Current Electricity Class 12 Numericals 2026-27 | CBSE Physics
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Current Electricity Class 12 Physics — Numericals with Step-by-Step Solutions
This page is a focused numerical-practice resource for CBSE Class 12 Physics 2026–27, Chapter 3: Current Electricity. The problems progress from direct formula application to multi-step cell, Kirchhoff and Wheatstone-bridge situations.
Scope note: The numerical set follows the current CBSE 2026–27 Chapter 3 scope used by Learn Revise Hub: current and drift velocity, mobility, Ohm's law, V–I behaviour, electrical energy and power, resistivity and conductivity, temperature dependence, emf and internal resistance, cells in series and parallel, Kirchhoff's rules and Wheatstone bridge. The questions below are original practice problems, not claimed official PYQs.
Q1. A current of 2.5 A flows through a conductor for 4 minutes. Find the charge that passes through any cross-section of the conductor.
Solution:
Use Q = It.
Time = 4 × 60 = 240 s.
Q = 2.5 × 240 = 600 C
Answer: 600 C
Q2. A charge of 900 C passes through a conductor in 5 minutes. Find the average current.
Solution:
I = Q/t
t = 5 × 60 = 300 s.
I = 900/300 = 3 A
Answer: 3 A
Q3. A resistor of resistance 8 Ω is connected across a 24 V source. Find the current through it.
Solution:
Using Ohm's law:
I = V/R = 24/8 = 3 A
Answer: 3 A
Q4. A wire has resistance 12 Ω, length 3 m and cross-sectional area 2 × 10−6 m². Find its resistivity.
Solution:
From R = ρL/A,
ρ = RA/L
ρ = (12)(2 × 10−6)/3 = 8 × 10−6 Ω m
Answer: 8 × 10−6 Ω m
Q5. A conductor has resistivity 1.5 × 10−6 Ω m, length 2 m and cross-sectional area 3 × 10−6 m². Find its resistance.
Solution:
R = ρL/A
R = (1.5 × 10−6 × 2)/(3 × 10−6) = 1 Ω
Answer: 1 Ω
Q6. A wire has resistance R. It is replaced by a wire of the same material and length but with twice the cross-sectional area. Find the new resistance.
Solution:
Since R = ρL/A and A' = 2A:
R' = ρL/(2A) = R/2
Answer: R/2
Q7. A wire is stretched uniformly so that its length becomes twice its original length while its volume remains constant. If its original resistance is 5 Ω and resistivity remains unchanged, find the new resistance.
Solution:
Constant volume means LA = L'A'. Since L' = 2L, A' = A/2.
R' = ρ(2L)/(A/2) = 4ρL/A = 4R
R' = 4 × 5 = 20 Ω
Answer: 20 Ω
2. Drift Velocity, Mobility and Microscopic Numericals
Q8. A conductor carries 4 A current. Its cross-sectional area is 2 × 10−6 m² and the number density of free electrons is 5 × 1028 m−3. Taking e = 1.6 × 10−19 C, find the drift speed.
Solution:
Use I = neAvd.
vd = I/(neA)
vd = 4/[(5 × 1028)(1.6 × 10−19)(2 × 10−6)]
vd = 2.5 × 10−4 m s−1
Answer: 2.5 × 10−4 m s−1
Q9. A metal conductor has drift velocity 3 × 10−4 m s−1 in an electric field of 0.6 V m−1. Find the mobility of the electrons.
Solution:
μ = vd/E
μ = (3 × 10−4)/(0.6) = 5 × 10−4 m² V−1 s−1
Answer: 5 × 10−4 m² V−1 s−1
Q10. A conductor has carrier density 8 × 1028 m−3, charge-carrier mobility 4 × 10−3 m² V−1 s−1, and electric field 0.5 V m−1. Find the current density. Take e = 1.6 × 10−19 C.
Solution:
Use J = neμE.
J = (8 × 1028)(1.6 × 10−19)(4 × 10−3)(0.5)
J = 2.56 × 107 A m−2
Answer: 2.56 × 107 A m−2
Q11. A conductor carries 1.6 A current through an area of 1 × 10−6 m². Find its current density.
Solution:
J = I/A = 1.6/(1 × 10−6) = 1.6 × 106 A m−2
Answer: 1.6 × 106 A m−2
Q12. A conductor has drift velocity 2 × 10−4 m s−1 in an electric field of 0.4 V m−1. Find the mobility. If the field is doubled while mobility remains unchanged, find the new drift velocity.
Solution:
Initial mobility:
μ = vd/E = (2 × 10−4)/0.4 = 5 × 10−4 m² V−1 s−1
New field = 0.8 V m−1.
v'd = μE' = (5 × 10−4)(0.8) = 4 × 10−4 m s−1
Answers: μ = 5 × 10−4 m² V−1 s−1; v'd = 4 × 10−4 m s−1.
3. Power, Energy and Temperature Numericals
Q13. A 6 Ω resistor is connected across a 12 V supply. Find the current and electrical power consumed.
Solution:
I = V/R = 12/6 = 2 A
P = VI = 12 × 2 = 24 W
Answer: I = 2 A; P = 24 W
Q14. A 10 Ω resistor carries 3 A current for 2 minutes. Find the electrical energy converted into heat.
Solution:
W = I²Rt
t = 120 s.
W = 3² × 10 × 120 = 10,800 J
Answer: 1.08 × 104 J
Q15. A 100 W electrical device operates for 5 hours. Find the energy consumed in kWh and joules.
Solution:
Power = 0.1 kW.
Energy = 0.1 × 5 = 0.5 kWh
Since 1 kWh = 3.6 × 106 J:
W = 0.5 × 3.6 × 106 = 1.8 × 106 J
Answer: 0.5 kWh = 1.8 × 106 J
Q16. A resistor has resistance 20 Ω at 20°C. Its temperature coefficient of resistance is 0.004 K−1. Find its resistance at 70°C, assuming the linear approximation is valid.
Solution:
RT = R0[1 + α(T − T0)]
R70 = 20[1 + 0.004(70 − 20)]
= 20(1.20) = 24 Ω
Answer: 24 Ω
Q17. A resistor is connected first across 10 V and then across 20 V. If its resistance remains constant at 5 Ω, find the ratio of the powers consumed in the two cases.
Solution:
At constant R, P = V²/R.
P2/P1 = (20/10)² = 4
Answer: P2 : P1 = 4 : 1
4. EMF, Internal Resistance and Cells
Q18. A cell has emf 12 V and internal resistance 2 Ω. It is connected to an external resistance of 4 Ω. Find the current and terminal voltage.
Solution:
I = ε/(R+r) = 12/(4+2) = 2 A
V = IR = 2 × 4 = 8 V
Check: ε − Ir = 12 − 2×2 = 8 V.
Answer: I = 2 A; V = 8 V
Q19. A cell has emf 2.0 V. When it supplies a current of 0.5 A to an external circuit, its terminal voltage is 1.8 V. Find its internal resistance.
Solution:
For a discharging cell:
V = ε − Ir
Therefore:
r = (ε − V)/I = (2.0 − 1.8)/0.5 = 0.4 Ω
Answer: 0.4 Ω
Q20. A cell of emf 6 V and internal resistance 1 Ω is connected to a variable external resistance R. Find the current when R = 2 Ω and when R = 5 Ω.
Solution:
For R = 2 Ω:
I1 = 6/(2+1) = 2 A
For R = 5 Ω:
I2 = 6/(5+1) = 1 A
Answer: 2 A and 1 A respectively.
Q21. Three identical cells, each of emf 2 V and internal resistance 0.5 Ω, are connected in series aiding to an external resistance of 4.5 Ω. Find the current and terminal voltage across the external resistance.
Solution:
εeq = 3×2 = 6 V
req = 3×0.5 = 1.5 Ω
I = 6/(4.5+1.5) = 1 A
External terminal voltage:
V = IR = 1×4.5 = 4.5 V
Answer: I = 1 A; V = 4.5 V
Q22. Four identical cells, each of emf 1.5 V and internal resistance 2 Ω, are connected in parallel. They supply an external resistance of 1 Ω. Find the current.
Solution:
For identical cells in parallel:
εeq = 1.5 V; req = 2/4 = 0.5 Ω
I = 1.5/(1+0.5) = 1 A
Answer: 1 A
Q23. A cell of emf 10 V and internal resistance 2 Ω is connected to an external resistance of 8 Ω. Find (i) current, (ii) power delivered to the external resistor, and (iii) power lost inside the cell.
Solution:
I = 10/(8+2) = 1 A
Pexternal = I²R = 1²×8 = 8 W
Pinternal = I²r = 1²×2 = 2 W
Answer: (i) 1 A, (ii) 8 W, (iii) 2 W
Q24. A cell is being charged with a current of 2 A. Its emf is 1.5 V and internal resistance is 0.5 Ω. Find the terminal potential difference during charging.
Solution:
For charging:
V = ε + Ir = 1.5 + (2)(0.5) = 2.5 V
Answer: 2.5 V
5. Kirchhoff's Rules Numericals
Q25. At a junction, currents of 5 A and 2 A enter. A current of 4 A leaves through one branch. Find the current in the remaining branch leaving the junction.
Solution:
Using Kirchhoff's junction rule:
Total entering = Total leaving
5 + 2 = 4 + I
I = 3 A
Answer: 3 A leaving the junction.
Q26. A single closed loop contains a 12 V cell and two resistors of 2 Ω and 4 Ω connected in series. Find the current using Kirchhoff's loop rule.
Solution:
Applying the loop rule:
12 − 2I − 4I = 0
6I = 12 ⇒ I = 2 A
Answer: 2 A
Q27. A loop contains a 10 V cell, a 4 Ω resistor and a 6 Ω resistor. If the assumed current is clockwise, write the loop equation and calculate the current.
Solution:
Taking the cell as a rise of 10 V and both resistor drops in the assumed current direction:
10 − 4I − 6I = 0
10 − 10I = 0 ⇒ I = 1 A
Answer: 1 A clockwise.
Q28. Two branches meet at a junction. A current I enters the junction, while 3 A and 5 A leave through two other branches. Find I.
Solution:
By the junction rule:
I = 3 + 5 = 8 A
Answer: 8 A entering the junction.
Q29. A current of 1.5 A is assumed clockwise in a circuit. After applying Kirchhoff's rules, the calculated value is −1.5 A. What is the actual current?
Solution:
The negative sign means the actual direction is opposite to the assumed direction.
Answer: 1.5 A anticlockwise.
Q30. A circuit loop contains a 15 V source and resistors 3 Ω and 2 Ω. A second source of 5 V opposes the 15 V source in the same loop. Find the current.
Solution:
Net emf = 15 − 5 = 10 V.
Total resistance = 3 + 2 = 5 Ω.
I = 10/5 = 2 A
Answer: 2 A in the direction driven by the 15 V source.
6. Wheatstone Bridge Numericals
Q31. A Wheatstone bridge is balanced when P = 2 Ω, Q = 4 Ω and R = 3 Ω. Find S.
Solution:
For a balanced bridge:
P/Q = R/S
2/4 = 3/S ⇒ S = 6 Ω
Answer: 6 Ω
Q32. In a balanced Wheatstone bridge, P = 5 Ω, Q = 10 Ω and S = 8 Ω. Find R.
Solution:
P/Q = R/S
5/10 = R/8
R = 4 Ω
Answer: 4 Ω
Q33. A bridge has P = 3 Ω, Q = 6 Ω, R = 4 Ω and S = 8 Ω. Determine whether it is balanced.
Solution:
Calculate the two ratios:
P/Q = 3/6 = 1/2
R/S = 4/8 = 1/2
The ratios are equal, so the bridge is balanced.
Answer: Yes, the bridge is balanced and galvanometer current is zero.
Q34. In a standard Wheatstone bridge, P = 4 Ω, Q = 5 Ω and R = 8 Ω. What value of S will make the bridge balanced?
Solution:
P/Q = R/S
4/5 = 8/S ⇒ S = 10 Ω
Answer: 10 Ω
7. Mixed Board-Level Problems
Q35. A wire carries a current of 2 A. Its cross-sectional area is 1.6 × 10−6 m² and free-electron density is 5 × 1028 m−3. Find the drift velocity. Take e = 1.6 × 10−19 C.
Solution:
vd = I/(neA)
vd = 2/[(5×1028)(1.6×10−19)(1.6×10−6)]
vd = 1.5625 × 10−4 m s−1
Answer: 1.56 × 10−4 m s−1 approximately.
Q36. A resistor has resistance 15 Ω at 20°C. Its resistance becomes 18 Ω at 70°C. Find its average temperature coefficient over this range.
Solution:
Use:
RT = R0[1 + α(T−T0)]
18 = 15[1 + α(50)]
18/15 = 1 + 50α
1.2 − 1 = 50α ⇒ α = 0.004 K−1
Answer: 4 × 10−3 K−1
Q37. A cell has emf 8 V. Its terminal voltage is 6 V when it supplies a current of 1 A. Find its internal resistance. If the external resistance is then changed so that current becomes 0.5 A, find the new terminal voltage.
Solution:
From the first condition:
6 = 8 − 1×r ⇒ r = 2 Ω
For the second condition:
V = ε − Ir = 8 − (0.5)(2) = 7 V
Answer: r = 2 Ω; new terminal voltage = 7 V.
Q38. A 20 V source supplies a current of 2 A to a circuit for 3 minutes. Find the electrical energy supplied by the source.
Solution:
W = VIt
t = 3×60 = 180 s.
W = 20×2×180 = 7200 J
Answer: 7.2 × 103 J
Q39. A cell of emf 12 V and internal resistance 1 Ω supplies 2 A. Find (i) terminal voltage, (ii) power supplied by the ideal emf, (iii) power dissipated internally.
Solution:
(i) V = ε − Ir = 12 − 2×1 = 10 V
(ii) Pemf = εI = 12×2 = 24 W
(iii) Pinternal = I²r = 2²×1 = 4 W
The remaining 20 W is delivered to the external circuit.
Answer: 10 V, 24 W and 4 W.
Q40. A balanced Wheatstone bridge has P = 2 Ω and Q = 5 Ω. If R = 6 Ω, find S. If S is instead changed to 10 Ω, state whether the bridge remains balanced.
Solution:
At balance:
P/Q = R/S
2/5 = 6/S ⇒ S = 15 Ω
Therefore, S must be 15 Ω. If S = 10 Ω, the ratio is not satisfied, so the bridge is not balanced.
Answer: Required S = 15 Ω; with S = 10 Ω, the bridge is unbalanced.
8. Current Electricity Numerical-Solving Checklist
Write the given quantities first. Include units.
Identify the physical model. Is it a direct current problem, microscopic-current problem, cell problem, Kirchhoff problem or Wheatstone bridge?
Write the governing equation before substituting numbers.
Convert time to seconds when using SI units.
Check powers of ten carefully in microscopic-current questions.
For a cell, decide whether it is supplying or being charged before choosing the sign in the terminal-voltage relation.
For Kirchhoff problems, choose current directions consistently. A negative answer means the actual direction is opposite to the assumed one.
For Wheatstone bridge questions, verify that the bridge is balanced before using the balance ratio.
Write the final numerical answer with its SI unit.
Check the magnitude. A quick order-of-magnitude check can catch many calculator or exponent errors.
Best practice: First solve each problem without looking at the solution. Then compare your equation, substitution, unit and final answer—not just the final number.
Source note: The chapter scope and unit-level marks were checked against the official CBSE 2026–27 Physics curriculum. The current CBSE Class XII SQP/MS page was also checked. The numerical problems and solutions on this page are original Learn Revise Hub practice material.
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