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CBSE · Class 12 · Physics

Alternating Current

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.
The peak value of a 110 V (rms) supply is:
  1. 110/√2 V
  2. 110 V
  3. 220 V
  4. 110√2 V
2.
The reactance of a 18 × 10−2 H inductor at an angular frequency of 950 rad/s is:
  1. 171 Ω
  2. 0.00585 Ω
  3. 342 Ω
  4. 17.1 Ω
3.
The reactance of a 1 μF capacitor at an angular frequency of 800 rad/s is:
  1. 12,500 Ω
  2. 0.0008 Ω
  3. 625 Ω
  4. 1,250 Ω
4.
In a series LCR circuit, R = 16 Ω, X_L = 64 Ω and X_C = 52 Ω. The impedance is:
  1. 12 Ω
  2. 132 Ω
  3. 20 Ω
  4. 28 Ω
5.
The rms value of an alternating current with peak I0 is:
  1. I0/√2
  2. 2I0
  3. I0√2
  4. I0/2
6.
The power factor of a series circuit with R = 6 Ω, X_L = 10 Ω and X_C = 18 Ω is:
  1. 3/5
  2. 1
  3. 4/5
  4. 3/4
7.
A series circuit has L = 1.5 H and C = 24 μF. Its resonant angular frequency is:
  1. 166.67 rad/s
  2. 333.33 rad/s
  3. 83.33 rad/s
  4. 1,666.67 rad/s
8.
Assertion (A): In a series LCR circuit with L = 0.1 H and C = 40 μF, the current is greatest at ω = 500 rad/s.
Reason (R): At resonance X_L = X_C, so the impedance equals R and is at its smallest.
  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 150 V (rms) supply drives a series circuit with R = 12 Ω, X_L = 58 Ω and X_C = 49 Ω. Find the current and the voltage across each element.
10.
For a series circuit with R = 20 Ω, X_L = 30 Ω and X_C = 51 Ω across 290 V (rms), find the power factor and the average power.
11.
An ideal transformer has 700 primary turns and 1500 secondary turns, with 2200 V across the primary. Find the secondary voltage, and the primary current when the secondary supplies 7 A.
12.
A series circuit has R = 70 Ω, L = 1 H and C = 4 μF. Find the resonant angular frequency and the quality factor.
13.
A 20 μF capacitor is connected to a 220 V, 50 Hz supply. Find its reactance and the rms current.
14.
A pure inductor of 14 × 10−1 H is connected to a 220 V, 60 Hz supply. Find its reactance and the rms current.
15.
A transformer takes 14 A at 220 V and is 85% efficient. Find the output power.
16.
Find the phase angle between current and voltage in a series circuit with R = 10 Ω, X_L = 45 Ω and X_C = 69 Ω, and say which leads.
Section C — Word problems4 × 4 = 16 marks
17.
A series LCR circuit has R = 80 Ω, X_L = 15 Ω and X_C = 99 Ω across a 464 V (rms) supply. Find the impedance, the current, the power factor, the average power and the voltage across each element.
18.
A 90 V (rms) variable-frequency supply drives a series circuit with R = 90 Ω, L = 1.2 H and C = 30 μF. Find the resonant angular frequency, the current at resonance, and the voltage across the inductor then.
19.
A step-down transformer changes 220 V to 11 V and has 1,600 turns in its primary. It supplies 1200 W to a load. Assuming it is ideal, find the number of secondary turns and the currents in both coils.
20.
A power station sends 600 kW along lines of total resistance 8 Ω at 22,000 V. Find the current and the power lost in the lines. What is the loss if the voltage is first stepped up 10 times?
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Physics

Answer key

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

QAnswerWorking
1(d) 110√2 VV0 = √2 V_rms
2(a) 171 ΩX_L = ωL
3(d) 1,250 ΩX_C = 1/(ωC)
4(c) 20 ΩZ = √(R2 + (X_L − X_C)2) = √(162 + 122)
5(a) I0/√2—
6(a) 3/5cos φ = R/Z = 6/10
7(a) 166.67 rad/sω0 = 1/√(LC) = 1/√(1.5 × 24 × 10−6)
8(a) Both A and R are true, and R is the correct explanation of A.A is true: ω0 = 1/√(LC) = 500 rad/s. R is true and explains A.
9I = 10 A; V_R = 120 V, V_L = 580 V, V_C = 490 VZ = 15 Ω
1020/29; 2,000 WP = V_rms I_rms cos φ = I2R
114,714.29 V; 15 AV_s/V_p = N_s/N_p; V_pI_p = V_sI_s
12500 rad/s; Q = 7.14ω0 = 1/√(LC); Q = ω0L/R
13500/π Ω; 4.4π × 10−1 AX_C = 1/(2πfC); I = V/X_C
141.68π × 102 Ω; 55/(42π) AX_L = 2πfL; I = V/X_L
152,618 WP_out = 85% of 220 × 14
16tan φ = 12/5, φ = tan−1(12/5); the current leads the voltagetan φ = |X_L − X_C|/R; Z = 26 Ω
17Z = 116 Ω; I = 4 A; cos φ = 20/29; P = 1,280 W; V_R = 320 V, V_L = 60 V, V_C = 396 VZ = √(R2 + (X_L − X_C)2)
18166.67 rad/s; 1 A; 200 V (the same across the capacitor)at resonance Z = R; V_L = I ω0L
1980 turns; secondary 109.09 A; primary 5.45 AN_s/N_p = V_s/V_p; P = VI on each side
2027.27 A; 5.95 kW; 0.0595 kW (100 times smaller)I = P/V; loss = I2R