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In an AC circuit, an alternating voltage of amplitude U0 = 100 V is applied. The phase difference between current and voltage is phi = pi/4. The resistance is R = 100 ohm and the inductor is resistance-free. The angular frequency is omega = 100 rad/s. Which of the following statements are correct?

1. The amplitude of current is sqrt(2) A.
2. The average power dissipated is 50 W.
3. The inductance is 1 H.
4. The amplitude of current through the inductance is 0.5 A.

Correct answer: The inductance is 1 H.

Solution

With phi = pi/4, XL = R = 100 ohm, Z = 100*sqrt(2) ohm. I0 = 100/(100*sqrt(2)) = 1/sqrt(2) A. Average power = (1/2)*I0²*R = 25 W. Inductance L = 1 H (correct). Statement A says amplitude = sqrt(2) A which is WRONG (it's 1/sqrt(2)). Statement B says 50 W which is WRONG (it's 25 W). Statement C says L = 1 H which is CORRECT. Statement D says 0.5 A through inductor - in series circuit, same current flows, amplitude is 1/sqrt(2) ≈ 0.707 A, not 0.5 A.

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