The current and voltage in an a. c. circuit are given as
$I = 50 \sin (50t) mA$
$V = 100 \sin (50t+π/3) volt$
The power dissipated in the circuit is

1.25 W

The correct answer is Option (2) → 1.25 W

Given:

$ I = 50\sin(50t)\,\text{mA} = 50 \times 10^{-3}\sin(50t)\,\text{A} $

$ V = 100\sin(50t + \pi/3)\,\text{V} $

rms values:

$ I_{rms} = \frac{I_{0}}{\sqrt{2}} = \frac{50 \times 10^{-3}}{\sqrt{2}} = \frac{0.05}{\sqrt{2}}\,\text{A} $

$ V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{100}{\sqrt{2}}\,\text{V} $

Phase difference: $ \phi = \pi/3 $

Average power dissipated:

$ P = V_{rms} I_{rms} \cos\phi $

$ P = \left(\frac{100}{\sqrt{2}}\right)\left(\frac{0.05}{\sqrt{2}}\right)\cos\left(\frac{\pi}{3}\right) $

$ P = \left(\frac{100 \times 0.05}{2}\right)\left(\frac{1}{2}\right) $

$ P = (2.5)(0.5) = 1.25 \,\text{W} $

Power dissipated = 1.25 W