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Solutions

Overview

Step-by-step NCERT solutions for Current Electricity (Chapter 3, CBSE Class 12 Physics) — every question and answer worked out in full, not just the final result. You can also read the Current Electricity textbook chapter.

Solved

What these solutions cover

All 9 questions in Current Electricity are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

  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?
  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?
  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?
  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 Ω. What is the resistivity of the material at the temperature of the experiment?
  5. 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.
  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.
  7. Determine the current in each branch of the network shown in Fig. 3.20. The network is a diamond with node B at top, A at left, C at right, and D at bottom. Resistors: A–B = 10 Ω, B–C = 5 Ω, B–D = 5 Ω (middle bridge), A–D = 5 Ω, C–D = 10 Ω. A 10 V battery in series with a 10 Ω resistor is connected across A and C along the bottom.
  8. 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?
  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.
Sample

Worked examples

3 questions from Current Electricity, solved step by step.

Exercises

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?

Given: emf ε = 12 V, internal resistance r = 0.4 Ω.

Maximum current is drawn when external resistance R = 0 (short-circuit condition):

I_max = ε / r = 12 / 0.4 = 30 A.

Note: In practice, drawing this current would damage the battery. Batteries specify a maximum safe discharge current well below this theoretical limit.

Answer: Maximum current I_max = 30 A

Exercises

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?

Given: ε = 10 V, r = 3 Ω, I = 0.5 A.

From Ohm's law for the full circuit: ε = I(R + r)

⇒ R + r = ε / I = 10 / 0.5 = 20 Ω

⇒ R = 20 - r = 20 - 3 = 17 Ω

Terminal voltage (= potential difference across R):

V = ε - I r = 10 - (0.5)(3) = 10 - 1.5 = 8.5 V

Alternatively, V = I R = 0.5 × 17 = 8.5 V. ✓

Answer: Resistance R = 17 Ω; Terminal voltage = 8.5 V

Exercises

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?

Given: R₀ = 100 Ω at T₀ = 27.0 °C, RT = 117 Ω, α = 1.70 × 10⁻⁴ °C⁻¹.

Using the temperature-resistance relation:

RT = R₀ [1 + α(T - T₀)]

⇒ T - T₀ = (RT/R₀ - 1) / α

= (117/100 - 1) / (1.70 × 10⁻⁴)

= 0.17 / (1.70 × 10⁻⁴)

= 1000 °C

⇒ T = T₀ + 1000 = 27.0 + 1000 = 1027 °C

Answer: Temperature of the element T = 1027 °C

All 9 questions in Current Electricity are solved step by step in the Solutions PDF above.

Keep solving

More solutions in Physics Part I

01Electric Charges and Fields02Electrostatic Potential and Capacitance04Moving Charges and Magnetism05Magnetism and Matter06Electromagnetic Induction07Alternating Current08Electromagnetic Waves
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More NCERT Solutions for Class 12

Read the Current Electricity textbook chapter / PDF, or browse all CBSE Class 12 Physics solutions.

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