Question

In a AC circuit the voltage and current are described by $V=200\sin (319t-\frac{\pi }{6})volts$ and $i=50\sin (314t+\frac{\pi }{6})mA$ respectively. The average power dissipated in the circuit is

• A. 2.5 watts
• B. 5.0 watts
• C. 10.0 watts
• D. 50.0 watts

Answer 1

Answer: (A)

2.5 watts

Comparing the current with $i=i_o\sin (wt+{\delta }_1)$
we get $i_o=50mA$ and ${\delta }_1=\frac{\pi }{6}$
Comparing the voltage with $V=V_o\sin (wt+{\delta }_2)$
we get $V_o=200$volts and ${\delta }_2=\frac{-\pi }{6}$
Phase difference between current and voltage $\varphi ={\delta }_1-{\delta }_2=\frac{\pi }{3}$
Average power dissipated in the circuit $P_{av}=\frac{V_o{i}_o}{2}\cos \varphi$
$⟹P_{av}=\frac{200\times 50\times {10}^{-3}}{2}\times \cos \frac{\pi }{3}=2.5$ watts