Class 12th Physics NCERT Solutions for Chapter 3 (Current Electricity) are available here. You can also download these solutions with the help of download link given at the end of this chapter. Some important topics of this chapter are Electric Current, Electric Currents in Conductors, Ohm’s Law, Drift of Electrons and The Origin of Resistivity, Resistivity of Various Materials, Temperature Dependence of Resistivity, Electrical Energy, Power, Combination of Resistors – Series and Parallel, Cells, EMF, Internal Resistance, Cells in Series and in Parallel, Kirchhoff’s Rules, Wheatstone Bridge, Meter Bridge, Potentiometer. Most of questions and solutions of this chapter of 12th Physics NCERT textbook are based on these topics.
NCERT Solutions for Class 12 Physics ‒ Chapter 3: Current Electricity, Chapter 3 are given below
Question3.1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery? 3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution3.1:
Given:
E = 12 V
r = 0.4 ohm
R = 0 ohm
From Ohm’s law
V = IR
I = E / (R + r)
Or I = 12 / 0.4 = 30 A.
NCERT Exemplar: CBSE Class 12 Physics – Chapter 3
Question 3.2: A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution3.2:
Given:
E = 10 V
r = 3 ohms
I = 0.5 A
Unknown resistor = R ohm
We know
E =IR
10 = (0.5) × (3 + R)
Or R = 17 ohms
Again by Ohm’s law
Terminal voltage, V = IR = 0.5 x 17 = 8.5 V.
Question3.3: (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution3.3:
Given:
R_{1} = 1 ohm
R_{2} = 2 ohms
R_{3} = 3 ohms
(a) For series combination
R = R1 + R2 + R3 = 6 ohms
(b) E = 12 V
Using Ohm’s law
I = 12 / 6 = 2 A
Therefore potential drops
V_{1} = IR_{1} = 2 x 1 = 2 V
Similarly,
V_{2} = 4 V
V_{3} = 6 V.
Question3.4: (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Solution3.4:
Given:
R_{1 }= 2 ohms
R_{2} = 4 ohms
R_{3} = 5 ohms
(a) For parallel configuration
(1/R) = (1/R_{1}) + (1/R_{2}) + (1/R_{3})
Substituting the values and solving, we get
R = 20 / 19 ohms
(b) By Ohm’s law
I = V / R = 19 A
Current in first resistor
I_{1} = V / R = 20 / 2 = 10 A
Current in second resistor
I_{2} = 5 A
Current in third resistor
I_{3} = 4 A.
Question3.5: At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^{–4} °C^{–1}.
Solution3.5:
Given:
T_{1} = 27^{o} C
R_{1} = 100 ohms
R_{2} = 117 ohms
α = 1.7 x 10^{-4} ^{o}C^{‒1}
R_{2} = R_{1}.[1 + α.(T_{2 }‒ T_{1})]
Substituting the values and solving, we get
T_{2} = 1027^{o }C.
Question3.6: A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^{–7} m^{2}, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Solution3.6:
l = 15 m
A = 6x10-17 m2
R = 5 ohms
Using the relation ��=��(��/��)
Substituting the values given ��= 2x10^{-7} ohm-meter.
Question3.7: A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Solution3.7:
R_{1 }= 2.1 ohms
T_{1} = 27.5 ^{o}C
R_{2} = 2.7 ohms
T_{2 }= 100 ^{o}C
We know
R_{2} = R_{1}.[1 + α.(T_{2} - T_{1})]
Or (2.7 / 2.1) – 1 = α.(100 – 27.5)
Or α = 3.94 x 10^{-3} /^{o}C.
Question3.8: A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^{–4} °C^{–1}.
Solution3.8:
Given:
V = 230 V
I_{1} = 3.2 A
I_{2} = 2.8 A
T_{1} = 27 ^{o}C
α = 1.7x10^{‒4} /^{o}C
Before settlement
R_{1} = 230 / 3.2 = 71.88 ohms
After settlement
R_{2} = 230 / 2.8 = 82.14 ohms
Using relation
R_{2} = R_{1}.[1 – α.(T_{2} – T_{1})]
Or 0.14 / α = T_{2} – 27
Or T_{2} = 867 ^{o}C.
Question3.9: Determine the current in each branch of the network shown in Fig. 3.30:
Solution3.9:
Redrawing the figure
Applying Kirchhoff’s rule in loop EADCE
-10I – 5 I + 5 I_{1} – 10 I + 10 I_{2} + 10 = 0
Or -25I + 5 I_{1} + 10 I_{2} = -10
Or 5I – I_{1} – 2 I_{2} = 2……………………………………….(1)
Now applying Kirchhoff’s rule in loop EABDCE
-10 I – 15 I_{1} + 5 I_{2} – 10 I +10 I_{2} + 10 = 0
Or 20 I + 15 I_{1} – 15 I_{2} = 10
Or 4 I +3 I_{1} – 3 I_{2} = 2…………………………………….(2)
In loop EABCE
-10 I – 10 I_{1} – 5 I_{2} + 10 = 0
Or 2 I + 2 I_{1} + I_{2} = 2…………………………………………(3)
Solving simultaneously above equations, we get
I = 0.59 A
I_{1} = 0.24 A
I_{2} = 0.35 A
Therefore
Current in branch AB = I_{1} = 0.24 A
Current in branch BC = I_{2} = 0.35 A
Current in branch BD = I_{1} – I_{2} = -0.11 A
Current in branch AD = I – I_{1} = 0.35 A
Current in branch DC = I – I_{2} = 0.24 A
Total Current = 0.59 A
Question3.10: (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Solution3.10:
(a) From meter bridge properties
39.5 / 60.5 = X / 12.5
Or
X = 8.16 ohms
The thick copper strips are used to minimise resistance of the connection not accounted for to avoid otherwise appearing error in the final result.
(b) Now according to question
X = 12.5 ohms
Y = 8.2 ohms
So new balance point
Z / (100 – Z) = 12.5 / 8.2
Or z = 60.5 cm from A
(c) Then the bridge will no longer act as Meter Bridge. The galvanometer will not show any current.
Question3.11: A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Solution3.11:
Given:
E = 8 V
r = 0.5 ohm
V = 120 V
R = 15.5 ohms
Current in circuit by applying Kirchhoff’s law in loop
-16I – 8 + 120 = 0
Or I = 7 A
Now terminal voltage
V = E + Ir = 8 + (7).(0.5) = 11.5 V.
Question3.12: In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution3.12:
We know for potentiometer
E_{1} = �� l_{1}
Or 1.25 = ��.35
Or ��= 0.036 V / cm
Now
E_{2 }= �� l_{2}
Or E_{2} = 0.036 x 63 = 2.25 V.
Question3. 13: The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10^{28} m^{–3}. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^{–6} m^{2} and it is carrying a current of 3.0 A.
Solution3.13:
Given
Free electrons = 8.5x1028 /m3
l = 3 m
A = 2x10^{-6} m^{2}
I = 3 A
e = 1.6 × 10^{-19} C
Volume of given specimen = 6 × 10^{-6} m^{3}
So,
Total electrons in the specimen
n = (6 × 10^{-6}) . (8.5 x 10^{28})
= 5.1x10^{23}
Charge accompanied by them
Q = ne
= (5.1x10^{23}).(1.6x10^{-19})
= 81711 C
Now we know by definition of current
I = Q / t
So
t = 81711 / 3
Or t = 27237
Making it approximate value
t = 2.7x10^{4} seconds.
Download NCERT Solutions for CBSE Class 12 Physics, Chapter 3: Current Electricity
DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article.
CBSE Class 12 Physics NCERT Exemplar Solutions: Chapter 11 – Dual Nature of Radiation and Matter
CBSE Class 12 Physics NCERT Exemplar Solutions: Chapter 10 – Wave Optics
CBSE Class 12 Physics NCERT Exemplar Solutions: Chapter 8 – Electromagnetic Waves
NCERT Exemplar Solutions for CBSE Class 12 Physics: Chapter 13 – Nuclei