Explanation:
C Given,
$\mathrm{V}=50 \sin (50 \mathrm{t}) \mathrm{V}$
$\mathrm{I}=100 \times 10^{-3} \sin \left(50 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{A}$
Then, $\mathrm{V}_{0}=50, \mathrm{I}_{0}=100 \times 10^{-3} \mathrm{~A}, \phi=\frac{\pi}{3}$
We know that, power dissipated in the circuit,
$\mathrm{P}=\frac{\mathrm{I}_{0} \mathrm{~V}_{0}}{2} \times \cos \phi$
$\mathrm{P}=\frac{50 \times 100 \times 10^{-3}}{2} \times \cos \frac{\pi}{3}$
$\mathrm{P}=1.25 \mathrm{~W}$