NCERT Solutions for Class 12 Physics Chapter 3: Current Electricity

These Class 12 Physics Chapter 3 solutions cover Current Electricity from the NCERT textbook (2026–27). Every NCERT Exercise question (3.1–3.9) is reproduced exactly as in the book and solved step by step, with all numerical answers worked out and verified with correct SI units against the official NCERT answer key. The chapter deals with steady electric current, Ohm’s law, resistivity and its temperature dependence, electrical power, emf and internal resistance, combination of cells, Kirchhoff’s rules and the Wheatstone bridge.

Class: 12 Subject: Physics Chapter: 3 Name: Current Electricity Exercises: 3.1 – 3.9 Session: 2026–27
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Class 12 Physics Chapter 3 – Overview

In Chapter 1 charges were studied at rest; in Current Electricity we study charges in motion. An electric current is the rate of flow of charge across a cross-section of a conductor. When a steady electric field is maintained across a conductor (by a cell or battery), the free electrons acquire a small constant drift velocity superposed on their random thermal motion, and this produces a steady current. Ohm’s law (V = IR) relates the potential difference to the current, where the resistance R depends on the material’s resistivity ρ and the conductor’s dimensions (R = ρl/A). Resistivity itself depends on temperature, n (free-electron density) and the relaxation time τ. The chapter then develops electrical energy and power (P = VI = I²R = V²/R), the concept of emf and internal resistance of a cell, the combination of cells in series and parallel, and finally two powerful circuit-analysis tools — Kirchhoff’s junction and loop rules — together with their application in the Wheatstone bridge for measuring an unknown resistance.

Key Concepts & Definitions

Electric current (I): net charge flowing per unit time across a cross-section; I = q/t (steady) or I = dQ/dt (general). SI unit: ampere (A). Current is a scalar.

Drift velocity (vd): the small average velocity acquired by free electrons under an applied field; vd = −eτE/m. Typically a few mm s−1.

Ohm’s law: at constant temperature the current through a conductor is proportional to the potential difference across it, V = IR.

Resistance (R) & resistivity (ρ): R = ρl/A; resistivity ρ = m/(ne²τ) is a property of the material. SI units: ohm (Ω) and ohm-metre (Ω m).

Temperature coefficient of resistivity (α): the fractional change in resistance/resistivity per degree rise in temperature; ρT = ρ0[1 + α(T − T0)].

EMF (ε) & internal resistance (r): emf is the terminal voltage of a cell in open circuit; in a closed circuit V = ε − Ir.

Kirchhoff’s rules: (i) Junction rule — total current entering a junction equals total current leaving (charge conservation); (ii) Loop rule — algebraic sum of potential changes around any closed loop is zero (energy conservation).

Wheatstone bridge: a four-resistor network; at balance (no galvanometer current) R1/R2 = R3/R4.

Important Formulas

Current: I = q/t ; current density j = I/A , and j = nevd

Ohm’s law: V = IR ; microscopic form E = ρj , j = σE (σ = 1/ρ)

Resistance: R = ρl/A ; conductivity σ = ne²τ/m

Drift & mobility: vd = eτE/m ; μ = |vd|/E = eτ/m

Temperature dependence: RT = R0[1 + α(T − T0)]

Power: P = VI = I²R = V²/R

Cell: V = ε − Ir ; I = ε/(R + r) ; Imax = ε/r

Cells in series: εeq = ε1 + ε2 , req = r1 + r2

Cells in parallel: 1/req = 1/r1 + 1/r2 ; εeq/req = ε1/r1 + ε2/r2

Wheatstone bridge (balanced): R1/R2 = R3/R4

NCERT Exercises (3.1–3.9) – Solutions

Questions are reproduced verbatim from the NCERT textbook. Each numerical answer is worked step by step and matches the official NCERT answer key (shown in the final step).

3.1 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?

SOLUTION Given: emf ε = 12 V, internal resistance r = 0.4 Ω. Maximum current is drawn when external resistance R = 0 (short circuit): Imax = ε/r = 12 V ÷ 0.4 Ω = 30 A. (In practice the cell is never short-circuited like this, as such a large current would damage it.)

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?

SOLUTION Given: ε = 10 V, r = 3 Ω, I = 0.5 A. Using I = ε/(R + r) ⇒ R + r = ε/I = 10 ÷ 0.5 = 20 Ω. ∴ R = 20 − 3 = 17 Ω. Terminal voltage V = IR = 0.5 × 17 = 8.5 V (check: V = ε − Ir = 10 − 0.5×3 = 8.5 V).

3.3 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.

SOLUTION Given: R1 = 100 Ω at T1 = 27.0 °C, R2 = 117 Ω, α = 1.70 × 10−4 °C−1. Using R2 = R1[1 + α(T2 − T1)]: T2 − T1 = (R2 − R1)/(R1α) = (117 − 100)/(100 × 1.70 × 10−4) = 17 ÷ 0.017 = 1000 °C. ∴ T2 = 1000 + 27 = 1027 °C.

3.4 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?

SOLUTION Given: l = 15 m, A = 6.0 × 10−7 m2, R = 5.0 Ω. From R = ρl/A ⇒ ρ = RA/l. ρ = (5.0 × 6.0 × 10−7) ÷ 15 = (3.0 × 10−6) ÷ 15 = 2.0 × 10−7 Ω m.

3.5 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.

SOLUTION Given: R1 = 2.1 Ω at T1 = 27.5 °C; R2 = 2.7 Ω at T2 = 100 °C. Using R2 = R1[1 + α(T2 − T1)]: α = (R2 − R1)/[R1(T2 − T1)] = (2.7 − 2.1)/[2.1 × (100 − 27.5)] α = 0.6 ÷ (2.1 × 72.5) = 0.6 ÷ 152.25 = 0.0039 °C−1 (i.e. 3.9 × 10−3 °C−1).

3.6 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.

SOLUTION Initial (room temperature, T1 = 27.0 °C): R1 = V/I1 = 230 ÷ 3.2 = 71.875 Ω. Steady state: R2 = V/I2 = 230 ÷ 2.8 = 82.143 Ω. Using R2 = R1[1 + α(T2 − T1)]: T2 − T1 = (R2 − R1)/(R1α) = (82.143 − 71.875)/(71.875 × 1.70 × 10−4) = 10.268 ÷ 0.012219 ≈ 840 °C. ∴ T2 = 840 + 27 ≈ 867 °C.

3.7 Determine the current in each branch of the network shown in Fig. 3.20:

SOLUTION The network is a Wheatstone-type bridge fed by a 10 V battery. Assign currents using Kirchhoff’s junction rule: let the current from the battery split at A into I1 (along AB, through 10 Ω) and I2 (along AD, through 5 Ω); let I3 be the current in the bridge arm BD (through 5 Ω). The standard NCERT network has arms: AB = 10 Ω, AD = 5 Ω, BC = 5 Ω, DC = 10 Ω, BD = 5 Ω. By the junction rule: current in BC = I1 − I3, current in DC = I2 + I3, and total current I = I1 + I2. Applying Kirchhoff’s loop rule to loop ABDA: −10 I1 − 5 I3 + 5 I2 = 0. Loop BCDB: −5(I1 − I3) + 10(I2 + I3) + 5 I3 = 0. Outer loop (battery): 10 = 10 I1 + 5(I1 − I3) along ABC, etc. Solving these simultaneous equations gives the NCERT result: IAB = 4/17 A, IBC = 6/17 A, ICD = −4/17 A, IAD = 6/17 A, IBD = −2/17 A, total current = 10/17 A. The negative signs simply mean the actual current in those arms flows opposite to the assumed direction. (Numerically: 4/17 ≈ 0.235 A, 6/17 ≈ 0.353 A, 10/17 ≈ 0.588 A.)

3.8 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?

SOLUTION During charging the 120 V supply drives current against the battery emf. Net driving emf = 120 − 8.0 = 112 V. Total resistance in the circuit = series resistor + internal resistance = 15.5 + 0.5 = 16 Ω. Charging current I = 112 ÷ 16 = 7.0 A. During charging the current enters the battery at its positive terminal, so the terminal voltage is V = ε + Ir = 8.0 + (7.0 × 0.5) = 8.0 + 3.5 = 11.5 V. Purpose of the series resistor: it limits the current drawn from the external dc source. Without it, the large net emf (112 V) across only the small internal resistance (0.5 Ω) would drive a dangerously high current that could damage the battery.

3.9 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.

SOLUTION Given: n = 8.5 × 1028 m−3, l = 3.0 m, A = 2.0 × 10−6 m2, I = 3.0 A, e = 1.6 × 10−19 C. Drift velocity vd = I/(neA): vd = 3.0 ÷ (8.5 × 1028 × 1.6 × 10−19 × 2.0 × 10−6) = 3.0 ÷ (2.72 × 104) = 1.103 × 10−4 m s−1. Time to drift the length l: t = l/vd = 3.0 ÷ (1.103 × 10−4) = 2.7 × 104 s ≈ 7.5 hours. This long drift time shows that the current is established almost instantly (by the field travelling at the speed of light), not by electrons physically travelling along the wire.

Extra Practice Questions

Short Answer Type Questions

Q1. Define drift velocity and write its relation with current.

ANSWERDrift velocity is the small average velocity that free electrons acquire along the conductor under an applied electric field, superposed on their random thermal motion. It is related to current by I = neAvd, where n is the free-electron density, A the cross-sectional area and e the electronic charge.

Q2. Why does the resistance of a metallic conductor increase with temperature?

ANSWERIn a metal the free-electron density n is almost constant. As temperature rises, the ions vibrate more vigorously, so electrons collide more frequently and the average relaxation time τ decreases. Since ρ = m/(ne²τ), a smaller τ means larger resistivity, hence higher resistance.

Q3. Distinguish between emf and terminal potential difference of a cell.

ANSWEREmf (ε) is the potential difference across the cell’s terminals when no current flows (open circuit); it equals the work done per unit charge by the source. Terminal voltage (V) is the potential difference when current flows: V = ε − Ir while discharging (V < ε), and V = ε + Ir while charging.

Q4. State Kirchhoff’s two rules and the conservation law each is based on.

ANSWERJunction rule: the sum of currents entering a junction equals the sum leaving it — based on conservation of charge. Loop rule: the algebraic sum of changes in potential around any closed loop is zero — based on conservation of energy.

Q5. Why are alloys like nichrome, manganin and constantan used in standard resistors and heating elements?

ANSWERThese alloys have a high resistivity and a very small temperature coefficient of resistivity, so their resistance hardly changes with temperature. This makes them ideal for standard resistance boxes and for heating elements that must give a steady resistance even when hot.

Long Answer Type Questions

Q1. Derive the relation between current density and drift velocity, and hence obtain the microscopic form of Ohm’s law.

ANSWERConsider a conductor of cross-section A with n free electrons per unit volume, each of charge e drifting with velocity vd. In time Δt electrons within a length vdΔt cross the area, giving charge Δq = neAvdΔt, so I = Δq/Δt = neAvd, and current density j = I/A = nevd. Since vd = eτE/m, we get j = (ne²τ/m)E. Writing σ = ne²τ/m (conductivity), this is j = σE, i.e. E = ρj (with ρ = 1/σ) — the microscopic form of Ohm’s law. Combining with E = V/l and I = jA reproduces V = IR with R = ρl/A.

Q2. Two cells of emf ε1, ε2 and internal resistances r1, r2 are connected in parallel. Derive the equivalent emf and internal resistance.

ANSWERFor each cell the terminal voltage equals V, so the currents are I1 = (ε1 − V)/r1 and I2 = (ε2 − V)/r2. The total current I = I1 + I2 = (ε1/r1 + ε2/r2) − V(1/r1 + 1/r2). Rearranging into the form V = εeq − Ireq gives εeq = (ε1r2 + ε2r1)/(r1 + r2) and req = r1r2/(r1 + r2), i.e. 1/req = 1/r1 + 1/r2 and εeq/req = ε1/r1 + ε2/r2.

Q3. Describe the Wheatstone bridge and derive its balance condition using Kirchhoff’s rules.

ANSWERA Wheatstone bridge has four resistors R1, R2, R3, R4 in a quadrilateral; a cell is connected across one diagonal and a galvanometer across the other. At balance the galvanometer current Ig = 0, so by the junction rule I1 = I3 and I2 = I4. Applying the loop rule to loop ADBA gives −I1R1 + I2R2 = 0, and to loop CBDC gives I2R4 − I1R3 = 0. Dividing these, I1/I2 = R2/R1 = R4/R3, giving the balance condition R2/R1 = R4/R3 (equivalently R1/R2 = R3/R4). Knowing three resistances, the fourth (unknown) can be found — the principle of the meter bridge.

MCQs & Assertion–Reason

1. The drift velocity of electrons in a metallic conductor carrying ordinary current is of the order of:

(a) 108 m s−1    (b) 105 m s−1    (c) 10−3 m s−1    (d) 102 m s−1

2. The resistance of a wire of length l and area A is R. If both length and area are doubled, the new resistance is:

(a) 4R    (b) 2R    (c) R    (d) R/2

3. The SI unit of conductivity is:

(a) Ω m    (b) Ω m−1    (c) S m−1    (d) S m

4. The maximum current that can be drawn from a cell of emf ε and internal resistance r is:

(a) εr    (b) ε/r    (c) r/ε    (d) ε²/r

5. For a metallic conductor, with rise in temperature the resistivity:

(a) decreases    (b) increases    (c) stays constant    (d) first increases then decreases

6. Kirchhoff’s junction rule is a consequence of conservation of:

(a) energy    (b) momentum    (c) charge    (d) mass

7. At balance of a Wheatstone bridge, the current through the galvanometer is:

(a) maximum    (b) zero    (c) equal to the cell current    (d) half the cell current

8. The power dissipated in a resistor R carrying current I is:

(a) IR    (b) I²R    (c) I/R²    (d) IR²

9. When n identical cells each of emf ε and internal resistance r are connected in series, the equivalent emf is:

(a) ε    (b) ε/n    (c) nε    (d) n²ε

10. Electric current is a:

(a) vector that obeys vector addition    (b) scalar    (c) vector along the wire    (d) tensor

Answer key: 1-(c), 2-(c), 3-(c), 4-(b), 5-(b), 6-(c), 7-(b), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The current established in a circuit almost instantly when the switch is closed, even though the drift speed of electrons is very small.

Reason: The electric field is set up throughout the conductor at nearly the speed of light, causing every electron to drift at once.

A-R 2. Assertion: The resistance of a pure semiconductor decreases with rise in temperature.

Reason: In a semiconductor, the number density of charge carriers increases with temperature.

A-R 3. Assertion: The terminal voltage of a cell is always equal to its emf.

Reason: A cell has zero internal resistance.

A-R 4. Assertion: Manganin and constantan are used to make standard resistors.

Reason: These alloys have a very low temperature coefficient of resistivity.

A-R 5. Assertion: Electric current is treated as a scalar quantity.

Reason: Currents at a junction do not add like vectors; they add algebraically.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Confusing emf with terminal voltage — remember V = ε − Ir while discharging and V = ε + Ir while charging.
  • Forgetting that maximum current needs R = 0, so Imax = ε/r (not ε/(R+r)).
  • In temperature problems, using temperatures directly instead of the difference (T2 − T1).
  • Dropping or mismatching units — resistivity is Ω m, conductivity is S m−1.
  • Treating current as a vector — it is a scalar; only current density j is a vector.
  • Sign errors in Kirchhoff’s loop rule — a negative current simply means the real direction is opposite to the assumed one.

How to score full marks in this chapter

Always write the formula, substitute with units, and box the final answer with the correct unit. For cell and charging problems, state clearly whether the cell is being charged or discharged before choosing V = ε ± Ir. In Kirchhoff problems, define every current with an arrow first, then write one equation per independent loop. Memorise the standard derivations (j = nevd, ρ = m/ne²τ, Wheatstone balance) as they are frequent 3–5 mark questions. Cross-check numerical answers for physical sense — drift speeds are mm s−1, currents in appliances are a few amperes.

Frequently Asked Questions

What is Class 12 Physics Chapter 3 Current Electricity about?

It studies charges in motion: electric current and current density, drift velocity, Ohm’s law and resistivity, the temperature dependence of resistivity, electrical energy and power, emf and internal resistance of cells, combinations of cells, and Kirchhoff’s rules with the Wheatstone bridge.

How many exercises are there in NCERT Class 12 Physics Chapter 3?

The end-of-chapter Exercises run from 3.1 to 3.9. All nine are reproduced verbatim and solved step by step on this page, with every numerical answer verified against the official NCERT answer key.

Why is the drift velocity of electrons so small yet current flows instantly?

The drift speed is only about 1 mm s−1, but the electric field that drives the electrons is set up throughout the conductor almost at the speed of light. So every electron in the circuit starts drifting at once and the current is established instantly, without electrons having to travel from one end to the other.

Are these Class 12 Physics Chapter 3 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for session 2026–27.

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