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Current Electricity Class 12 Important Questions 2026-27 | CBSE Physics
Current Electricity Class 12 Physics — Important Questions with Answers
This practice set is designed for CBSE Class 12 Physics 2026–27. It focuses on the concepts, explanations, derivations and numerical situations that students should be able to handle from the current Chapter 3 scope.
Important syllabus note: These are original CBSE-style practice questions, not claimed official CBSE previous-year questions. The 2026–27 CBSE curriculum lists Current Electricity as Chapter 3 under Unit II and assigns 17 marks to Unit II; CBSE does not publish a fixed Chapter-3-only mark allocation. The questions below therefore use marks as practice formats, not as a prediction of the actual paper.
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1. 1-Mark Concept Questions
Q1. What is electric current? Write its SI unit.
Answer: Electric current is the rate of flow of electric charge through a cross-section of a conductor. I = dQ/dt Its SI unit is ampere (A).
Q2. In a metallic conductor, in which direction does electron drift occur relative to the electric field?
Answer: Electron drift is opposite to the direction of the electric field.
Q3. Define mobility of charge carriers.
Answer: Mobility is the magnitude of drift velocity acquired per unit applied electric field. μ = vd/E
Q4. Write the relation between current and drift velocity in a metallic conductor.
Answer: I = neAvd, where n is the number density of charge carriers, e is the magnitude of charge, A is cross-sectional area and vd is drift velocity.
Q5. State Ohm's law.
Answer: At constant physical conditions such as temperature, the current through a conductor is directly proportional to the potential difference across it: V = IR.
Q6. What is the SI unit of resistivity?
Answer: Ohm metre (Ω m).
Q7. What is conductivity in terms of resistivity?
Answer: Conductivity is the reciprocal of resistivity: σ = 1/ρ.
Q8. What happens to the resistance of a typical metallic conductor when its temperature increases?
Answer: Over the usual approximately linear temperature range, its resistance increases.
Q9. A cell has emf ε and internal resistance r. Write its terminal voltage while delivering current I.
Answer: V = ε − Ir.
Q10. What is the balanced condition of a Wheatstone bridge?
Answer: For the standard bridge arrangement, if the four arms are P, Q, R and S, the balance condition is P/Q = R/S. At balance, no current flows through the galvanometer.
2. 2-Mark Short-Answer Questions
Q11. Why does a metallic conductor not carry a net current when no electric field is applied, even though its electrons are moving?
Answer: Free electrons have random thermal motion. Their velocities are distributed in different directions, so the average velocity and hence the net drift velocity are zero. Therefore there is no net current.
Q12. Distinguish between resistance and resistivity.
Answer: Resistance is the opposition offered by a particular conductor and depends on its material, length, cross-sectional area and physical conditions. Resistivity is a material property at specified physical conditions and is related by R = ρL/A. Their SI units are Ω and Ω m respectively.
Q13. Why can a conductor have very small electron drift speed but still carry a measurable current?
Answer: Current depends on the product I = neAvd. Even when drift velocity is small, a large number density of charge carriers and a finite cross-sectional area can produce a measurable current.
Q14. A conductor has its length doubled while its cross-sectional area remains unchanged. What happens to its resistance if temperature and material remain unchanged?
Answer: Since R = ρL/A, doubling L doubles R. Therefore the new resistance is 2R.
Q15. State two differences between emf and terminal potential difference of a cell.
Answer: EMF is the energy supplied by the source per unit charge and is the open-circuit potential difference of an idealised cell. Terminal potential difference is the potential difference available at the terminals under the operating condition. When a cell supplies current I, V = ε − Ir, so terminal voltage is less than emf for a real cell with internal resistance.
Q16. What do the junction rule and loop rule of Kirchhoff express physically?
Answer: The junction rule expresses conservation of charge. The loop rule expresses conservation of energy in an electrical circuit.
Q17. A cell is being charged rather than supplying current. How does the terminal-voltage relation change?
Answer: For charging current I, the terminal potential difference is V = ε + Ir, assuming the usual sign convention for the charging direction.
Q18. Why is the current through the galvanometer zero when a Wheatstone bridge is balanced?
Answer: At balance, the two points connected to the galvanometer are at the same potential. Hence the potential difference across the galvanometer is zero and its current is zero.
3. 3-Mark Questions and Numericals
Q19. Derive the relation I = neAvd.
Answer:
Consider a conductor of cross-sectional area A. In time Δt, charge carriers with drift speed vd travel a distance vdΔt.
Volume of the conductor segment swept by the carriers = AvdΔt.
Number of charge carriers in this volume = nAvdΔt.
Total charge = nAvdΔt × e.
Therefore, I = Q/Δt = neAvd.
Q20. A wire carries a current of 2.0 A. How much charge passes through a cross-section in 5 minutes?
Solution: Q = It = 2.0 × (5 × 60) = 600 C. Therefore, Q = 600 C.
Q21. A resistor of 6 Ω is connected across a 12 V supply. Find the current and power.
Solution:
I = V/R = 12/6 = 2 A
P = VI = 12 × 2 = 24 W
Therefore, I = 2 A and P = 24 W.
Q22. A conductor has resistance 8 Ω, length 2 m and cross-sectional area 1 × 10−6 m². Calculate its resistivity.
Solution:
ρ = RA/L = (8 × 1 × 10−6)/2 = 4 × 10−6 Ω m
Therefore, ρ = 4 × 10−6 Ω m.
Q23. A metallic resistor has resistance 10 Ω at 20°C. Its temperature coefficient of resistance is 0.004 K−1. Find its resistance at 70°C, assuming the linear relation remains valid.
Solution:
RT = R0[1 + α(T − T0)]
R70 = 10[1 + 0.004(70 − 20)] = 10(1.20) = 12 Ω
Therefore, R70 = 12 Ω.
Q24. A cell of emf 12 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. Find the current and terminal voltage.
Solution:
I = ε/(R+r) = 12/(5+1) = 2 A
V = IR = 2 × 5 = 10 V
Therefore, I = 2 A and terminal voltage = 10 V.
Q25. Three identical cells, each of emf 2 V and internal resistance 0.5 Ω, are connected in series with an external resistance of 4.5 Ω. Find the current.
Solution:
εeq = 3 × 2 = 6 V
req = 3 × 0.5 = 1.5 Ω
I = 6/(4.5 + 1.5) = 1 A
Therefore, I = 1 A.
Q26. A Wheatstone bridge has P = 2 Ω, Q = 4 Ω and R = 3 Ω. Find S for a balanced bridge.
Solution:
For balance, P/Q = R/S.
Thus, 2/4 = 3/S ⇒ S = 6 Ω.
Therefore, S = 6 Ω.
Q27. In a circuit, a branch current calculated using Kirchhoff's rules is −0.8 A. What does the negative sign mean?
Answer: The actual current is 0.8 A in the direction opposite to the direction initially assumed for that branch.
4. 5-Mark / Multi-Step Questions
Q28. A cell of emf 10 V and internal resistance 2 Ω is connected to an external resistor of 8 Ω. (a) Find the current. (b) Find terminal voltage. (c) Find power delivered to the external resistor. (d) Find power lost inside the cell.
Solution:
(a) I = ε/(R+r) = 10/(8+2) = 1 A
(b) V = IR = 1 × 8 = 8 V
(c) Pexternal = I²R = 1² × 8 = 8 W
(d) Pinternal = I²r = 1² × 2 = 2 W
Answers: 1 A, 8 V, 8 W and 2 W.
Q29. Explain how Kirchhoff's rules can be used to solve a two-loop circuit containing cells and resistors.
Answer:
- Assign a current to each independent branch; directions may initially be chosen arbitrarily.
- Apply the junction rule at a suitable junction: the algebraic sum of currents is zero.
- Select independent closed loops.
- Apply the loop rule to each loop, using consistent signs for emf and resistor potential changes.
- Solve the resulting simultaneous equations.
- Interpret any negative current as a reversal of the assumed direction.
The method follows conservation of charge at junctions and conservation of energy around closed loops.
Q30. Explain the principle of a balanced Wheatstone bridge and derive its balance relation in words.
Answer:
At balance, no current flows through the galvanometer, so the two galvanometer junctions are at the same potential. The potential drops along the two branches therefore occur in the same ratio as their corresponding resistances. For the standard labelling P, Q, R and S, the balanced condition is:
P/Q = R/Sor equivalently PS = QR. This relation can also be obtained by applying Kirchhoff's loop rule to the balanced network.
Q31. A wire is stretched so that its length becomes twice its original length while its volume remains constant. Assuming resistivity does not change, determine the new resistance in terms of the original resistance.
Solution:
Constant volume gives LA = L'A'. If L' = 2L, then A' = A/2.
Since R = ρL/A,
R' = ρ(2L)/(A/2) = 4ρL/A = 4RTherefore, the new resistance is 4R.
5. Rapid Revision Questions
| Question | Answer in one line |
|---|---|
| Current formula? | I = dQ/dt |
| Current density? | J = I/A |
| Current–drift relation? | I = neAvd |
| Mobility? | μ = vd/E |
| Ohm's law? | V = IR under constant physical conditions. |
| Resistance of a uniform wire? | R = ρL/A |
| Conductivity? | σ = 1/ρ |
| Power? | P = VI = I²R = V²/R |
| Cell delivering current? | V = ε − Ir |
| Cell being charged? | V = ε + Ir |
| Series identical cells? | εeq = nε, req = nr |
| Parallel identical cells? | εeq = ε, req = r/n |
| Kirchhoff junction rule? | ΣI = 0 |
| Kirchhoff loop rule? | ΣΔV = 0 |
| Wheatstone balance? | P/Q = R/S |
Self-test: Cover the right-hand column and reproduce each answer from memory. Then solve Q20–Q26 without looking at the solutions.
6. Answer-Writing Mistakes to Avoid
- Writing a formula without defining the symbols in a derivation-based answer.
- Forgetting units in numerical answers.
- Using Celsius differences incorrectly; a temperature interval in °C has the same numerical size as the interval in K.
- Mixing up emf and terminal voltage.
- Using the delivering-current sign for a charging cell.
- Changing a Kirchhoff current direction midway through the calculation without changing the equations consistently.
- Ignoring the meaning of a negative current obtained from simultaneous equations.
- Applying the Wheatstone balance relation when the bridge is not balanced.
- Calling a practice question a PYQ without verifying its official source and year.
7. Continue Current Electricity Preparation
The detailed notes for this chapter are already published. More practice formats will be added one at a time so that every internal link points to a verified live page.
8. Official CBSE Resources
- CBSE Physics Curriculum 2026–27
- CBSE Class XII Sample Question Papers & Marking Schemes 2026–27
- CBSE Question Paper Archive
Source note: The syllabus scope is based on the official CBSE 2026–27 Physics curriculum. The practice questions and answers on this page are original Learn Revise Hub material. External websites were reviewed for topic coverage and content gaps, but their topic lists were not treated as authoritative where they included material outside the current CBSE Chapter 3 scope.
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