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A sinusoidal voltage V = V0*sin(omega*t) is applied to a series LR circuit with L = 35 mH and R = 11 ohm. Given V_rms = 220 V, frequency f = omega/(2*pi) = 50 Hz, and pi = 22/7, find the amplitude (peak value) of the steady-state current.

1. About 20 A
2. About 14 A
3. About 31 A
4. About 28 A

Correct answer: About 20 A

Solution

X_L = 2*pi*f*L = 2*(22/7)*50*0.035 = 11 ohm. Impedance Z = sqrt(R² + X_L²) = sqrt(11² + 11²) = 11*sqrt(2) ohm. Peak voltage V0 = sqrt(2)*220 = 311 V. Current amplitude I0 = V0/Z = 311/(11*1.414) = 311/15.56 = 20 A. (The phase difference is tan(phi) = X_L/R = 1, so phi = 45 deg, with current lagging the voltage.)

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