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An inductor 20 mH, a capacitor 50 μF and a resistor 40Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is:

1. 0.51 W
2. 0.67 W
3. 0.76 W
4. 0.89 W

Correct answer: 0.67 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). Using ω = 340 rad/s, L = 20 mH, and C = 50 μF, we find XL = 6.8 Ω and XC = 58.8 Ω. Thus, Z = √(40² + (6.8 - 58.8)²) ≈ 57.2 Ω. The RMS current is I = Vrms/Z = (10/√2)/57.2 ≈ 0.124 A. Finally, P = I²R ≈ (0.124)² × 40 ≈ 0.67 W.

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