Physics · Current electricity · chapter 3 of 14

Class 12 Physics Chapter 3: Current Electricity worksheet with answer key

A 20-mark worksheet on Current Electricity for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

The SI unit of resistivity is

(a) S m⁻¹(b) Ω m⁻¹(c) Ω m⁻²(d) Ω m

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Answer: (d) Ω m

From R = ρl/A, ρ = RA/l, with unit Ω m² / m = Ω m. S m⁻¹ is the unit of conductivity.

Question 2

A heater is rated 1.0 kW at 220 V. Assuming its resistance does not change, the power it draws when connected to a 110 V supply is

(a) 125 W(b) 500 W(c) 1000 W(d) 250 W

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Answer: (d) 250 W

R = V²/P is fixed, so P ∝ V². Halving V gives P = 1000 × (110/220)² = 250 W. Taking P ∝ V gives the wrong 500 W.

Question 3

A steady current of 2.0 A flows through a wire of uniform cross-section 0.50 mm². The magnitude of the current density in the wire is

(a) 8.0 × 10⁶ A m⁻²(b) 1.0 × 10⁶ A m⁻²(c) 0.25 × 10⁶ A m⁻²(d) 4.0 × 10⁶ A m⁻²

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Answer: (d) 4.0 × 10⁶ A m⁻²

j = I/A = 2.0 A / (0.50 × 10⁻⁶ m²) = 4.0 × 10⁶ A m⁻².

Question 4

In a conductor placed in a uniform electric field of 0.60 V m⁻¹, the charge carriers drift at 0.30 mm s⁻¹. Their mobility is

(a) 500 cm² V⁻¹ s⁻¹(b) 0.050 cm² V⁻¹ s⁻¹(c) 5.0 cm² V⁻¹ s⁻¹(d) 0.20 cm² V⁻¹ s⁻¹

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Answer: (c) 5.0 cm² V⁻¹ s⁻¹

μ = v_d/E = (0.30 × 10⁻³ m s⁻¹)/(0.60 V m⁻¹) = 5.0 × 10⁻⁴ m² V⁻¹ s⁻¹. Since 1 m² V⁻¹ s⁻¹ = 10⁴ cm² V⁻¹ s⁻¹, μ = 5.0 cm² V⁻¹ s⁻¹.

Question 5

A steady potential difference is maintained across a metal wire. The drift velocity of its free electrons is

(a) along the field, and increases steadily with time(b) opposite to the field, and constant in time(c) opposite to the field, and equal to the electrons' thermal speed(d) zero, since the random velocities of the electrons average to zero

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Answer: (b) opposite to the field, and constant in time

v_d = −(eE/m)τ: the minus sign puts it opposite to E, and it does not depend on time even though each electron accelerates between collisions. The random velocities average to zero, so the drift is due to the field alone and is far smaller than the thermal speed.

Question 6 · 2 marks

Ohm's law is not a fundamental law of nature. State two different ways in which a device can depart from it, naming a device or material for each.

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(i) V stops being proportional to I, or the relation depends on the sign of V, as in a diode (rectifier), where reversing V does not give the same magnitude of current. (ii) The relation is not unique, more than one V for the same I, as in GaAs.

one departure with an example (diode: non-linear or sign-dependent) (1)
a second departure with an example (GaAs: non-unique relation) (1)

Question 7 · 2 marks

State the condition for balance of a Wheatstone bridge with arms P, Q, R and S. Why does the balance point not depend on the resistance of the galvanometer?

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Balance: P/Q = R/S (with P, Q in one pair of adjacent arms and R, S in the other). At balance no current flows through the galvanometer, so its resistance carries no potential drop and does not enter the condition.

P/Q = R/S (1)
I_g = 0 at balance, so the galvanometer's resistance plays no part (1)

Question 8 · 2 marks

A copper wire (resistivity 1.7 × 10⁻⁸ Ω m) carries a current density of 2.0 × 10⁶ A m⁻². Find the electric field inside the wire and state its direction relative to the current density.

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E = ρj = 1.7 × 10⁻⁸ × 2.0 × 10⁶ = 3.4 × 10⁻² V m⁻¹. Since j = σE with σ positive, E points along j.

E = ρj = 3.4 × 10⁻² V m⁻¹ (1)
E is along the current density (1)

Question 9 · 3 marks

Three identical cells, each of emf 2.0 V and internal resistance 0.60 Ω, are joined in parallel across a 1.8 Ω resistor. (a) Find the emf and internal resistance of the equivalent cell. (b) Find the current in the resistor and in each cell. (c) Find the current in the resistor if the three cells are instead joined in series.

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(a) 1/r_eq = 3/0.60, r_eq = 0.20 Ω; ε_eq/r_eq = 3 × 2.0/0.60, ε_eq = 2.0 V. (b) I = 2.0/(1.8 + 0.20) = 1.0 A, so each cell gives 1/3 A. (c) Series: ε = 6.0 V, r = 1.8 Ω, I = 6.0/3.6 = 1.67 A.

ε_eq = 2.0 V and r_eq = 0.20 Ω (1)
1.0 A in the resistor, 1/3 A in each cell (1)
series: 6.0/3.6 = 1.67 A (1)

Question 10 · 3 marks

A coil of metal X (temperature coefficient of resistivity 5.0 × 10⁻³ °C⁻¹, referred to 20 °C) and a manganin coil Y each have a resistance of 10.0 Ω at 20 °C. Treat the resistance of Y as constant and ignore changes in dimensions. (a) At what temperature is the resistance of X 1.5 times that of Y? (b) What is the resistance of X and Y in parallel at that temperature?

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(a) R_X = 10.0[1 + 5.0 × 10⁻³(T − 20)] = 15.0 Ω gives T − 20 = 100, so T = 120 °C. Y stays at 10.0 Ω because manganin's resistivity depends very little on temperature. (b) 15.0 × 10.0/25.0 = 6.0 Ω.

R_T = R₀[1 + α(T − T₀)] set to 15.0 Ω (1)
T = 120 °C (1)
parallel resistance 6.0 Ω (1)

Question 11 · 3 marks

Electrons in a conductor of cross-section A, with n free electrons per unit volume and relaxation time τ, are in a uniform field E. Starting from the drift velocity, show that the conductivity is σ = ne²τ/m.

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Each electron has acceleration −eE/m; averaging over electrons, with average time τ since the last collision and zero average random velocity, v_d = −eEτ/m. In time Δt the electrons within |v_d|Δt of an area A cross it: charge neA|v_d|Δt, so I = neA|v_d| = (e²Aτn/m)E. Then j = I/A = (ne²τ/m)E; comparing with j = σE, σ = ne²τ/m.

v_d = −eEτ/m from averaging (1)
I = neAv_d leading to I = (ne²τA/m)E (1)
j = (ne²τ/m)E compared with j = σE (1)

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