First slide

Power in a.c. circuit

Question

An inductor $20 m H,$ a capacitor $50 μ F$ and a resistor $40 Ω$ are connected in series across a source of emf $V = 10 \sin 340 t .$ The power loss in A.C. circuit is

Difficult

A

$0.76 W$
B

$0.89 W$
C

$0.46 W$
D

$0.67 W$

Solution

$Here, L = 20 mH = 20 \times 10^{- 3} H, C = 50 μ F = 50 \times 10^{- 6} F$
$\begin{array}{l} R = 40 Ω, V = 10 \sin 340 t = V_{0} \sin ω t \\ ω = 340 rad s^{- 1}, V_{0} = 10 V \\ X_{L} = ω L = 340 \times 20 \times 10^{- 3} = 6.8 Ω \end{array}$
$\begin{array}{l} X_{C} = \frac{1}{ω C} = \frac{1}{340 \times 50 \times 10^{- 6}} = \frac{10^{4}}{34 \times 5} = 58.82 Ω \\ Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} = \sqrt{(40)^{2} + (58.82 - 6.8)^{2}} \\ = \sqrt{(40)^{2} + (52.02)^{2}} = 65.62 Ω \end{array}$
The peak current in the circuit is
$I_{0} = \frac{V_{0}}{Z} = \frac{10}{65.62} A,$
$\cos ϕ = \frac{R}{Z} = (\frac{40}{65.62})$
Power loss in A.C. circuit
$\begin{array}{l} = V_{rms} I_{rms} \cos ϕ = \frac{1}{2} V_{0} I_{0} \cos ϕ \\ = \frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62} = 0.46 W \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App