Current Electricity Class 12 Physics Chapter Test 2026-27

Current Electricity Class 12 Physics Chapter Test 2026–27
An original, exam-focused chapter test covering the complete Current Electricity scope prescribed by CBSE for 2026–27.

Important: This is original CBSE-style practice, not an official CBSE question paper or a collection of verified previous-year questions. The test covers the 2026–27 Chapter 3 scope: current, drift velocity, mobility, current density, Ohm's law, V–I characteristics, power and energy, resistivity and conductivity, temperature dependence, emf/internal resistance, cells, Kirchhoff's rules and Wheatstone bridge. CBSE assigns 17 marks to Unit II Current Electricity at unit level.

Test Instructions

• Suggested time: 60 minutes.
• Maximum practice score: 40 marks.
• Attempt the questions before opening the answer section.
• Use SI units unless another unit is explicitly stated.
• Show working for numerical and circuit-based questions.
• Use g = 9.8 m s⁻² only if a question requires it; most questions here do not.

Section A — MCQs

10 questions × 1 mark = 10 marks

1. A charge of 12 C passes through a conductor in 3 s. The average current is:

A. 2 A

B. 3 A

C. 4 A

D. 36 A

Answer: C — I = Q/t = 12/3 = 4 A.

2. The drift speed of electrons in a metallic conductor is generally:

A. comparable to the speed of light

B. much smaller than their random thermal speed

C. always zero when current flows

D. independent of electric field

Answer: B — Drift speed is the small average directed component produced by the applied field.

3. A wire has resistance R. If its length is doubled while its area and material remain unchanged, its resistance becomes:

A. R/4

B. R/2

C. 2R

D. 4R

Answer: C — R = ρL/A, so R is directly proportional to L.

4. The SI unit of resistivity is:

A. Ω

B. Ω m

C. Ω/m

D. S m

Answer: B.

5. For a fixed voltage across a resistor, the power dissipated is:

A. directly proportional to R

B. inversely proportional to R

C. independent of R

D. proportional to R²

Answer: B — P = V²/R.

6. A cell has emf ε and internal resistance r. When it supplies current I, its terminal voltage is:

A. ε + Ir

B. ε − Ir

C. Ir − ε

D. ε/I − r

Answer: B for discharge.

7. For identical cells connected in parallel, the equivalent emf is:

A. nε

B. ε/n

C. ε

D. zero

Answer: C — identical parallel cells have the same emf as one cell.

8. Kirchhoff's junction rule follows from conservation of:

A. energy

B. charge

C. momentum

D. mass only

Answer: B.

9. At balance in a Wheatstone bridge:

A. the galvanometer current is maximum

B. the galvanometer current is zero

C. all four resistances must be equal

D. the battery current is zero

Answer: B.

10. If the radius of a uniform wire is doubled, with length and material unchanged, its resistance becomes:

A. 4R

B. 2R

C. R/2

D. R/4

Answer: D — area becomes four times, so R becomes one-fourth.

Section B — Short Answer & Concept Application

5 questions × 2 marks = 10 marks

11. Define current density. A current of 3 A is uniformly distributed through a cross-sectional area of 2 mm². Find its current density.

Answer: J = I/A. Since 2 mm² = 2 × 10⁻⁶ m², J = 3/(2 × 10⁻⁶) = 1.5 × 10⁶ A m⁻².

12. State Ohm's law. What condition must be maintained for the law to apply?

Answer: At constant physical conditions, especially temperature, the potential difference across an ohmic conductor is directly proportional to current: V = IR. The resistance must remain constant over the measurement.

13. Distinguish between emf and terminal potential difference of a discharging cell.

Answer: Emf ε is the energy supplied by the source per unit charge. During discharge, terminal voltage is V = ε − Ir, so it is less than ε when I and r are non-zero.

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

Answer: For many metals, increased temperature increases lattice vibrations, which increases opposition to electron drift. This gives a positive temperature coefficient over the relevant range.

15. State Kirchhoff's two rules and the conservation principle behind each.

Answer: Junction rule: algebraic sum of currents at a junction is zero — conservation of charge. Loop rule: algebraic sum of potential changes around a closed loop is zero — conservation of energy.

Section C — Numericals

5 questions × 3 marks = 15 marks

16. A wire of resistance 8 Ω carries a current of 2.5 A. Find (i) the potential difference across it and (ii) the power dissipated.

Solution: V = IR = 2.5 × 8 = 20 V. P = I²R = (2.5)² × 8 = 50 W.

17. A wire of length 2 m and cross-sectional area 1.5 × 10⁻⁶ m² has resistance 4 Ω. Find its resistivity.

Solution: ρ = RA/L = (4 × 1.5 × 10⁻⁶)/2 = 3.0 × 10⁻⁶ Ω m.

18. A cell of emf 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. Find the circuit current and terminal voltage.

Solution: I = ε/(R+r) = 12/6 = 2 A. V = IR = 2 × 5 = 10 V. Also ε − Ir = 12 − 2 = 10 V.

19. Two identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in series aiding to a 2 Ω external resistor. Find the current.

Solution: εeq = 3 V and req = 1 Ω. I = 3/(2+1) = 1 A.

20. A current of 4 A flows through a resistor for 5 minutes. If the resistance is 3 Ω, calculate the electrical energy converted into heat.

Solution: t = 300 s. W = I²Rt = 16 × 3 × 300 = 14,400 J = 14.4 kJ.

Section D — Higher-Order / Circuit Reasoning

3 questions × 5 marks = 15 marks

21. Explain the relation between current, number density of carriers, cross-sectional area and drift velocity in a metallic conductor. A wire carries 1.6 A and has area 2 × 10⁻⁶ m². If n = 5 × 10²⁸ m⁻³, calculate the drift speed of electrons. Take e = 1.6 × 10⁻¹⁹ C.

Solution: I = neAv_d. Therefore v_d = I/(neA) = 1.6 / [(5 × 10²⁸)(1.6 × 10⁻¹⁹)(2 × 10⁻⁶)] = 1.0 × 10⁻⁴ m s⁻¹. Current increases with carrier density, charge magnitude, area and drift speed.

22. A circuit contains two resistors of 4 Ω and 6 Ω connected in parallel across a 12 V source. Find (i) equivalent resistance, (ii) total current, (iii) current through each resistor, and (iv) total power.

Solution: 1/Req = 1/4 + 1/6 = 5/12, so Req = 2.4 Ω. Total current = 12/2.4 = 5 A. Branch currents: I₄ = 12/4 = 3 A, I₆ = 12/6 = 2 A. Total power = VI = 12 × 5 = 60 W.

23. In a Wheatstone bridge, the four arms are P = 2 Ω, Q = 3 Ω, R = 4 Ω and S = 6 Ω. Determine whether the bridge is balanced. If it is balanced, state the galvanometer current.

Solution: P/Q = 2/3 and R/S = 4/6 = 2/3. Hence the bridge is balanced. Therefore the galvanometer current is zero.

Answer Summary

Question	Expected answer	Marks
1–10	C, B, C, B, B, B, C, B, B, D	10
11	J = 1.5 × 10⁶ A m⁻²	2
12	Ohm's law + constant physical conditions	2
13	ε vs V = ε − Ir	2
14	Positive temperature coefficient of metals	2
15	Junction + loop rules and conservation laws	2
16	20 V; 50 W	3
17	3.0 × 10⁻⁶ Ω m	3
18	2 A; 10 V	3
19	1 A	3
20	14,400 J	3
21	vd = 1.0 × 10⁻⁴ m s⁻¹	5
22	2.4 Ω; 5 A; 3 A & 2 A; 60 W	5
23	Balanced; IG = 0	5

Score Interpretation

Score	Revision signal
34–40	Strong chapter control; focus on speed, presentation and mixed application.
28–33	Good base; revise formulas and revisit the questions where working was weak.
20–27	Concepts are developing; revise Notes and Formula Sheet before another timed test.
Below 20	Rebuild the chapter systematically: concepts → formulas → basic numericals → application.

Continue Your Current Electricity Preparation

Notes Important Questions MCQs Numericals Case-Based Questions Assertion–Reason Questions Formula Sheet & Quick Revision Chapter 1 Hub Chapter 2 Hub

Recommended sequence: Take this test without looking at the answers → record your score → revisit only the weak topics → retake the relevant practice set.