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Class 12 Physics NCERT Solutions

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Current Electricity Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 3, Current Electricity — 9 questions from 3.1 to 3.9, each worked through step by step in the CBSE marking pattern. Ohm's law, resistivity and its temperature dependence, combinations of resistors, cells, Kirchhoff's rules and the Wheatstone bridge.

Class:12Subject:PhysicsChapter:3
4 Key Formulas24 Practice MCQs
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 3?

Chapter 3 carries 9 exercise questions, numbered 3.1 to 3.9. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Current Electricity covers Ohm's law, the emf–terminal-voltage connection, resistivity and temperature, and Kirchhoff's rules for networks. Boards test the standard forms hard: I = E/(R + r), V = E – Ir, R = R₀(1 + αΔT), ρ = RA/l, and loop-and-junction analysis of bridges. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Define the current direction on every branch before writing loop equations — the junction rule then fixes the remaining currents. Absolute temperature in kelvin is not needed here, but ΔT must be in the same degree unit as α. When a battery is being charged the current is forced backwards through it, so terminal voltage becomes E + Ir, not E – Ir. And always name the quantity you are conserving (charge at a junction, potential around a loop).
02

NCERT Exercise 3.1 — Maximum Current Drawn From a Car Battery

1Exercise question

Step-by-step solution

  1. 1The current is maximum when the external resistance is zero (short circuit).
  2. 2I_max = E/r = 12/0.4 = 30 A.

Final answer

I_max = 30 A (this is the short-circuit current).

03

NCERT Exercise 3.2 — Load Resistance and Terminal Voltage From a Known Current

1Exercise question

Step-by-step solution

  1. 1By Ohm's law for the whole circuit: I = E/(R + r).
  2. 20.5 = 10/(R + 3) → R + 3 = 20 → R = 17 Ω.
  3. 3Terminal voltage: V = E – Ir = 10 – (0.5)(3) = 8.5 V.
  4. 4Check through the external resistor: V = IR = (0.5)(17) = 8.5 V, the same value.

Final answer

R = 17 Ω; terminal voltage = 8.5 V.

04

NCERT Exercise 3.3 — Temperature of a Heating Element From Its Resistance Rise

1Exercise question

Step-by-step solution

  1. 1Use R = R₀(1 + αΔT), giving ΔT = (R/R₀ – 1)/α.
  2. 2ΔT = (117/100 – 1)/(1.70 × 10⁻⁴) = 0.17/1.70 × 10⁻⁴ = 1000 °C.
  3. 3Steady temperature = T₀ + ΔT = 27 + 1000 = 1027 °C.

Final answer

Temperature of the element = 1027 °C (≈ 1.0 × 10³ °C above room temperature).

05

NCERT Exercise 3.4 — Resistivity of a Wire From R, Length and Area

1Exercise question

Step-by-step solution

  1. 1R = ρl/A, so ρ = RA/l.
  2. 2ρ = (5.0 × 6.0 × 10⁻⁷)/15 = (3.0 × 10⁻⁶)/15 = 2.0 × 10⁻⁷ Ω m.

Final answer

ρ = 2.0 × 10⁻⁷ Ω m.

06

NCERT Exercise 3.5 — Temperature Coefficient of Silver

1Exercise question

Step-by-step solution

  1. 1α = (R₂ – R₁)/[R₁(T₂ – T₁)] for small spreads where R depends linearly on T.
  2. 2α = (2.7 – 2.1)/[2.1 × (100 – 27.5)] = 0.6/(2.1 × 72.5).
  3. 3α = 0.6/152.25 = 3.9 × 10⁻³ °C⁻¹.

Final answer

α ≈ 3.9 × 10⁻³ °C⁻¹.

07

NCERT Exercise 3.6 — Steady Temperature of a Nichrome Heater

1Exercise question

Step-by-step solution

  1. 1Resistance at room temperature: R₂₇ = 230/3.2 = 71.875 Ω.
  2. 2Steady resistance: R = 230/2.8 = 82.14 Ω.
  3. 3R = R₀(1 + αΔT) → ΔT = (R/R₀ – 1)/α.
  4. 4ΔT = (82.14/71.875 – 1)/(1.70 × 10⁻⁴) = (1.1429 – 1)/1.70 × 10⁻⁴ = 0.1429/1.70 × 10⁻⁴ ≈ 840 °C.
  5. 5Steady temperature = 27 + 840 = 867 °C.

Final answer

Steady temperature ≈ 8.4 × 10² °C above room temperature, i.e. about 867 °C.

08

NCERT Exercise 3.7 — Current in Every Branch of a Bridge Network (Kirchhoff)

1Exercise question

Step-by-step solution

  1. 1Label the currents: I₁ = current supplied by the battery (through the 10 Ω feed resistor), I₂ = current in AB, I₃ = current in AD, I₄ = current in the diagonal BD (labelled from B to D).
  2. 2Then by the junction rule: current in BC = I₂ – I₄ and current in CD = I₃ + I₄.
  3. 3Loop ABDA: 10I₂ + 5I₄ – 5I₃ = 0 → I₃ = 2I₂ + I₄. ... (1)
  4. 4Loop BCDB: 5(I₂ – I₄) – 10(I₃ + I₄) – 5I₄ = 0 → I₂ = 2I₃ + 4I₄. ... (2)
  5. 5From (1) and (2): I₃ = 2(2I₃ + 4I₄) + I₄ → –3I₃ = 9I₄ → I₃ = –3I₄.
  6. 6Using (1): –3I₄ = 2I₂ + I₄ → I₂ = –2I₄.
  7. 7Outer loop A–B–C–feed–A: –10 + 10I₁ + 10I₂ + 5(I₂ – I₄) = 0 and I₁ = I₂ + I₃, giving 5I₂ + 2I₃ – I₄ = 2.
  8. 8Substitute I₂ = –2I₄ and I₃ = –3I₄: 5(–2I₄) + 2(–3I₄) – I₄ = 2 → –17I₄ = 2 → I₄ = –2/17 A.
  9. 9Hence I₃ = 6/17 A, I₂ = 4/17 A, and I₁ = I₂ + I₃ = 10/17 A.
  10. 10Branch currents: AB = I₂ = 4/17 A; AD = I₃ = 6/17 A; BC = I₂ – I₄ = 4/17 + 2/17 = 6/17 A; CD = I₃ + I₄ = 6/17 – 2/17 = 4/17 A; diagonal BD = |I₄| = 2/17 A (flowing opposite to the assumed B→D label, i.e. from D to B).

Final answer

Supply current 10/17 A; AB = 4/17 A, AD = 6/17 A, BC = 6/17 A, CD = 4/17 A, BD = 2/17 A (directed from D to B against the label).

09

NCERT Exercise 3.8 — Terminal Voltage of a Storage Battery Being Charged

1Exercise question

Step-by-step solution

  1. 1During charging the current is forced into the battery against its emf: I = (120 – 8.0)/(15.5 + 0.5) = 112/16 = 7 A.
  2. 2Terminal voltage = E + Ir = 8.0 + (7)(0.5) = 11.5 V.
  3. 3Cross-check across the series resistor: 120 – (7)(15.5) = 120 – 108.5 = 11.5 V, the same value.
  4. 4The series resistor limits the charging current — without it the initial surge would be (120 – 8)/0.5 = 224 A, far too large for the battery.

Final answer

Terminal voltage during charging = 11.5 V; the series resistor limits (damps) the charging current to a safe value.

10

NCERT Exercise 3.9 — Drift Time of an Electron Across a Copper Wire

1Exercise question

Step-by-step solution

  1. 1Drift speed: v_d = I/(n e A).
  2. 2v_d = 3.0/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶) = 3.0/(2.72 × 10⁴) = 1.10 × 10⁻⁴ m s⁻¹.
  3. 3Time to drift 3.0 m: t = l/v_d = 3.0/(1.10 × 10⁻⁴) = 2.7 × 10⁴ s.

Final answer

t ≈ 2.7 × 10⁴ s (about 7.6 hours) — a striking illustration of how slow the actual drift of electrons is.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Ohm's law

Resistance of a wire

Temperature dependence of resistance

Terminal potential of a cell

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Kirchhoff's junction rule conserves charge and the loop rule conserves energy — set up the junction rule first, since it fixes the sign convention for the loop equations.
  • In a balanced Wheatstone bridge no current flows through the galvanometer, so the four resistances satisfy R1/R2 = R3/R4 and the bridge can be treated as two separate loops.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 3 (Current Electricity)?

There are 9 exercise questions in this chapter, numbered 3.1 to 3.9. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Current Electricity Class 12 Physics?

The formulas this chapter's questions actually turn on are: Ohm's law, Resistance of a wire, Temperature dependence of resistance, Terminal potential of a cell. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Current Electricity important for JEE Main and NEET?

Very important — current electricity carries the heaviest numerical weight in Class 12 boards, and potential-divider, combination and Wheatstone-bridge problems appear in every JEE Main and NEET paper.

Interactive Quiz

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