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.
The following graph shows the current versus voltage values of some unknown conductor. What is the resistance of this conductor?
a) 2 ohm
b) 4 ohm
c) 8 ohm
d) 1 ohm
Answer:
a) 2 ohm
Solution:
V = IR
R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
From graph,
\(\text { Slope }=R=\frac{\Delta V}{\Delta I}=\frac{(40)}{(20)}=\frac{4}{2}=2 \Omega\)
Question 2.
A 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
a) πΩ
b) \(\frac{\pi}{2} \Omega\)
c) 2πΩ
d) \(\frac{\pi}{4} \Omega\)
Answer:
a) πΩ
Solution:
Circumference of circle = 2πr = 2 × π × 1 = 2π
Resistance of wire = 2 × 2π
Resistance of each section = \(\frac{4 \pi}{2}\) = 2π
Equivalent Resistance = \(\begin{array}{l}
\frac{2 \pi \times 2 \pi}{2 \pi+2 \pi} \\
\frac{4 \pi^{2}}{4 \pi}=\pi \text { ohm }
\end{array}\)
Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
a) 400 W
b) 2 W
c) 480 W
d) 240 W
Ans :
c) 480 W
Solution:
P = VI
= \(\frac{V^{2}}{R}=\frac{240 \times 24 \not b}{12 \phi}\)
= 240 × 2 = 480W
P = 480W
Question 4.
A carbon resistor of (47 ± 4.7 ) kΩ to be marked with rings of different colours for its identification. The colour code sequence will be
a) Yellow — Green — Violet — Gold
b) Yellow — Violet — Orange — Silver
c) Violet — Yellow — Orange — Silver
d) Green — Orange — Violet — Gold
Answer:
b) Yellow — Violet — Orange — Silver
Solution:
Colour code:
B B R O Y G B V G W
0 1 2 3 4 5 6 7 8 9
4 — yellow
7 — violet
1kΩ = 103 = Orange
b) is correct answer.
Question 5.
What is the value of resistance of the following resistor?
a) 100 kΩ
b) 10 kΩ
c) 1 kΩ
d) 1000 kΩ
Answer:
a) 100 kΩ
Solution:
Brown – 1
Black – 0
Yellow – 10^{4}
10 × 10^{4} = 100 × 10^{3} = 100 kΩ
Question 6.
Two wires of A and B with circular crosssection are made up of the same material with equal lengths. Suppose R_{A} = 3R_{B}, then what is the ratio of radius of wire A to that of B?
a) 3
b) √3
c) \(\frac{1}{\sqrt{3}}\)
d) \(\frac{1}{2}\)
Answer:
c) \(\frac{1}{\sqrt{3}}\)
Solution:
Given R_{A} = 3_{B}
\(\begin{array}{l}
\qquad \rho=\frac{\pi r^{2} R}{l} \\
R \alpha \frac{1}{r^{2}} \\
R_{A}=\frac{1}{r_{1}^{2}} \quad R_{B}=\frac{1}{r_{2}^{2}} \\
\frac{1}{r_{1}^{2}}=3 \frac{1}{r_{2}^{2}} \Rightarrow \frac{r_{1}^{2}}{r_{2}^{2}}=\frac{1}{3} \\
\frac{r_{1}}{r_{2}}=\frac{1}{\sqrt{3}}
\end{array}\)
Question 7.
A wire connected to a power supply of 230 V has power dissipation P_{1}. Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case, power dissipation is P_{2}. The ratio of \(\frac{P_{2}}{P_{1}}\) is
a) 1
b) 2
c) 3
d) 4
Answer:
d) 4
Solution:
Question 8.
In 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
a) R
b) 2R
c) \(\frac{\mathrm{R}}{4}\)
d) \(\frac{R}{2}\)
Answer:
c) \(\frac{\mathrm{R}}{4}\)
Solution:
\(P=\frac{V^{2}}{R}\)
India, V = 220V, P = 60W, R = R
USA, V = 110V, P’ = 60W, R’ = ?
∴ R = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}\)
P = P’
∴ \(\mathrm{R}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}}=\frac{110 \times 110}{220 \times 220}=\frac{1}{4} \quad {\mathrm{R}^{\prime}=\frac{\mathrm{R}}{4}}\)
Question 9.
In 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
a) 14 A
b) 8 A
c) 10 A
d) 12 A
Answer:
d) 12A
Solution:
Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P?
a) 1.5 Ω
b) 2.5 Ω
c) 3.5 Ω
d) 4.5 Ω
Answer:
c) 3.5 Ω
Solution:
By Kirchhoff’s voltage law,
9 = (3 × 1) + (2.5 × 1) + (P × 1)
9 =3 + 2.5 + P
9 = 5.5 + P
P = 9 – 5.5
P = 3.5 Ω
Question 11.
What is the current drawn out from the battery?
a) 1 A
b) 2 A
c) 3 A
d) 4 A
Solution:
Question 12.
The temperature deficient of resistance of a wire is 0.00125 per°C At 20°C, its resistance is 1Ω. The resistance of the wire will be 2Ω at
a) 800°C
b) 700°C
c) 850°C
d) 820°C
Answer:
d) 820°C
Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
a) 0.2 Ω
b) 0.5 Ω
c) 0.8 Ω
d) 1.0 Ω
Answer:
b) 0.5 Ω
Solution:
\(r=\left(\frac{EV}{V}\right) R\)
E = 2.1 V
V = IR = 0.2 × 10
V = 2 V
\(\begin{aligned}
r &=\left(\frac{2.12}{2}\right) 10 \\
&=\frac{0.1}{2} \times 10
\end{aligned}\)
r = 0.5 Ω
Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
a) each of them increases
b) each of them decreases
c) copper increases and germanium decreases
d) copper decreases and germanium increases
Answer:
d) copper decreases and germanium increases
Solution:
Copper 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.
Question 15.
In Joule’s heating law, when R and t are constant, if the H is taken along the y axis and I^{2} along the xaxis, the graph is _______
a) straight line
b) parabola
c) circle
d) ellipse
Answer:
a) straight line
Solution:
H = I^{2}RT
H α I^{2}
∴ The graph is a straight line.
II. Short Answer Questions:
Question 1.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\vec { j } \) . \(\vec { A } \)
The dot product of two vector quantity is a scalar form. Hence, the current is called a scalar quantity.
Question 2.
Define Current density.
Answer:
The current density is defined as the current per unit area of the crosssection of the conductor.
J = \(\mathrm{I} / \mathrm{A}\)
S.I unit is A/m^{2}
Question 3.
Distinguish between drift velocity and mobility.
Answer:
Question 4.
State microscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s Law relates voltage, current, and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.
V = IR.
Question 5.
State macroscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s law is V = IR
Where V → Potential difference
I → Current
R → Resistance.
Question 6.
What are the ohmic and nonohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be nonohmic.
Question 7.
Define electrical resistivity.
Answer:
The 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 crosssection.
\(\rho=\frac{R A}{L}\)
Its unit is ohmmeter (Ωm)
Question 8.
Define temperature coefficient of resistance.
Answer:
The ability of certain metals, their compounds, and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.
Question 9.
What are superconductors?
Answer:
 The resistance of certain materials becomes zero below a certain temperature, known as critical or transition temperature.
 The materials that exhibit this property are known as superconductors and the phenomenon is known as superconductivity.
Question 10.
What are electric power and electric energy?
Answer:
Electric Energy  Electric Power 
Electric potential energy gained by the charge carries as they move through potential difference v is dw = v dQ  Electric Power is the rate at which energy is transformed.
P = dw/dt 
dw = P dt  P = VI 
Unit: 1 kWh  Unit: Watt 
Question 11.
Derive the expression for power P = VI in an electrical circuit.
Answer:
Power is defined as the rate at which the electrical potential energy is delivered.
\(\begin{array}{l}
Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
P = [latex]\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I^{2}R
P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).
Question 13.
State Kirchhoff’s current rule.
Answer:
 Kirchhoff’s current rule states that the algebraic sum of the currents at any junction of a circuit is zero.
i.e Σ_{junction} i = 0  It is a statement of conservation of electric charge.
Question 14.
State Kirchhoff’s voltage rule.
Answer:
It 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.
Question 15.
State the principle of the potentiometer.
Answer:
The emf (E) of the cell is directly proportional to the balancing length (l)
ξ ∝ l
ξ = Irl
Where I →current
r → resistance per unit length of the wire.
Question 16.
What do you mean by the internal resistance of a cell?
Answer:
The 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.
Question 17.
State Joule’s law of heating.
Answer:
The heat developed in an electrical circuit due to the flow of current varies directly as
 the square of the current
 the resistance of the circuit and
 the time of flow H = I^{2}Rt
Question 18.
What is the Seebeck effect?
Answer:
Seebeck 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.
Question 19.
What is the Thomson effect?
Answer:
 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.
 This is known as the Thomson effect. Thomson effect is also reversible.
Question 20.
What is the Peltier effect?
Answer:
When 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.
Question 21.
State the applications of the Seebeck effect.
Answer:
 The Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
 This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
 The Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.
III. Long Answer Questions:
Question 1.
Describe the microscopic model of current and obtain the general form of Ohm’s law.
Answer:
 XY is a conductor of an area of crosssection A
 E is the applied electric field.
 n is the number of electrons per unit volume with the same drift velocity (v_{d})
Let electrons move through a distance dx in time interval dt.
Microscopic model of the current
∴ \(v d=\frac{d x}{d t}\)
dx = v_{d} . dt …………(1)
Electrons available in the given volume
= volume × number per unit volume
= A. dx × n
= A . v_{d} dt × n [From (1)]
Total charge in volume dQ = charge × number of electrons
dQ = (e) (Av_{d} dt) n
Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
 Consider a segment of wire of length l and area of cross section A
 Electric field is created when a potential difference V is applied
 If ‘E’ is uniform, then, V = El
 Current density, J = σ E = σ\(\frac{\mathrm{V}}{l}\)
Since J = \(\frac{\mathrm{I}}{\mathrm{A}}\)
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
Rearranging the above equation, Current through the conductor
Limitations:
There are certain materials and devices where the proportionality of V and I does not hold good.
Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
Resistors in series:
 Let V_{1}, V_{2}, V_{3} be the potential difference across R_{1}, R_{2}, R_{3}
 In the series network, the current is the same.
 The net potential difference V = V_{1} + V_{2} + V_{3}
 By Ohm’s law: V_{1} = IR_{1}, V_{2} = IR_{2}, V_{3}= IR_{3}
 The equivalent resistance is R_{S} and V = IR_{S} in series.
 IR_{S} = IR_{1} + IR_{2} + IR_{3}
 R_{S} = R_{1} + R_{2} + R_{3}
 The equivalent resistance is the sum of the individual resistances.
Resistors in Parallel:
 In a parallel network, the potential difference across each resistor is the same.
 current I is divided into I_{1}, I_{2}, I_{3}, same that
I = I_{1} + I_{2} + I_{3} ………..(1)
By ohm’s law
 In parallel network, the reciprocal of the equivalent resistance is equal to the sum of the reciprocal of the resistance of the individual resistors.
Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
 The emf of the cell ξ is found by connecting a high resistance voltmeter across it. Without connecting the external resistance R.
 Since the voltmeter draws very little current for deflection, the circuit is called an open circuit.
 Voltmeter reading gives the emf of the cell.
 Resistance ‘R is included in the circuit and current I is established in the circuit.
 The potential drop across the resistor.
R is V = IR …………(1)  Due to internal resistance (r) the voltmeter reads a value V, less than the emf of cell (E).
 V = ξ – Ir
 Ir = ξ – V …………(2)
 Dividing (2) by (1)
 Internal resistance of the cell
\(\frac{\mathrm{Ir}}{\mathrm{IR}}=\frac{\xi\mathrm{V}}{\mathrm{V}}\)
\(\mathrm{r}=\left(\frac{\xi\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)  Since ξ, V and R are known, internal resistance r can be calculated
Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s rules are used to find current and voltage for more complex circuits. There are 2 rules.
 Kirchhoff’s current rule.
 Kirchhoff’s voltage rule.
Kirchhoff’s current rule:
The algebraic sum of the currents at any function of a circuit is zero.
Sign convention:
 Current flowing towards a junction is positive
 Current flowing away from a junction is negative.
 From the figure
I_{1} + (I_{2}) + (I_{3}) + I_{4} + I_{5} = 0 (or)
I_{1} + I_{4} + I_{5} = I_{2} + I_{3}
Kirchhoff’s second [voltage rule or Loop rule]:
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.
Sign convention:
 The product of current and resistance is positive if the current direction is followed (a)
 If current is opposite to the direction of the loop, then the product of current and resistance is negative (h)
 The emf is positive when moving from negative to the positive terminal (c) & (d).
Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
 Let P, Q, R, and S be the four resistances in the form of a bridge.
 A cell of emf E is connected between points A and C
 The current I is divided into I_{1}, I_{2}, I_{3,} and I_{4} across the four branches.
By Kirchhoff’s current rule  At junction B
I_{1} – I_{g} – I_{3} = 0 ………..(1)
 At junction D
I_{2} + I_{g} – I_{4} = 0 …………..(2)
By Kirchoff’s voltage rule  At closed path ABDA
I_{1}P + I_{g}G – I_{2}R = 0 ……….(3)  At closed path ABCDA
I_{1}P + I_{3}Q – I_{4}S – I_{2}R = 0 ………(4)  When the galvanometer shows zero deflection, the points B and D are at the same potential, and I_{g} = 0
 Substitute I_{g} = 0 in (1), (2) and (3)
I_{1} = I_{3} …………….(5)
I_{2} = I_{4} …………….(6)
I_{1}P = I_{2}R …………….(7)  Substitute (5) and (6) in (4)
I_{1}P + I_{1}Q – I_{2}S – I_{2}R = 0
I_{1}(P + Q) = I_{2}(R + S) …………….(8)
Dividing (8) by (7),
This is the condition of bridge balance.
Question 7.
Explain the determination of unknown resistance using a meter bridge.
Answer:
 It consists of a uniform manganin wire AB of 1meter length.
 The wire is stretched along a meter scale on a wooden board between 2 copper strips C and D
 Between the 2 copper strips, another copper strip E is mounted to enclose gaps G_{1} and G_{2}
 An unknown resistance P is connected in G_{1}
 A standard resistance Q is connected in G_{2}.
 A jockey is connected to E through a galvanometer (G) and a high resistance (HR)
 The exact position of the jockey can be read on the scale.
 A Lechianche cell and a key (k) are connected across the ends of the wire.
 By measuring the radius r and length
 l of the wire P_{1} then specific resistance of the wire is found.
P = Resistance × \(\frac{A}{l}\)  If l is unknown resistance specific Resistance
\(P=\frac{P \cdot \pi r^{2}}{l}\)
 The jockey is adjusted for zero deflection. Let the point be J.
 Lengths AJ and JB replace resistances R and S in Wheatstone’s bridge
 Then, \(\frac{P}{Q}=\frac{R}{S}=\frac{\left.R^{\prime} A\right]}{R^{\prime} J B}\)
Where R’ → resistance per unit length of wire
\(\frac{P}{Q}=\frac{A J}{J B}=\frac{l_{1}}{l_{2}}\)
\(P=Q \cdot \frac{l_{1}}{l_{2}}\)  Due to the soldering of copper strips, there occurs an error called end resistance.
 This can be eliminated by taking another set of readings by interchanging P and Q.
 By measuring the radius a and length l of the wire P, then specific resistance of the wire is found
\(\rho=\text { Resistance } \times \frac{A}{l}\)
If \(\rho\) is unknown resistance equation becomes
\(\rho=\frac{P \cdot \pi a^{2}}{l}\)
Question 8.
How the emf of two cells are compared using a potentiometer?
Answer:
The primary circuit consists of battery (Bt), Key (k), and rheostat (Rh)
 The end C is connected to M of a DPPT switch.
 The terminal N is connected to the jockey (J) through a galvanometer (G) and high resistance (HR)
 cell 1 is connected between M_{1} and N_{1}
 cell 2 is connected between M_{2} and N_{2}
 The jockey is adjusted for zero deflection.
 The DPDT has pressed towards ξ_{1} The potential difference across balancing length l_{1} is Irl_{1}
 ξ_{1} = Irl_{1} ………..(1)
 DPDT switch is pressed towards ξ_{2}
 ξ_{2} = Irl_{2} ………..(2)
Comparison of EMF of two cells  Dividing (1) by (2)
\(\frac{\xi 1}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)  If ξ_{1} is known, then unknown emf
\(\xi_{2}=\xi_{1} \cdot \frac{l_{2}}{l_{1}}\)  The experiment can be repeated several times by changing the current emf of flowing through it.
IV. Numeric Problems:
Question 1.
The 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?
Answer:
Question 2.
Lightning is a very good example of a natural current. In typical lightning, there is 10^{9} J energy transfer across the potential difference of 5 × 10^{7} V during a time interval of 0.2 s. Using this information, estimate the following quantities:
(a) the total amount of charge transferred between cloud and ground
(b) the current in the lightning bolt
(c) the power delivered in 0.2 s.
Answer:
Given data: E = 10^{9}J, V = 5 × 10^{7} V, t = 0.2s
Question 3.
A copper wire of 10^{6} m^{2} area of crosssection, carries a current of 2 A. If the number of electrons per cubic meter is 8 × 10^{28}, calculate the current density and average drift velocity.
Answer:
Given data:
A = 10^{6} m^{2}, I = 2A, n = 8 × 10^{28}
Formula:
Current density, J = \(J=\frac{I}{A}\)
Question 4.
The resistance of a nichrome wire at 0°C is 10Ω. If its temperature coefficient of resistivity of nichrome is 0.004/ °C, find its resistance of the wire at boiling point of water. Comment on the result.
Answer:
Given data:
T_{1} = 10°C, R_{0} = 10Ω
The boiling point of water T = 100°C
α = 0.0041°C
R_{T}=?
Formula:
R_{T} = R0 [l + α(T – T_{0})]
R_{T} = 10 [1+0.004(100 – 20)]
= 10(1 + 0.32)
= 10 (1.32)
RT = 13.2Ω
As the temperature increase the resistance of the wire also increases.
Question 5.
The rod given in the figure is made up of two different materials.
Both have square crosssections of 3 mm side. The resistivity of the first material is 4 × 10^{3}Ω and that of the second material has a resistivity of 5 × 10^{3}Ωm. What is the resistivity of the rod between its ends?
Answer:
Question 6.
Three 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.
(a) Calculate the current in the circuit when S is open and closed
(b) What happens to the intensities of the bulbs A, B, and C.
(c) Calculate the voltage across the three bulbs when S is open and closed
(d) Calculate the power delivered to the circuit when S is opened and closed
(e) Does the power delivered to the circuit decreases, increases or remain the same?
Answer:
Question 7.
An electronics hobbyist is building a radio which requires 150Ω in her circuit, but she has only 220Ω, 79Ω, and 92Ω resistors available. How can she connect the available resistors to get the desired value of resistance?
Answer:
Available resistances = 220Ω, 79Ω 92Ω
Case I:
If 3 resistors are connected in series, then
R_{S} = R_{1} + R_{2} + R_{3} = 220 + 79 + 92 = 391
This value is greater than the required resistance so it is not possible.
Case II:
If 3 resistors are connected in parallel, then
Question 8.
A cell supplies a current of 0.9 A through a 2Ω resistor and a current of 0.3 A through a 7Ω resistor. Calculate the internal resistance of the cell.
Answer:
ξ = 0.3(7 + r)
Since ‘ξ’ is constant
0.9(2+r) = 0.3(7+r)
1.8 + 0.9r = 2.1 + 0.3r
O.6r = 0.3
r = \(\frac{0.3}{0.6}\)
r = \(\frac{1}{2}\) = O.5Ω
Question 9.
Calculate the currents in the following circuit.
Answer:
At junction B, applying current law,
I_{1} – I_{2} – I_{3} = 0
I_{1} = I_{2} + I_{3} ………..(1)
Kirchhoff’s voltage law in loop ABEFA
100I_{3} + 100I_{1} = 15
Using (1), we get,
100I_{3} + 100 (I_{2} + I_{3}) = 15
100I_{3} + 100I_{2} + 100_{3} = 15
100I_{2} + 200_{3} = 15 ……………(2)
Voltage law in loop BCDEB
100I_{2} – 100I_{3} = 9
100I_{3} – 100I_{2} = 9 ……………(3)
Adding (1) & (2), we get
\(\begin{array}{l}
200 \mathrm{I}_{3}+100 \mathrm{I}_{2}=15 \\
100 \mathrm{I}_{3}100 \mathrm{I}_{2}=9 \\
300 \mathrm{I}_{3} \quad=24 \quad \mathrm{I}_{3}=\frac{24}{300}=0.08
\end{array}\)
I_{3} = 0.08A
(3) → 100I_{3} – 100I_{2} = 9
Substituting the value of I_{3}, we get
100 × 0.08 – 100I_{2} = 9
8 – 100I_{2} = 9
100I_{2} = 1
\(\Rightarrow \mathrm{I}_{2}=\frac{1}{100}=0.01 \mathrm{~A}\)
I_{2} = 0.01A
From (1),
I_{1} = I_{2} + I_{3}
= 0.01 + 0.08
I_{1} = 0.07A
Question 10.
A potentiometer wire has a length of 4 m and a resistance of 20Ω. It is connected in series with a resistance of 2980Ω and a cell of emf 4 V. Calculate the potential along the wire.
Answer:
Given data:
l = 4m of R = 20Ω
In series with R’ = 2980Ω
E = 4 V
Question 11.
Determine the current flowing through the galvanometer (G) as shown in the figure.
Answer:
I_{2} = I – I_{1}
Current flowing through the circuit I = 2A
Applying Kirchhoffs II law to PQSP
5I_{1} + 10I_{g} – 15I_{2} = 0
5I_{1} + 10I_{g} – 15(I – I_{1}) = 0
5I_{1} + 10I_{g} – 15I + 15I_{1} = 0
20I_{1} + 10I_{g} = 15 × I
20I_{1} + 10I_{g} = 30
2I_{1} + I_{g} = 3 ………..(1)
Applying Kirchhoffs II law to QRSQ
10(I_{1} – I_{g}) – 20 (I_{2} + I_{g}) – 10 I_{g} = 0
10I_{1} – 10 I_{g} – 20 (I – I_{1} + I_{g}) – 10 I_{g} = 0
10I_{1} – 10 I_{g} – 20I + 20I_{1} – 20I_{g} – 10I_{g} = 0
30I_{1} – 40I_{g} = 20I
30I_{1} – 40I_{g} = 20 × 2
3I_{1} – 4I_{g} = 4 ………..(2)
(1) × 3 ⇒ 6I_{1} + 3I_{g} = 9 …………..(3)
(2) × 2 ⇒ 6I_{1} – 8I_{g} = 8 ……………(4)
Solving (3) and (4)
+ 11 I_{g} = 1
\(\mathrm{I}_{\mathrm{g}}=\frac{1}{11} \mathrm{~A}\)
Question 12.
Two cells each of 5V are connected in series across an 8Ω resistor and three parallel resistors of 4Ω, 6Ω, and 12Ω. Draw a circuit diagram for the above arrangement. Calculate
(i) the current drawn from the cell
(ii) current through each resistor
Answer:
Circuit Diagram:
i) Current drawn from the cell
E_{eq} = 5+5 = 10V
R_{eff} = R_{s}+ R_{p}
R_{s} = 8Ω
\(\frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{18+12+6}{72}=\frac{36}{72}\)
∴R_{p} = \(\frac{72}{36}\) = 2Ω
R_{eff} = 8 + 2 = 10Ω
\(I=\frac{E_{e g}}{R_{e f f}}=\frac{10}{10}=1 A\)
∴ I = 1A
Voltage drop V= IR =1 × 2 = 2V
V = 2V
ii) Current through each resistor.
∴ current in 4Ω resistor
I = \(\frac{2}{4}\) = O.5A
I = O.5A
current in 6Ω resistor, I = \(\frac{2}{6}\) = 0.33A
I = 0.33A
current in 12Ω resistor, I = \(\frac{2}{12}\) = \(\frac{1}{6}\) = 0.17A
I = 0.17A
Question 13.
Four 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  Q  R  S  
P removed  *  on  on  on 
Q removed  on  *  on  off 
R removed  off  off  *  off 
S removed  on  off  on  * 
Draw the circuit diagram for these bulbs.
Answer:
Question 14.
In 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?
Answer:
E_{1} = 1.25 V
l_{1} = 35 cm
E_{2} =?
l_{2} = 63cm
img 81
Emf of the second cell is 2.25 V
Part II:
12th Physics Guide Current Electricity Additional Questions and Answers
I(a). Match the following:
a. current  i. coulomb 
b. Drift velocity  ii. m^{2}/Vs 
c. charge  iii. m/s 
d. mobility  iv. Ampere 
Answer:
a. iv
b. iii
c. i
d. ii
I(b). Match the following:
a. microscopic form of Ohm’s law  i. V = IR 
b. microscopic form of Ohm’s law  ii. a \(\tau\) 
c. Drift velocity (v_{d})  iii.\(\overrightarrow{\mathrm{J}}=\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}} \overrightarrow{\mathrm{E}}\) 
d. Current density (J)  iv. nev_{d} 
Answer:
a. iii
b. i
c. ii
d. iv
I(c). Match the following:
a. Heating element  i. 3000° 
b. Fuse wire  ii. Copper 
c. carbon are furnace  iii. Copper 
d. Electrical lamp  iv. Nichrome 
Answer:
a. iv
b. iii
c. i
d. ii
I(d). Match the following:
a. Power  i. Ohmmetre 
b. Energy  ii. Watt 
c. electrical resistivity  iii. Ω_{1} m Ω_{1} 
d. electrical conductivity  iv. Joule 
Answer:
a. ii
b. iv
c. i
d. iii
II. Fill in the Blanks.
Question 1.
The graph plotted with current against the voltage that obeys Ohm’s law is _______.
Answer:
Straight line
Question 2.
Current is a _______ quantity.
Answer:
Scalar
Question 3.
________ is used in complicated circuits.
Answer:
Kirchhoff’s laws
Question 4.
Mercury exhibits superconductor behaviour at _______.
Answer:
4.2K
III. Choose the Odd One Out:
Question 1.
a) Current
b) current density
c) drift velocity
d) Electric field
Answer:
a) Current – it is scalar and others are vector.
Question 2.
a) Current
b) resistance
c) time
d) electromotive force
Answer:
d) electromotive force – because heat dissipated depends only on the first 3 factors.
Question 3.
a) Current rule
b) Voltage rule
c) Wheatstone’s bridge
d) Joule’s law
Answer:
d) Joule’s law – Because the first 3 comes under Kirchhoff’s laws.
Question 4.
a) 1kWh
b) 1Js^{1
}c) 1000Wh
d) 3.6 × 10^{6}J
Answer:
b) 1Js^{1} – Others are equivalent of 1 unit of electrical energy
IV. Choose the Incorrect Pair:
Question 1.
a) Ohmic material – Straight line
b) non – Ohmic material – nonlinear
c) macroscope form – V = IR
d) macroscope form – V = El
Answer:
d) macroscope form – V = El
Question 2.
a) Electrical resistivity – Ωm
b) Electrical conductivity – Ω^{1} m^{1}
c) Internal resistance – Ωm^{1}
d) Electrical resistance – Ω
Answer:
c) Internal resistance – Ωm^{1}
Question 3.
a) Resistors in series – R_{s} = R_{1} + R_{2}
b) Resistors in parallel – R_{p} = \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
c) Internal resistance – r = \(\left(\frac{\mathrm{E}\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)
d) Energy – E = VI
Answer:
b) Resistors in parallel – R = R_{p} = \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
Question 4.
a) Electromotive force – force acting
b) Electric energy – work done
c) Electric power – rate of change in electric energy
d) Internal resistance – resistance offered
Answer:
a) Electromotive force – force acting because emf is the potential difference
V. Choose the Correct Pair:
Question 1.
a) Household appliances – Parallel network
b) HousehoLd appliances – Series network
c) \(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) – R_{1} + R_{2}
d) RR_{p} – \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
Answer:
a) Household appliances — Parallel network
Question 2.
Colour code in carbon resistor
a) Brown ring – 0
b) Yellow ring – 1
c) Silver ring – 10%
d) Colourless – 5%
Answer:
a) Silver – 10%
Question 3.
a) Voltmeter – used to measure current
b) Ammeter – used to measure voltage
c) Multimeter – used to measure voltage current and resistance
Answer:
c) Multimeter – used to measure voltage current and resistance
Question 4.
a) Electric heater – tungsten
b) Electric fuse – trippers
c) Electric furnace – Nichrorne
d) Electric lamp – Carbon arc
Answer:
b) Electric fuse – trippers
VI. Assertion and Reason:
Question 1.
A: Fuses are connected in series to protect electric devices.
R: It melts and breaks the circuit if the current exceeds a certain value.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
b) R explains A
Question 2.
A: Positive charge flows from higher electric potential to lower electric potential.
R: These positive charges constitute the flow of current.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
c) A is correct, R is wrong
Question 3.
A: Current is a scalar quantity.
R: Current does not obey the law of vector addition.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
b) R explains A
Question 4.
A: A semiconductor with a negative temperature coefficient of resistance is called a thermopile.
R: The value of a remains the same for all materials
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
d) Both A and R are wrong
VII. Choose the Correct Statement:
Question 1.
a) All carbon resistors have a tolerance value.
b) The positive charges constitute the electric current
c) Allessandro Volta invented the telephone
d) Ohm’s law is used for complicated circuits.
Answer:
a) All carbon resistors have a tolerance value
Question 2.
a) The innermost electrons in the atom is called the free electrons.
b) All atoms are neutral with an equal number of protons and electrons.
c) Conventional current flows from negative to positive charges.
d) The outermost electrons are core electrons.
Answer:
b) All atoms are neutral with an equal number of protons and electrons
Question 3.
a) Charging the battery on my mobile.
b) My mobile phone battery has no charge
c) Battery has no charge.
d) My mobile is charging.
Answer:
d) My mobile is charging.
Question 4.
a) Ohmic materials do not have constant resistance.
b) The V – I graph for ohmic materials is nonlinear.
c) Nonohmic materials do not have constant resistance.
d) The V – I graph for nonohmic materials is linear. constant resistance.
Answer:
c) Nonohmic materials do not have constant resistance.
VIII. Choose the Incorrect Statement:
Question 1.
a) The value of the equivalent resistance in series connection is lesser than each individual resistance.
b) The value of equivalent resistance in the parallel network is lesser than individual resistance.
c) The value of equivalent resistance in the series network is R_{s} = R_{1} + R_{2}
d) The value of equivalent resistance in parallel network is \(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
Answer:
a) The value of the equivalent resistance in series connection is lesser than each individual resistance.
Question 2.
a) Superconductivity was discovered by H.K. Onnes.
b) The resistance of certain materials becomes zero.
c) That temperature is called critical temperature.
d) Experiments showed Ag exhibits superconductivity at 4.2K
Answer:
d) Experiments showed Ag exhibits superconductivity at 4.2K
Question 3.
a) Electric power produced by a resistor is I“R.
b) It depends on the square of the current.
c) If the current is doubled, the power will increase by 8 times.
d) The SI unit of power is a watt.
Answer:
c) If the current is doubled, the power will increase by 8 times.
Question 4.
a) Kirchhoff’s current law obeys the laws of conservation of charges
b) Kirchhoff’s voltage law obeys the laws of conservation of energy
c) Important application of Kirchhoff s rules is Wheatstone’s bridge
d) This rule is used for simple electrical circuits.
Answer:
d) This rule is used for simple electrical circuits.
IX. Choose the best answer:
Question 1.
When 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 crosssection, the drift velocity of the electrons will be
(a) \(\frac { v }{ 4 }\)
(b) \(\frac { v }{ 2 }\)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
V_{d} = \(\frac { 1 }{ nAe }\); v’_{d} = \(\frac { 2I }{ (2A)ne }\) = v_{d}
Question 2.
The resistance of the wire varies inversely as _______.
a) Area of the crosssection
b) Resistivity
c) Length
d) Temperature
Answer:
a) Area of the crosssection
Question 3.
The curve representing Ohms law is a _______.
a) linear
b) cosme function
c) parabola
d) Hyperbola
Answer:
a) linear
Question 4.
A 10 m long wire of resistance 20Ω is connected in series with a battery of emf 3V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = \(\frac { 20 }{ 3 }\) x 3 = 2 V
Potential gradient = \(\frac { v }{ l }\) = \(\frac { 2 }{ 10 }\) = 0.2 V/m
Question 5.
To produce an electric current what is the requirement?
a) A voltage source
b) a source of energy that moves charges
c) an electric field
d) Any of the above
Answer:
d) Any of the above
Question 6.
In Joule’s heating law, when R and t are constant, if the H is taken along the yaxis and I along the xaxis, the graph is:
a) straight line
b) parabola
c) circle
d) ellipse
Answer:
a) straight line
Question 7.
A series circuit consists of 3 resistors with 140Ω, 250Ω, and 220Ω. The total resistance is _______.
a) 330Ω
b) 610Ω
c) 720Ω
d) None of the above
Answer:
b) 610Ω
Question 8.
A cell has an emf of 1.5 V. When shortcircuited, it gives a current of 3A. The internal resistance of the cell is
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) \(\frac { 1 }{ 4.5 }\) Ω
Answer:
(a) 0.5 Ω
Hint:
r = \(\frac { ξ }{ I }\) = \(\frac { 1.5 }{ 3 }\) = 0.5 Ω.
Question 9.
The instrument used for measuring electric current is ________.
a) galvanometer
b) ammeter
c) voltmeter
d) potentiometer
Answer:
b) ammeter
Question 10.
What is the most commonly used conductor in electronics?
a) copper
b) aluminum
c) gold
d) silver
Answer:
a) copper
Question 11.
Kirchhoff’s two laws for electrical circuits are manifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy
Question 12.
Why does a circuit require a battery?
a) measure current
b) maintain a potential difference
c) oppose the current
d) measure potential
Answer:
b) maintain a potential difference
Question 13.
In the above circuit, the equivalent resistance between A and B is
a) \(\frac{20}{3}\) Ω
b)10Ω
c)16Ω
d)20Ω
Answer:
c) 16Ω
Question 14.
A flow of 10^{7} electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10^{26} A
(b) 1.6 x 10^{12} A
(c) 1.6 x 10^{12} A
(d) 1.6 x 10^{26} A
Answer:
(c) 1.6 x 10^{12} A
Hint:
I = \(\frac { Q }{ t }\) = \(\frac{10^{7} \times 1.6 \times 10^{19}}{1}\) = 1.6 x 10^{12} A.
Question 15.
A short circuit has _______.
a) no resistance
b) no conductance
c) low current
d) None of the above
Answer:
b) no conductance
Question 16.
A current of 2A flows through a 12V bulb then calculates the resistance?
a) 6Ω
b) 0.16Ω
c) 32Ω
d) 0.32Ω
Answer:
c) 32Ω
Question 17.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel
Question 18.
The resistors each on resistance 1Ω is connected as shown in the figure. The resultant resistance between A and B is
a) \(\frac{11}{5} \Omega\)
b) \(\frac{5}{11} \Omega\)
c) \(\frac{6}{5} \Omega\)
d) \(\frac{5}{6} \Omega\)
Answer:
b) \(\frac{5}{11} \Omega\)
Question 19.
The rings in the carbon resistor are red red red silver. What is its value?
a) 22 × 10^{2} ± 10%
b) 22 × 10^{3} ± 10%
c) 22 × 10^{2} ± 20%
d) 22 × 10^{3} ± 20%
Answer:
a) 22 × 10^{2} ± 10%
Question 20.
The electrical resistivity of a thin copper wire and a thick copper rod are respectively ρ_{1}m and ρ_{2}m then _______
a) \(\rho_{1}>\rho_{2}\)
b) \(\rho_{2}^{>} \rho_{1}\)
c) \(\rho_{1}=\rho_{2}\)
d) \(\frac{\rho_{2}}{\rho_{1}}=\infty\)
Answer:
c) \(\rho_{1}=\rho_{2}\)
Question 21.
A material with a negative temperature coefficient of resistance is called _______.
a) metal
b) alloy
c) thermistor
d) thermometer
Answer:
c) thermistor
Question 22.
Kirchoff’s first law is a consequence of conservation of ________
a) current
b) charges
c) energy
d) power
Answer:
b) charges
Question 23.
Condition for bridge balance of Wheatstone’s bridge is _______.
a) \(\frac{P}{Q}=\frac{S}{R}\)
b) \(\frac{P}{Q}=R S\)
c) \(\frac{P}{Q}=\frac{R}{S}\)
d) \(\frac{Q}{P}=\frac{R}{S}\)
Answer:
c) \(\frac{P}{Q}=\frac{R}{S}\)
Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) \(\frac { E }{ 5 }\)
(c) E
(d) \(\frac { 5E }{ 2 }\)
Answer:
(c) E
Question 25.
Nichrome wire is used as the heating element because it has ________
a) low specific resistance
b) low melting point
c) high specific resistance
d) high conductivity
Answer:
c) high specific resistance
Question 26.
Peltier effect is the converse of ________
a) Joule effect
b) Raman effect
c) Thomson effect
d) Seebeck effect
Answer:
d) Seebeck effect
Question 27.
The potential difference between the points A and B in the given figure is ________.
a) 3V
b) +3V
c) +6V
d) +9V
Answer:
d) +9V
Question 28.
Potentiometer measures potential more accurately because
(a) It measures potential in the open circuit.
(b) It uses a sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measures potential in the open circuit.
X. Two Mark Questions:
Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any crosssection of a conductor in time t
then, I = \(\frac { Q }{ t }\).
Question 2.
Define 1 ampere.
Answer:
1A of current is equivalent to 1 Coulomb of charge passing through a perpendicular crosssection in 1 second.
Question 3.
Does lightning produce an electric current? How?
Answer:
 The lightning bolt produces an enormous electric current in a short time.
 During lightning, a very high potential difference is created between the clouds and ground so charges flow between the clouds and ground.
Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials becomes zero below a certain temperature T_{c}. This temperature is known as critical temperature or transition temperature.
Question 5.
Define mobility.
Answer:
 The mobility of the electron is defined as the magnitude of the drift velocity per unit electric field.
\(\mu=\frac{\left\vec{V}_{d}\right}{E}\)  The SI unit of mobility is \(\frac{\mathrm{m}^{2}}{\mathrm{~V}_{\mathrm{S}}}\)
Question 6.
Define resistance.
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor.
Question 7.
Three bulbs 40w, 60w, and 100w are connected to 220 V mains which bulb will glow brightly. If they are joined in series?
Answer:
From the relation = \(\frac{\mathrm{V}^{2}}{\mathrm{p}}\)
It follows that resistance of 40w bulb will be maximum.
Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:
 Insulator
 Semiconductor.
Question 9.
Define equivalent resistance when resistors are connected in series.
Answer:
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual) resistances.
R_{S} = R_{1} + R_{2} + R_{3}
Question 10.
Define equivalent resistance in a parallel resistance network.
Answer:
When 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
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)
Question 11.
Household appliances are always connected in parallel, why?
Answer:
Household appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.
Question 12.
What is a carbon resistor?
Answer:
 Carbon 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.
Question 13.
What is a multimeter?
Answer:
A 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.
Question 14.
What is a thermistor?
Answer:
A semiconductor with a negative temperature coefficient of resistance is called a thermistor.
Question 15.
Define the internal resistance of a battery.
Answer:
A real battery is made of electrodes and electrolytes, there is resistance to the flow of charges within the battery.
Question 16.
The headlights of a car turn dim on starting the car, why?
Answer:
When the car engine is started with headlights turned on, they sometimes become dim. This is due to the internal resistance of the car battery.
Question 17.
Why is a potentiometer preferred over a voltmeter for comparison of emf of cells?
Answer:
The 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.
Question 18.
What are the factors that affect the heat produced in an electrical circuit?
Answer:
 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
 In the absence of any other external effect, this energy is spent heating the conductor.
The amount of heat(H) produced is H = VIt
For a resistance R,
H = I^{2}Rt
This is known as Joule’s law of heating.
The heat developed in an electrical circuit due to the flow of current varies directly as

 the square of the current
 the resistance of the circuit and
 the time of flow.
Question 19.
Why nichrome is used as a heating element?
Answer:
Nichrome has a high specific resistance arid can be heated to very high temperatures without oxidation.
Question 20.
What is the use of trippers in houses?
Answer:
Whenever 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.
Question 21.
What are the applications of the heating effect of current?
Answer:
Electric discharge lamps, electric welding, and electric arc also utilize the heating effect of current
Question 22.
What is the thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as the thermoelectric effect.
Question 23.
A wire of resistance 10Ω is stretched uniformly, to thrice its original length calculate the resistance of the stretched wire.
Answer:
R = 10Ω ⇒ R = \(\frac{\rho l}{A}\)
Original length = l
New length = l’ =3l
Original area = A
If length increases, area decreases
Question 24.
Calculate the internal resistance of a 2.1V cell which gives a current of 0.2A through a resistance of 10Ω
Answer:
\(\mathrm{r}=\left(\frac{\xi\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)
V = IR = 0.2 × 10 = 2V
\(r=\left(\frac{2.12}{2}\right) 10\) = 0.5Ω
r = 0.5Ω
Question 25.
Differentiate between the Joule heating effect and the Peltier effect.
Answer:
Joule heating effect  Peltier effect 
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.  1. An electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. 
Question 26.
What is the transition or critical temperature in superconductivity?
Answer:
The 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.
Question 27.
Write the formula for electrical conductivity and write its unit?
Answer:
 Electrical conductivity,
\(\sigma=\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}}\)  It’s unit is Ω^{1} m^{1}
Question 28.
Calculate the current in the wire, if a charge of 180C flows through a wire for 1 minute.
Answer:
Current I = \(\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{180}{60}\) = 3A
I = 3A
Question 29.
A parallel combination of two cells of emf’s E_{1} and E_{2}, internal resistances r_{1} and r_{2} are used to supply current to load of resistance R – Write the expression for the current through the load in terms E_{1}, E_{2}, r_{1,} and r_{2}.
Answer:
Question 30.
A wire with uniform crosssection A, length l, and resistance R is bent into a complete circle. Calculate the resistance between two diametrically opposite points.
Answer:
The resistance between two diametrically opposite points is R/4Ω.
XI. Three Mark Questions.
Question 1.
Derive a relation between drift velocity and mobility.
Answer:
 The drift velocity is the average velocity acquired by the electrons inside the conductor
 The average time between successive collisions is called the mean free time denoted by τ.
 The acceleration \(\overrightarrow{\mathrm{a}\) experienced by the electron in an electric field \(\overrightarrow{\mathrm{E}\) is given by
Question 2.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon 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.
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 = 10^{3} (orange) and tolerance = 5% (gold). The value of resistance = 56 x 10^{3} Q or 56 kΩ with a tolerance value of 5%.
Question 3.
Draw the current versus voltage graph for the ohmic and nonohmic devices. Give one example for each.
Answer:
 Ohmic devices – Conductors.
 Nonohmic devices – Diode.
Question 4.
Repairing the electrical connection with the wet skin is always dangerous why?
Answer:
 The human body contains a large amount of water which has a low resistance of around 200Ω and the dry skin has high resistance of around 500 kΩ.
 But when the skin is wet, the resistance is reduced to around 1000Ω. This is the reason, repairing the electrical connection with the wet skin is always dangerous.
Question 5.
Under what condition, is the heat produced in an electric circuit
(i) directly proportional and
(ii) inversely proportional to the resistance of the circuit?
Answer:
H = I^{2}Rt
Its constant current is flowing through a circuit then heat produced is directly proportional to resistance R
But H = I^{2}Rt
\(\begin{array}{l}
I=\frac{V}{R} \\
\text { then } H=\frac{V^{2}}{R^{2}} \times k t=\frac{V^{2}}{R} t
\end{array}\)
If a constant voltage is flowing through a circuit then heat produced is inversely proportional to resistance R.
Question 6.
How are batteries formed?
Answer:
 An electric cell converts chemical energy into electrical energy to produce electricity. It contains two electrodes immersed in an electrolyte.
 Several electric cells connected together form a battery.
 When a cell or battery is connected to a circuit, electrons flow from the negative terminal to the positive terminal through the circuit.
 By using chemical reactions, a battery produces a potential difference across its terminals.
 This potential difference provides the energy to move the electrons through the circuit.
Question 7.
How will you determine the current in a circuit when 2 or more cells are connected in series.
Answer:
 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.
 The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.
 Suppose n cells, each of emf ξ volts and internal resistance r ohms is connected in series with an external resistance R
 The total emf of the battery = nξ
 The total resistance in the circuit = nr + R. By Ohm’s law, the current in the circuit is
 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.
Question 8.
What is end resistance and how will you eliminate it.
Answer:
 The bridge wire is soldered at the ends of the copper strips.
 Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances.
 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.
Question 9.
Write a note on the electric furnace.
Answer:
 Furnaces are used to manufacture a large number of technologically important materials such as steel, silicon carbide, quartz, gallium arsenide, etc.
 To produce temperatures up to 1500°C, a molybdenumnichrome wire wound on a silica tube is used.
 Carbon arc furnaces produce temperatures up to 3000 °C.
Question 10.
Compare the emf and potential difference
Answer:
Emf  Potential difference 
1. The difference of potentials between the 2 terminals of a cell in an open circuit is called emf.  The difference in potentials between any 2 points in a closed circuit is called the potential difference. 
2. The emf is independent of the external resistance  It is proportional to the resistance between any 2 points. 
Question 11.
The colours of the carbon resistors are orange, orange, orange. What is the value of the resistor?
Answer:
 The first orange ring corresponds to 3. The next orange ring corresponds to 3.
 The next orange ring corresponds to 3, that is the 3rd orange ring corresponds to 10^{3}.
 There is no coloured ring at the end. So tolerance is 20%.
 Total resistance is 33 × 10^{3}Ω ± 20%
Question 12.
In the given circuit, the heat produced per second in 6Ω resistor is 32 J. Then calculate the heat produced per second in 3Ω resistor.
Answer:
Question 13.
Find the value of unknown resistance x in the given circuit.
Answer:
P.d across 1Ω = P.d across xaxis
6 × 1 = (11 – 9)xaxis
6 = 2x
[∴11 = 3 + 6 + x, x = (11 – 9)]
x = 3Ω
Question 14.
Aluminum 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.
Answer:
Question 15.
The resistance of a metal wire of length AB to 2Ω. Another wire of length PQ of the same material with twice the diameter of AB is found to have the same resistance of 2Ω. what is the length of PQ?
Answer:
Question 16.
A 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?
Answer:
Question 17.
Derive an expression for Joule’s law.
Answer:
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 = Vit
In the absence of any other external effect, this energy is spent heating the conductor.
The amount of heat(H) produced is
H = Vit
For a resistance R,
H = I^{2}Rt
This is known as Joule’s law of heating. The heat developed in an electrical circuit due to the flow of current varies directly as
 the square of the current
 the resistance of the circuit and
 the time of flow.
Question 18.
Write a note on electric fuses.
Answer:
 Fuses are connected in series in a circuit to protect the electric devices from the heat developed by the passage of excessive current.
 It is a short length of a wire made of a low melting point material.
 It melts and breaks the circuit if the current exceeds a certain value.
 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.
Question 19.
Distinguish between Positive and Negative Thomson effect.
Answer:
Positive Thomson effect  Negative Thomson Effect 
1. Heat is absorbed first and then evolved  Heat is evolved first and then absorbed 
2. Heat is transferred in the direction of current  Heat is transferred in the direction opposite to the direction of the current. 
3. Eg. Silver, Zinc, cadmium  Eg. Platinum, nickel, cobalt, mercury 
Question 20.
V = I graph for a metallic wire at two different temperatures T_{1} and T_{2} is shown in the Fig which of the two temperatures T_{1} and T_{2} is higher and why?
Answer:
If the same potential difference V is applied current equal to I_{1} and I_{2} will flow through the wire, when its temperature is kept at T_{1} and T_{2} respectively since R = V/I it follows that resistance of the wire is more at temperature T_{2} further, as in case of the metals the resistance increases with rising of temperature T_{2} > T_{1}.
Question 21.
Which material is used for the meter bridge wire and why?
Answer:
The 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.
XII. Five Mark Questions:
Question 1.
Establish a relation between drift velocity and current density.
Answer:
 XY is a conductor of the area of crosssection A
 E is the applied electric field.
 n is the number of electrons per unit volume with the same drift velocity (vd)
 Let electrons move through a distance dx in time interval dt.
\(\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} ; \mathrm{d} \mathrm{x}=\mathrm{v}_{\mathrm{d}} \mathrm{dt} \ldots \ldots \ldots \ldots \ldots(1)\)
 Electrons available in the given volume = volume × number per unit volume
= Adx × n
= A v_{d} × n[from (1)]
Total charge in volume dQ = charge × number of electrons
dQ = (e)(A v_{d})n
Hence the current,
\(\begin{array}{l}
I=\frac{d Q}{d t} \\
I=\frac{n e A v_{d} d t}{d t}
\end{array}\)
I = neAv_{d}dt.
Question 2.
Deduce ohm’s law using the concept of current density?
Answer:
 Consider a segment of wire of length l and area of cross section A.
 Electric field is created when a potential difference V is applied.
 If ‘E’ is uniform, then, V = El
 Current density, J = σE = σ\(\frac{V}{l}\)
 Since J = \(\frac{I}{A}\)
\(\frac{I}{A}=\sigma \frac{V}{l}\)
Rearranging the above equation,
\(\mathrm{V}=\mathrm{I}\left(\frac{I}{\sigma A}\right)\)
Where \(\left(\frac{I}{\sigma A}\right)\) – resistance of conductor (R)
:. R α \(\left(\frac{I}{\sigma A}\right)\)
:. The macroscopic form of ohm’s law is
V = IR
V α I.
Question 3.
Explain how the internal resistance of a cell can be determined using a potentiometer.
Answer:
Measurement of internal resistance
 To measure the internal resistance of a cell, the circuit connections are made as shown in Figure.
 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 K_{1} which forms the primary circuit.
Measurement of internal resistance:
 The positive terminal of the cell ξ whose internal resistance is to be determined is also connected to the end C of the wire.
 The negative terminal of the cell £ is connected to a jockey through a galvanometer and a high resistance.
 A resistance box R and key K_{2} are connected across the cell ξ
 With K_{2} 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
ξ α l_{1} ………..(1)  With the help of resistance box R, introduce resistance, say R and then key K_{2} is closed.
 The current passing through the cell and the resistance R is given by
\(I=\frac{\xi}{R+r}\)  The potential difference across R is
\(V=\frac{\xi R}{R+r}\)  When this potential difference is balanced on the potentiometer wire, let l_{2} be the balancing length, then
\(\mathrm{r}=\mathrm{R}\left(\frac{l_{1}l_{2}}{l_{2}}\right)\)  The internal resistance of the cell is not constant but increases with the increase of external resistance connected across its terminals.
Question 4.
Define electric power and electric potential energy. A potential difference V is applied across a resistance R. Find a relation for the power consumed.
Answer:
1. The electrical power P, is the rate at which the electrical potential energy is delivered.
2. If the charge q, flows between two points having a potential difference V, then the work done in moving the charge is dW=VdQ.
i.e dW=V dQ
3. Consider a circuit in which a battery of voltage V is connected to the resistor.
4. 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.
5. 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.
6. 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.
7. This process occurs continuously till the battery is connected in the circuit.
8. The electrical power P, is the rate at which the electrical potential energy is delivered,
\(P=\frac{d W}{d t}\)
W.K.T dW = V dQ
\(P=\frac{d}{d t}(V d Q)=V\left(\frac{d Q}{d t}\right)\)
P = VI (∴\(I=\frac{d Q}{d t}\))
SI unit = Watt
9. The bigger units of electric power are kilowatt = 1KW = 10^{3} W
Megawatt = IMW = 10^{6} W
10.Expression for electric power in terms of I and R. P = I^{2}R (∴ V=IR)
11. Expression for electric power interms of V and R.
\(\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\left(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\right)\)
Question 5.
Explain the principle of potentiometer.
Answer:
Potentiometer:
 A steady current is maintained across the wire CD by a battery.
 The battery, key, and potentiometer wire are connected in series from the primary circuit.
 The positive terminal of a primary cell of emf ξ 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.
 Let contact be made at any point J on the wire by a jockey.
 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.
 CJ is the balancing length L.
 The potential difference across CJ is equal to IrL
 ξ = IrL
 Where I – current, r – resistance per unit length of the wire.
 Since I and r are constants, ξ α l
 The emf of the cell is directly proportional to the balancing length.
Question 6.
Explain the Thomson effect.
Answer:
Thomson effect:
If 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.
Positive Thomson effect:
 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.
 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.
 A similar effect is observed in metals like silver, zinc, and cadmium.
Negative Thomson effect:
 When the copper bar is replaced by an iron bar, heat is evolved along CA and absorbed along BC.
 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.
 A similar effect is observed in metals like platinum, nickel, cobalt, and mercury.
Question 7.
Give a short account on carbon resistor
Answer:
 Carbon 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.
 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%.
 While reading the colour code, hold the resistor with colour bands to your left. Resistors never start with a metallic band on the left.
Question 8a.
Define electric power
Answer:
Electric power is the rate at which electrical potential energy is delivered.
(i.e) \(\frac{\mathrm{dW}}{\mathrm{dt}}=\mathrm{P}=\mathrm{Vi}\)
unit → watt
Question 8b.
An electric iron of resistance of 80Ω is operated at 200 V for 2 hours, find the electrical energy consumed.
Answer:
Given data: R = 80Ω t = 2hours v = 200V
P = VI
P = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
= \(\frac{200 \times 200}{80}=500 \mathrm{~W}\)
Energy consumed in 2 hour = Power in watt × time in hour
E = 500 × 2 = 1000 Wh
E = 1 kWh
Question 9.
To balance the Wheatstone’s bridge shown in the figure, determine an additional resistance that has to be connected with 15Ω
Answer:
Given data
P = 4.2Ω, Q = 3Ω, R = 2Ω, S = 15Ω
Question 10.
Determine the effective resistance of the given circuit between points A and B.
Answer:
The combined resistance between the points A and B is
R = R’ + R’ + R’
= 7.5 + 7.5 + 7.5
R = 22.5Ω
Question 11.
Find the heat energy produced in a resistance of 10Ω when 5A current flows through it for 5 minutes.
Answer:
R = 10Ω, I = 5A, t = 5 minutes = 5 x 60s
H = I^{2}Rt
= 5^{2} × 10 × 5 × 60
= 25 × 10 × 300
= 25 x 3000
H = 75000J or 75 KJ.
Question 12.
The resistance of a nichrome wire ar 0°C is 10Ω. If its temperature coefficient of resistance is 0.004/°C find its resistance at the boiling point of water. Comment on the result.
Answer:
Given data:
R_{0} = 10Ω, α = 0.004/°C, t = 100°C
R_{t} = R_{0} (1 + α t)
=10 (1 + 0.004 × 100)
= 10 (1 + 0.4)
= 10 (1.4)
R_{t} = 14Ω
As the temperature increases, the resistance of the wire increases.
Question 13.
When two resistors connected in series and parallel their equivalent resistances are 15Ω, 56/15 Ω respectively. Find the two resistances.
Answer:
R_{S} = R_{1} + R_{2} = 15Ω ………….(1)
\(R_{p}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{56}{15} \Omega\) ………..(2)
From equation (1) substituting R_{1} + R_{2} in equation (2)
\(\frac{R_{1} R_{2}}{15}=\frac{56}{15} \Omega\)
∴R_{1}R_{2} = 56
R_{2 } = \(\frac{56}{15}\) = 15 …………..(3)
Substituting for R_{2} in equation (1) from equation (3)
R_{1} = \(\frac{56}{\mathrm{R}_{1}}\) = 15
Then, \(\frac{R_{1}^{2}+56}{R_{1}}=15\)
R_{1}^{2} + 56 = 15R_{1}
R_{1}^{2} – 15 R_{1} + 56 = 0
The above equation can be solved using the simple mathematics
R_{1}^{2} – 8R_{1} – 7R_{1} + 56=0
R_{1} (R_{1} – 8) – 7 (R_{1} – 8) = 0
(R_{1} – 8) (R_{1} – 7) = 0
(R_{1} = 8Ω) or R_{1} = 7Ω
If R_{1} = 8Ω
using in equation (1)
8 + R_{2} = 15
R_{2} = 15 – 8 = 7Ω
R_{2} = 7Ω i.e, (when R_{1} = 8Ω; R_{2} = 7Ω)
If R_{1} = 7Ω,
Substituting the equation (1)
7 + R_{2} = 15
R_{2} = 8Ω i.e, (when R_{1} = 7Ω, R_{2} = 8Ω)
Question 14.
In the circuit, find the current through each branch of the circuit and the potential drop across the 10Ω resistor.
Answer:
ApIying kirchhof’s second rule for loop AFDCA
10 (I_{1} + I_{2})+ 2 I_{1} = 4
12 I_{1} + 10 I_{2} = 4 ……….(1)
For loop AFEBA
10 (I_{1}+ I_{2}) + 4 I_{2} = 5
FE I_{2}4V 5V A B
10 I_{1} + 14I_{2} = 5 …………(2)
(1) × 1o ⇒ 120 I_{1} + 100 I_{2} = 40 ………….(3)
(2) × 12 ⇒ 120 I_{1} + 168 I_{2} = 60 …………..(4)
(4) – (3) 68 I_{12} = 20
\(I_{2}=\frac{20}{68}=\frac{10}{34}=0.2941\)
I_{2} = 0.2941A
Substituting I_{2} in eqn (1),
12 I_{1} + 10(0.2941) = 4
12I_{1} = 4 – 2.941 = 1.059
I_{1} = \(\frac{1.059}{12}\) = 0.08825A
I_{1} = 0.08825A
The voltage drop across 10Ω resistor is
V = IR
V=(I_{1} + I_{2})R
= (0.008825 + 0.2941) × 10
V = 3.824 V
Question 15.
Obtain the condition for maximum current through a resistor, when a number of cells are connected cells in series
(i) in series and
(ii) in parallel suppose n cells, each of emf ξ volts and internal resistance r ohms is connected in series with on external resistance R as shown in the figure.
Answer:
Cells in series:
The total emf of the cells = nξ
The total internal resistance of the cells = r + r + r +……..n times = nr
Therefore, total resistance of the circuit = R+nr
If I is current flowing through the circuit then,
\(I=\frac{\text { total emf }}{\text { total resistance }}\)
\(I=\frac{n \xi}{n r+R}\)
total resistance nr + R
1) When r << R, I = \(\frac{n \xi}{R}\) = nI,
I_{1}, 5 the current due to a single cell
2) When r<< R, I = \(\frac{n \xi}{nr}\) ≈ \(\frac{\xi}{r}\) current due to single cell.
From 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.
Cells in parallel:
Let 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.
Let ξ be the emf and r the internal resistance of each cell.
The total emf of the cells = ξ
The total internal resistance of the cells is given by,
\(\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}
{\mathrm{r}}+\ldots . . \mathrm{n} \text { times }\)
\(\begin{aligned} \frac{1}{\mathrm{r}_{\mathrm{eq}}} &=\frac{\mathrm{n}}{\mathrm{r}} \\ \mathrm{r}_{\mathrm{eq}} &=\frac{\mathrm{r}}{\mathrm{n}} \end{aligned}\)
The total resistance of the circuit = R + \(\frac{\mathrm{r}}{\mathrm{n}}\)
If I is the current flowing through the circuit then,
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.
Question 16.
When 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?
Answer:
 X_{L} = 2πfL
 In the case of a DC source, f will be equal to zero.
∴ X_{L} also becomes zero.  When the resistance is less, more current 2A passes through the circuit.
 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.