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An inductor 20 mH, a capacitor 100 μF and a resistor 50Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is:

1. 0.79 W
2. 0.43 W
3. 1.13 W
4. 2.74 W

Correct answer: 0.79 W

Solution

The power loss in an AC circuit is given by P = I²R, where I is the RMS current. First, calculate the impedance Z = √(R² + (XL - XC)²), where XL = ωL and XC = 1/(ωC). Here, ω = 314 rad/s, L = 20 mH, and C = 100 μF. After finding Z, calculate I = Vrms/Z, where Vrms = Vpeak/√2. Finally, substitute I and R into P = I²R to get 0.79 W.

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