Alternating Current
ICSE · Class 12 · Physics
Most important questions from Alternating Current for ICSE Class 12 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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A 200 V sinusoidal voltage is applied to a series L-R circuit with resistance 10 ohm and inductance 0.8 H. The angular frequency is 300 rad/s. What is the peak current in the circuit?
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0.83 A
First find XL = omega L = 300 x 0.8 = 240 ohm. Then Z = sqrt(10^2 + 240^2) = 240.2 ohm approximately. The peak voltage is V0 = 200 V. So I0 = V0/Z = 200/240.2 = 0.83 A.
A 100 microfarad capacitor is in series with a 40 ohm resistor and connected to a 110 V, 60 Hz supply. What is the maximum current in the circuit?
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3.23 A
Find XC = 1/(2 pi f C) = 1/(2 x 3.14 x 60 x 100 x 10^-6) = 26.5 ohm. Then Z = sqrt(40^2 + 26.5^2) = 48 ohm approximately. Peak voltage is V0 = 110 x 1.414 = 155.5 V. So I0 = V0/Z = 155.5/48 = 3.23 A.
A 230 V, 50 Hz source is connected to a series circuit containing a 80 mH inductor and a 60 microfarad capacitor with negligible resistance. What is the current amplitude in the circuit?
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11.6 A
For a pure L-C circuit, Z = |XC - XL|. Here XL = 2 pi f L = 2 x 3.14 x 50 x 0.08 = 25.1 ohm. XC = 1/(2 pi f C) = 1/(2 x 3.14 x 50 x 60 x 10^-6) = 53.1 ohm. So Z = 53.1 - 25.1 = 28.0 ohm. Peak voltage V0 = 230 x 1.414 = 325.2 V. Therefore I0 = V0/Z = 325.2/28.0 = 11.6 A.
In a series L-C-R circuit, R = 100 ohm, L = 0.12 H, C = 480 nF, and the rms supply voltage is 230 V. What is the resonant frequency?
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663 Hz
Resonant frequency is f0 = (1/2 pi) sqrt(1/LC). Substituting L = 0.12 H and C = 480 x 10^-9 F gives f0 = (1/2 x 3.14) x sqrt(1/(0.12 x 480 x 10^-9)). This evaluates to about 663 Hz.
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