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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 :
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- 0.51 W
- 0.67 W
- 0.76 W
- 0.89 W
Correct Option: A
Given: L = 20 mH; C = 50 µF; R = 40 ΩV = 10 sin 340 t
∴Vruns = 10 / √2
| Xc = | 1 | = | 1 | = 58.8 Ω |
| ωc | 340 x 50 x 10-6 |
XL = ωL = 340 × 20 × 10–3 = 6.8 Ω
Impedance, Z = √R² + (Xc - Xc)²
= √40² + (58.8 - 6.8)²
= √4304 Ω
Power loss in A.C. circuit,
| P = i²rms R = |  | x 89 |  | ² R | Z |
| = | | 10/√2 | | ² x 40 = | 50 x 40 | ≈ 0.51 W | ||
| √4304 | 4304 |
