Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
Try each problem first, then open the solution to check your working.
The peak value of the mains voltage is 311 V. Find the rms value.
Given: V₀ = 311 V
Use: V_rms = V₀ / √2
V_rms = 311 / 1.414 = 220 V
Answer: 220 V
The rms current in a circuit is 5 A. Find the peak current.
Given: I_rms = 5 A
Use: i₀ = √2 I_rms
i₀ = 1.414 × 5 = 7.07 A
Answer: 7.07 A
A 100 W bulb is used on a 220 V mains supply. Find its resistance and rms current.
Given: P = 100 W; V = 220 V
Use: R = V²/P; I = P/V
R = 220² / 100 = 484 Ω
I = 100 / 220 = 0.455 A
Answer: 484 Ω; 0.45 A
A current of peak value 3 A passes through a 10 Ω resistor. Find the average power.
Given: i₀ = 3 A; R = 10 Ω
Use: P = ½ i₀² R
P = ½ × 9 × 10 = 45 W
Answer: 45 W
A voltage is v = 300 sin 100πt. Find its frequency, peak and rms value.
Given: v = 300 sin 100πt
Use: Compare with v₀ sin 2πft
2πf = 100π, so f = 50 Hz
Peak = 300 V; rms = 300/√2 = 212 V
Answer: 50 Hz; 300 V; 212 V
How long after the voltage passes through zero does a 50 Hz ac reach its peak?
Given: f = 50 Hz
Use: A quarter of a period
T = 1/50 = 20 ms
t = T/4 = 5 ms
Answer: 5 ms
A 20 mH inductor is connected to a 100 V, 50 Hz supply. Find its reactance and the rms current.
Given: L = 20 mH; V = 100 V; f = 50 Hz
Use: X_L = 2π f L; I = V / X_L
X_L = 2π × 50 × 0.02 = 6.28 Ω
I = 100 / 6.28 = 15.9 A
Answer: 6.28 Ω; 15.9 A
A 10 μF capacitor is connected to a 220 V, 50 Hz supply. Find its reactance and the rms current.
Given: C = 10 × 10⁻⁶ F; V = 220 V; f = 50 Hz
Use: X_C = 1 / 2π f C; I = V / X_C
X_C = 1 / (2π × 50 × 10⁻⁵) = 318 Ω
I = 220 / 318 = 0.69 A
Answer: 318 Ω; 0.69 A
A pure inductor has L = 0.2 H. Find its reactance at 500 Hz.
Given: L = 0.2 H; f = 500 Hz
Use: X_L = 2π f L
X_L = 2π × 500 × 0.2 = 628 Ω
Answer: 628 Ω
Find the reactance of a 0.1 μF capacitor at 1 kHz.
Given: C = 10⁻⁷ F; f = 1000 Hz
Use: X_C = 1 / 2π f C
X_C = 1 / (2π × 1000 × 10⁻⁷) = 1592 Ω
Answer: 1.59 kΩ
A circuit has X_L = 50 Ω and X_C = 80 Ω at 50 Hz. What are they at 100 Hz?
Given: X_L = 50 Ω; X_C = 80 Ω
Use: X_L ∝ f; X_C ∝ 1/f
X_L = 100 Ω
X_C = 40 Ω
Answer: X_L = 100 Ω; X_C = 40 Ω
A series circuit has R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω, connected to 100 V (rms). Find Z, I and the phase angle.
Given: R = 30; X_L = 80; X_C = 40 Ω; V = 100 V
Use: Z = √[R² + (X_L − X_C)²]; tan φ = (X_L − X_C)/R
Z = √(900 + 1600) = 50 Ω
I = 100 / 50 = 2 A
tan φ = 40/30 = 1.33, φ = 53° (current lags)
Answer: 50 Ω; 2 A; 53°
For the circuit above find the power factor and the average power.
Given: R = 30 Ω; Z = 50 Ω; I = 2 A
Use: cos φ = R/Z; P = V I cos φ
cos φ = 30/50 = 0.6
P = 100 × 2 × 0.6 = 120 W (= I²R = 4 × 30 ✓)
Answer: 0.6; 120 W
Find the resonant angular frequency and the frequency of a series circuit with L = 1 H and C = 100 μF.
Given: L = 1 H; C = 10⁻⁴ F
Use: ω₀ = 1/√(LC)
ω₀ = 1/√(10⁻⁴) = 100 rad/s
f₀ = 100 / 2π = 15.9 Hz
Answer: 100 rad/s; 15.9 Hz
A series circuit has R = 10 Ω, L = 1 H, C = 100 μF and is connected to 220 V (rms) at resonance. Find the current and the voltage across the inductor.
Given: R = 10 Ω; ω₀ = 100 rad/s; V = 220 V
Use: At resonance Z = R
I = 220 / 10 = 22 A
X_L = ω₀ L = 100 Ω
V_L = I X_L = 22 × 100 = 2200 V (equal to V_C and opposite)
Answer: 22 A; 2200 V
For R = 10 Ω, L = 1 H, C = 100 μF find the Q factor and the bandwidth.
Given: R = 10 Ω; L = 1 H; ω₀ = 100 rad/s
Use: Q = ω₀ L / R; Δω = R / L
Q = 100 × 1 / 10 = 10
Δω = 10 / 1 = 10 rad/s
Answer: Q = 10; bandwidth 10 rad/s
A series LCR circuit has R = 20 Ω and a 200 V (rms) supply at resonance. Find the average power.
Given: R = 20 Ω; V = 200 V
Use: P = V² / R
I = 200 / 20 = 10 A
P = I² R = 100 × 20 = 2000 W
Answer: 2000 W
Find the angular frequency of oscillation of a circuit with L = 10 mH and C = 1 μF.
Given: L = 10⁻² H; C = 10⁻⁶ F
Use: ω = 1/√(LC)
ω = 1 / √(10⁻⁸) = 10⁴ rad/s
Answer: 10⁴ rad/s
A transformer steps 2200 V down to 220 V. The primary has 1000 turns. Find the number of turns in the secondary.
Given: V_p = 2200 V; V_s = 220 V; N_p = 1000
Use: N_s = N_p V_s / V_p
N_s = 1000 × 220 / 2200 = 100
Answer: 100 turns
An ideal transformer takes 2 A at 220 V and delivers power at 22 V. Find the secondary current.
Given: I_p = 2 A; V_p = 220 V; V_s = 22 V
Use: V_p I_p = V_s I_s
I_s = 220 × 2 / 22 = 20 A
Answer: 20 A
A transformer takes 1000 W from the mains and delivers 900 W. Find its efficiency and the power lost.
Given: P_in = 1000 W; P_out = 900 W
Use: Efficiency = output / input
Efficiency = 900/1000 = 90 %
Loss = 100 W
Answer: 90 %; 100 W
100 kW is sent through a line of resistance 0.5 Ω at (a) 2200 V and (b) 22 000 V. Find the power lost in the line in each case.
Given: P = 10⁵ W; R = 0.5 Ω
Use: I = P/V; loss = I² R
(a) I = 10⁵ / 2200 = 45.5 A; loss = 45.5² × 0.5 = 1033 W
(b) I = 10⁵ / 22000 = 4.55 A; loss = 4.55² × 0.5 = 10.3 W
Answer: (a) about 1.03 kW; (b) about 10 W (100 times less)
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(a) I = V / R = 220 / 100 = 2.20 A.
(b) P = V I = 220 × 2.20 = 484 W.
(a) V_rms = V₀ / √2 = 300 / 1.414 = 212.1 V.
(b) i₀ = √2 I_rms = 1.414 × 10 = 14.1 A.
X_L = 2π f L = 2π × 50 × 44 × 10⁻³ = 13.82 Ω.
I = V / X_L = 220 / 13.82 = 15.9 A.
X_C = 1 / (2π f C) = 1 / (2π × 60 × 60 × 10⁻⁶) = 44.2 Ω.
I = V / X_C = 110 / 44.2 = 2.49 A.
Zero in both cases. For a pure inductor the current lags the voltage by 90° and for a pure capacitor it leads by 90°, so the power factor cos φ = 0 and P = V I cos φ = 0.
During one quarter of the cycle the element absorbs energy from the source (the inductor stores it in its magnetic field, the capacitor in its electric field) and during the next quarter it returns the same energy. No energy is dissipated.
ω = 1 / √(LC) = 1 / √(27 × 10⁻³ × 30 × 10⁻⁶) = 1 / √(8.1 × 10⁻⁷) = 1.11 × 10³ rad s⁻¹.
At the natural frequency (resonance) X_L = X_C, so Z = R = 20 Ω and cos φ = 1.
I_rms = V / R = 200 / 20 = 10 A. Average power P = I² R = 100 × 20 = 2000 W (= V²/R = 40000/20).
(a) ω₀ = 1 / √(LC) = 1 / √(5.0 × 80 × 10⁻⁶) = 1 / √(4 × 10⁻⁴) = 50 rad s⁻¹, so ν₀ = ω₀ / 2π = 7.96 Hz.
(b) At resonance Z = R = 40 Ω. The amplitude of current i₀ = V₀ / R = 230 √2 / 40 = 8.13 A (rms value 5.75 A).
(c) I_rms = 230 / 40 = 5.75 A. Across R: V_R = I R = 5.75 × 40 = 230 V. X_L = ω₀ L = 50 × 5.0 = 250 Ω and X_C = 1 / ω₀ C = 1 / (50 × 80 × 10⁻⁶) = 250 Ω. So V_L = 5.75 × 250 = 1437.5 V and V_C = 5.75 × 250 = 1437.5 V.
V_L leads the current by 90° and V_C lags it by 90°, so they are 180° out of phase and equal in magnitude. The drop across the LC combination is V_L − V_C = 1437.5 − 1437.5 = 0. The whole source voltage appears across R.