Question:
If a resistor of resistance \(30\,\Omega\) and an inductor of reactance \(40\,\Omega\) are connected in series to an AC source of peak voltage \(200\sqrt2\,\mathrm{V}\), then the average power loss over a complete cycle is
If a resistor of resistance \(30\,\Omega\) and an inductor of reactance \(40\,\Omega\) are connected in series to an AC source of peak voltage \(200\sqrt2\,\mathrm{V}\), then the average power loss over a complete cycle is
Show Hint
For a series \(RL\) circuit,
\[
\boxed{
P=I_{\rm rms}^2R
=\frac{V_{\rm rms}^2R}{R^2+X_L^2}
}
\]
Updated On: Jul 15, 2026
- \(120\,\mathrm{W}\)
- \(960\,\mathrm{W}\)
- \(480\,\mathrm{W}\)
- \(240\,\mathrm{W}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Calculate the impedance.
\[
Z=\sqrt{R^2+X_L^2}
=\sqrt{30^2+40^2}
=50\,\Omega.
\]
Step 2: Find the rms voltage and current. Given, \[ V_0=200\sqrt2\,\mathrm{V}. \] Hence, \[ V_{\rm rms} =\frac{V_0}{\sqrt2} =200\,\mathrm{V}. \] Current, \[ I_{\rm rms} =\frac{V_{\rm rms}}{Z} =\frac{200}{50} =4\,\mathrm{A}. \]
Step 3: Calculate the average power. \[ P = I_{\rm rms}^2R = 4^2\times30 = 480\,\mathrm{W}. \] Hence, \[ \boxed{480\,\mathrm{W}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
Step 2: Find the rms voltage and current. Given, \[ V_0=200\sqrt2\,\mathrm{V}. \] Hence, \[ V_{\rm rms} =\frac{V_0}{\sqrt2} =200\,\mathrm{V}. \] Current, \[ I_{\rm rms} =\frac{V_{\rm rms}}{Z} =\frac{200}{50} =4\,\mathrm{A}. \]
Step 3: Calculate the average power. \[ P = I_{\rm rms}^2R = 4^2\times30 = 480\,\mathrm{W}. \] Hence, \[ \boxed{480\,\mathrm{W}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
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