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A series AC circuit has an inductor (20 mH), a capacitor (120 microF) and a resistor (60 ohm), driven by a 24 V, 50 Hz source. How much energy is dissipated in the circuit over 60 s?

1. 2.26 * 10³ J
2. 3.39 * 10³ J
3. 5.65 * 10² J
4. 5.17 * 10² J

Correct answer: 5.17 * 10² J

Solution

With omega = 100*pi, X_L approx 6.28 ohm and X_C approx 26.5 ohm, giving Z approx 63.3 ohm. The rms current is about 0.379 A, average power P = I²*R approx 8.62 W, and over 60 s the energy is about 517 J = 5.17*10² J.

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