/p>\n

                                                                                      2) When r<< R, I = \\(\\frac{n \\xi}{nr}\\) \u2248 \\(\\frac{\\xi}{r}\\) current due to single cell.
                                                                                      \nFrom the above discussion it maybe concluded that in order to have maximum current, the cells should be connected in series, when the total internal resistance of the cells is negligible as compared to the external resistance in the circuit.<\/p>\n

                                                                                      Cells in parallel:
                                                                                      \nLet n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B as shown in figure.
                                                                                      \n\"Samacheer<\/p>\n

                                                                                      Let \u03be be the emf and r the internal resistance of each cell.
                                                                                      \nThe total emf of the cells = \u03be
                                                                                      \nThe total internal resistance of the cells is given by,
                                                                                      \n\\(\\frac{1}{\\mathrm{r}_{\\mathrm{eq}}}=\\frac{1}{\\mathrm{r}}+\\frac{1}{\\mathrm{r}}+\\frac{1}
                                                                                      \n{\\mathrm{r}}+\\ldots . . \\mathrm{n} \\text { times }\\)
                                                                                      \n\\(\\begin{aligned} \\frac{1}{\\mathrm{r}_{\\mathrm{eq}}} &=\\frac{\\mathrm{n}}{\\mathrm{r}} \\\\ \\mathrm{r}_{\\mathrm{eq}} &=\\frac{\\mathrm{r}}{\\mathrm{n}} \\end{aligned}\\)
                                                                                      \nThe total resistance of the circuit = R + \\(\\frac{\\mathrm{r}}{\\mathrm{n}}\\)
                                                                                      \nIf I is the current flowing through the circuit then,
                                                                                      \n\"Samacheer<\/p>\n

                                                                                      Form the above discussion, it may be concluded that in order to have maximum current, the cells should be connected in parallel, when the external resistance in the circuit is negligible as compared to the total internal resistance of the cells.<\/p>\n

                                                                                      Question 16.
                                                                                      \nWhen an inductor is connected to a 230 V d.c. source, a current of 2A passes through it. When the same conductor is connected to a 230 V, 50 HZ. a.c.source, the amount of current decreases (i.elA)). Why?
                                                                                      \nAnswer:<\/p>\n

                                                                                        \n
                                                                                      1. XL<\/sub> = 2\u03c0fL<\/li>\n
                                                                                      2. In the case of a DC source, f will be equal to zero.
                                                                                        \n\u2234 XL<\/sub> also becomes zero.<\/li>\n
                                                                                      3. When the resistance is less, more current 2A passes through the circuit.<\/li>\n
                                                                                      4. In the case of an AC source with f= 50 HZ, there will be some resistance added to the circuit, which reduces the overall current to 1A.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                                                                                        Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 2 Current Electricity Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity 12th Physics Guide Current Electricity Text Book Back Questions and Answers Part – 1 Text Book Evaluation I. Multiple choice questions: Question 1. …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[5],"tags":[],"class_list":["post-28383","post","type-post","status-publish","format-standard","hentry","category-class-12"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/28383"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=28383"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/28383\/revisions"}],"predecessor-version":[{"id":41434,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/28383\/revisions\/41434"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=28383"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=28383"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=28383"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}