Thank you for reporting, we will resolve it shortly
Q.
In an AC circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V=50 \sin (50 t) \vee I=50 \sin \left(50 t \frac{\pi}{3}\right) mA$
As we know that, the power dissipated in an AC
circuit is $P=\frac{E_{0} \dot{i} \cos \phi}{2}$
Here, $E_{0} =50\, V $
$ i_{0}=50 \,mA =50 \times 10^{-3} A \text { and } \phi=\frac{\pi}{3} $
$\left(\text { i.e. } \cos \frac{\pi}{3}=\frac{1}{2}\right) $
$ \therefore P =\frac{50 \times 50 \times 10^{-3} \times 1}{2 \times 2}=0.625 \,W $