Question

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

पर्याय

• 0.79 W

• 0.43 W

• 2.74 W

• 1.13 W

Answer 1

An inductor 20 mH, a capacitor 100 µF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is 0.79 W.

Explanation:

Average power in impedance Z = \[\sqrt{R^{2}+\left(\frac{\omega L-1)}{\omega C}\right)^{2}}\]

ωL = X_L =(314) × (20 × 10^-3) = 6.280 Ω

Χ_C = 1/Cω = 10⁶/(100 × 314) = 31.84 Ω

Solving for impedance Z = 56.15 Ω

\[I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{m}}{(\sqrt{2}\times Z)}\] = 0.1259 A

Hence, power loss in the circuit = I² × R = 0.0159 × 50 = 0.79 W