quare) values of voltage and current, and $ \phi $ is the phase difference between them.

Given:

Voltage $ E(t) = 200 \sin (50t) $ volts

Current $ I(t) = 100 \sin \left(50t + \frac{\pi}{3}\right) $ mA

Convert the peak values to RMS values:

For a sinusoidal function, the RMS value is given by:

$$ V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} $$

$$ I_{\text{RMS}} = \frac{I_{\text{peak}}}{\sqrt{2}} $$

Therefore:

$$ V_{\text{RMS}} = \frac{200}{\sqrt{2}} $$

$$ I_{\text{RMS}} = \frac{100}{\sqrt{2}} $$

The phase difference $ \phi = \frac{\pi}{3} $.

Calculate the power factor:

The power factor is given by $ \cos \phi $.

Given $ \phi = \frac{\pi}{3} $, we use:

$$ \cos \frac{\pi}{3} = \frac{1}{2} $$

Calculate the average power:

Substituting the values into the power formula:

$$ P_{\text{avg}} = \left(\frac{200}{\sqrt{2}}\right)\left(\frac{100}{\sqrt{2}}\right)\left(\frac{1}{2}\right) $$

$$ P_{\text{avg}} = \left(\frac{200 \times 100}{2}\right) \times \frac{1}{2} $$

$$ P_{\text{avg}} = 10000 \times \frac{1}{2} $$

$$ P_{\text{avg}} = 5000 \, \text{mW} = 5 \, \text{W} $$

Thus, the power dissipated in the circuit is 5 watts.

Option D: 5 watt

Alternating Current

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