NCERT Class 12 Physics Chapter 3 Current Electricity Solutions, NCERT Solutions For Class 12 Physics to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 12 Physics Chapter 3 Current Electricity Question Answer and select needs one.
NCERT Class 12 Physics Chapter 3 Current Electricity
Also, you can read the CBSE book online in these sections NCERT Class 12 Physics Chapter 3 Current Electricity Solutions by Expert Teachers as per NCERT (CBSE) Book guidelines. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Physics Chapter 3 Current Electricity Solutions for All Subjects, You can practice these here.
Current Electricity
Chapter: 3
| Part – I |
EXERCISE
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?
Ans: maximum current Imax is:
Imax = E / r
= 12 V / 0.4 Ω
= 30 A.
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?
Ans: VR = I × R
Given:
E = 10 V
r = 3 Ω
I = 0.5 A
Vr = I × r
Vr = 0.5 A × 3Ω
1.5 V.
Vterminal = E – Vr
= 10V – 1.5 V
= 8.5 V.
The resistance of the resistor R is:
R = VR/
= 8.5 V / 0.5 A
= 17Ω
3. 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.
Ans: Given:
R = R0 (1 + α (T – T0))
Here:
R = 117 Ω
R0 = 100 Ω
α = 1.70 × 10-4 °C-1
T0 = 27.0 °C
T = ?
Calculate T:
T = (R – R0) / (αR0) + T0
= (117 – 100) / (1.70 × 10-4 × 100) + 27.0
= 17 / 0.017 + 27.0
= 1000 + 27.0
= 1027.0 °C.
4. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment?
Ans: Given:
I = 15 m
A = 6.0 × 10-7 m2
R = 5.0 Ω
Here:
𝘱 = RA / L
= 5.0 × 6.0 × 10-7 / 15
= 2 × 10-7 Ωm.
5. A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver.
Ans: R2 = R1 [1 + a (T2 – T1)]
Given:
R1 = 2.1Ω at T1 = 27.5 °C
R2 = 2.7Ω at T2 = 100°C
Rearranging the formula to solve α:
α = (R2/R1-1) / T2 – T1
= (2.7 / 2.1 – 1) / 100 – 27.5
= 1.2857 – 1 / 72.5
= 0.2857 / 72.5
= 0.00394 °C-1.
6. 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.
Ans: Given:
V = 230 V
I1 = 3.2 A
I2 = 2.8 A
T1 = 27.0°C
α = 1.70 × 10–4 °C–1.
Initial R1:
R1 = 230 V / 3.2 A
= 71.875 Ω
Steady R2:
R2 = 230 V / 2.8 A
= 82.143 Ω
Steady Temperature:
T2 = 867.5 °C.
7. Determine the current in each branch of the network shown in Fig. 3.20:
Ans: Let I, I1, I2, I3 be the currents as shown in image, we apply Kirchhoff’s second rule to different loops.
8. A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Ans: Given:
E = 8.0 V
r = 0.5 Ω
Vsupply = 120 V
R = 15.5 Ω
Rtotal = r + R = 0.5 Ω + 15.5 Ω
= 16.0 Ω.
I = Vsupply – E / Rtotal
= 120 V – 8.0 V / 16.0 V
= 112 V / 16.0
= 7.0 A.
Vterminal = E + I × r
= 8.0 V + 7.0 A × 0.5 Ω
= 8.0 V + 3.5 V
= 11.5 V.
9. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 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 m2 and it is carrying a current of 3.0 A.
Ans: Here:
vd = I / n × A × e
Given:
I = 3.0 A
n = 8.5 × 1028 m-3
A = 2.0 × 10-6 m2
e = 1.6 × 10-19 C.
vd = 3.0 A / (8.5 × 1028 m-3 × 2.0 × 10-6m2 × 1.6 × 10-19 C)
vd = 3.0 / (2.7 × 104) m/s
vd = 1.1 × 10-4 m/s.
Calculate time:
t = L / vd
= 3.0 m / (1.1 × 10-4 m/s)
t = 2.7 × 104 s.
