Question

A 25 \(\Omega\) resistor and inductor in series with voltage \( V = 100\sin(100\pi t) \), impedance = 50 Ω. Find average power:

• A. 10 W
• B. 25 W
• C. 50 W
• D. 100 W

Hint:

In AC, average power is \( P = I^2 R \) using only resistive part.

Answer 1

The Correct Option is C

RMS voltage \( V_{\text{rms}} = \frac{100}{\sqrt{2}} \), impedance \( Z = 50 \, \Omega \)
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100/\sqrt{2}}{50} = \frac{2}{\sqrt{2}} = \sqrt{2} \, A \)
Power = \( I_{\text{rms}}^2 R = 2 \cdot 25 = 50 \, W \)