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CBSE · Class 12 · Physics

Current Electricity

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A current of 4 A flows for 21 s. The charge that passes is:
  1. 84 C
  2. 25 C
  3. 5.25 C
  4. 0.19 C
2.
A wire 18 m long with cross-section 8 mm2 has resistivity 37 × 10−8 Ω m. Its resistance is:
  1. 83.25 Ω
  2. 0.08325 Ω
  3. 0.8325 Ω
  4. 0.16 Ω
3.
The equivalent resistance of 2 Ω, 5 Ω in parallel is:
  1. 2.86 Ω
  2. 7 Ω
  3. 1.43 Ω
  4. 2.43 Ω
4.
A cell of emf 23 V and internal resistance 3 Ω is joined to a 7 Ω resistor. The terminal voltage is:
  1. 16.1 V
  2. 29.9 V
  3. 23 V
  4. 6.9 V
5.
Ohm's law states that, at a fixed temperature:
  1. V is proportional to I2
  2. V is proportional to I
  3. R is proportional to I
  4. V is proportional to 1/I
6.
The power used by a 45 Ω resistor across 230 V is:
  1. 2,351.11 W
  2. 5.11 W
  3. 10,350 W
  4. 1,175.56 W
7.
In a metre bridge, the balance point is 49 cm from the end where the unknown resistance X is joined; a 10 Ω resistor is in the other gap. X is:
  1. 10.41 Ω
  2. 20.82 Ω
  3. 20.41 Ω
  4. 9.61 Ω
8.
Assertion (A): A Wheatstone bridge with P = 10 Ω, Q = 60 Ω and R = 30 Ω balances when S = 5 Ω.
Reason (R): A Wheatstone bridge is balanced when P/Q = R/S.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A 7 Ω resistor is in series with 2 Ω and 7 Ω in parallel, across an ideal 20 V battery. Find the current through each resistor.
10.
6 identical cells, each of emf 4 V and internal resistance 1 Ω, drive a 7 Ω resistor. Find the current when the cells are joined (a) in series, (b) in parallel.
11.
A wire has resistance 85 Ω at 0 °C and temperature coefficient 8 × 10−3 per °C. Find its resistance at 70 °C.
12.
Two cells of emf 10 V and 6 V (no internal resistance) are joined through 4 Ω and 5 Ω to a common junction, with their negative terminals joined together, and a 4 Ω resistor runs from the junction to the negative side. Find the current in the 4 Ω resistor.
13.
In a metre bridge with X in the left gap and 17 Ω in the right gap, the null point is 63 cm from the left end. Find X. Where would the null point be if the two were swapped?
14.
A copper wire of cross-section 5 mm2 carries 1 A. Find the drift speed of the electrons if there are 8 × 1028 free electrons per m3. (e = 1.6 × 10−19 C)
15.
An 800 W appliance runs 3 hours a day for 16 days. Find the energy used in kWh and the cost at Rs 9 per unit.
16.
A wire of resistance 4 Ω is stretched uniformly to 4 times its length. Find its new resistance.
Section C — Word problems4 × 4 = 16 marks
17.
Cells of 11 V and 8 V (no internal resistance) are joined in parallel through resistors of 4 Ω and 2 Ω respectively, and the pair drives a 4 Ω load. Using Kirchhoff's rules, find the current through each cell and through the load.
18.
A cell of emf 9 V and internal resistance 1 Ω drives a load R. Find the power in the load when R = 11 Ω, and the value of R that takes the most power, with that power.
19.
A Wheatstone bridge has P from A to B, Q from B to C, R from A to D and S from D to C, with the cell across A and C and a galvanometer across B and D. P = 80 Ω, Q = 80 Ω and S = 90 Ω. Find R for balance. If R is then doubled, in which direction does current flow through the galvanometer?
20.
A home uses a 1300 W heater and 6 bulbs of 100 W, all for 8 hours a day on 230 V mains. Find the total current drawn, the energy used in 30 days, and the cost at Rs 7 per kWh.
AmbifulWorksheets
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CBSE · Class 12 · Physics

Answer key

Current Electricity · AW-12P-CEL-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 84 CQ = It
2(c) 0.8325 ΩR = ρL/A
3(c) 1.43 Ω1/R = Σ 1/Ri
4(a) 16.1 VV = E − Ir = IR
5(b) V is proportional to I—
6(d) 1,175.56 WP = V2/R
7(a) 10.41 ΩX/R = l/(100 − l) = 49/51
8(d) A is false, but R is true.A is false: P/Q = R/S gives S = QR/P = 180 Ω. R is true.
97 Ω: 2.34 A; 2 Ω: 1.82 A; 7 Ω: 0.52 AR = 7 + 1.56 = 8.56 Ω
10(a) 1.85 A; (b) 0.56 Aseries: nE/(R + nr); parallel: E/(R + r/n)
11132.6 ΩR = R0(1 + αΔT)
121.32 AKirchhoff: (10 − V)/4 + (6 − V)/5 = V/4 gives V = 5.29 V at the junction
13X = 28.95 Ω; 37 cm from the left endX/17 = 63/37
141.5625 × 10−3 m/sv = I/(neA) = 1 ÷ (8 × 1028 × 1.6 × 10−19 × 5 × 10−6)
1538.4 kWh; Rs 345.6800 × 3 × 16 ÷ 1000
1664 Ωvolume is unchanged, so A falls 4 times; R ∝ L/A grows 42 = 16 times
1711 V cell: 1.0625 A; 8 V cell: 0.625 A (a negative value means it is being charged); load: 1.6875 Ajunction potential V = 6.75 V from (11 − V)/4 + (8 − V)/2 = V/4
186.1875 W; R = 1 Ω; 20.25 WP = E2R/(R + r)2, largest when R = r
19R = 90 Ω; with R doubled the potential at D falls, so current flows from B to Dbalance: P/Q = R/S; V_D = V × S/(R + S) drops when R grows
208.26 A; 456 kWh; Rs 3,192I = P/V; energy = P × t