Question

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

पर्याय

• 0.51 W

• 0.67 W

• 0.76 W

• 0.89 W

Answer 1

An inductor 20 mH, a capacitor 50 µF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in AC circuit is 0.51 W.

Explanation:

X_L = ωL = 340 × 20 × 10^-3 = 68 × 10^-1 = 6.8 Ω

Now, X_C = \[\frac{1}{\omega C}=\frac{1}{(240\times50\times10^{-6})}\] = \[\frac {10000}{170}\] = 58.8 Ω

Now, Z = \[\sqrt{40^2+\left(58.8-6.8\right)^2}\] = \[\sqrt{40^2+(52)^2}\] = 65.6 Ω

Now, P = V²_rms \[\frac {R}{Z^2}\]

P = \[\left(\frac{10}{\sqrt{2}^2}\right)\times\frac{40}{(65.6\times65.6)}\]

P = 100 × \[\frac {40}{(2×4303.7)}\] = 0.47 = 0.51 W