Alternating Current
ICSE · Class 12 · Physics
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A series L-C-R circuit has resistance 20 ohm, inductance 1.5 H, and capacitance 35 microfarad. When the supply frequency equals the resonant frequency, what is the average power transferred to the circuit if the rms supply voltage is 200 V?
A 220 V, 50 Hz AC supply is applied to a 44 mH inductor. What is the rms current in the circuit?
A 110 V, 60 Hz AC supply is connected to a 60 microfarad capacitor. What is the rms current in the circuit?
A coil has inductance 0.50 H and resistance 100 ohm. It is connected to a 240 V, 50 Hz AC supply. What is the maximum current in the coil?
Sample Questions
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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