Question

In an $ AC $ circuit $ V $ and $ I $ are given below, then find the power dissipated in the circuit $ V = 50 \,sin(50t)v $ $ I=50\,sin\left(50t \frac{\pi}{3}\right)\,mA $

• A. 0.625 W
• B. 1.25 W
• C. 2.50 W
• D. 5.0 W

Answer 1

Answer: (A)

As we know that, the power dissipated in an $A C$ circuit is
$ P=\frac{E_{0} i_{0} \cos \phi}{2}$
Here $E_{0}=50 \,V $
$i_{0}=50 \,mA =50 \times 10^{-3} A \text { and } \phi=\frac{\pi}{3}$
$( { i.e. } \cos \frac{\pi}{3}=\frac{1}{2}) $
$\therefore P=\frac{50 \times 50 \times 10^{-3} \times 1}{2 \times 2}=0.625 \,W$