Question:
An inductor of reactance \(1\,\Omega\) and a resistor of resistance \(3\,\Omega\) are connected in series to the terminals of \(10\,\text{V}\) (rms) ac source. The power dissipated in the circuit is
An inductor of reactance \(1\,\Omega\) and a resistor of resistance \(3\,\Omega\) are connected in series to the terminals of \(10\,\text{V}\) (rms) ac source. The power dissipated in the circuit is
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In an \(RL\) circuit, only the resistor dissipates real power:
\[
P=I_{\text{rms}}^2R.
\]
The inductor stores and releases energy but does not dissipate average power.
Updated On: Jun 18, 2026
- \(33.3\,\text{W}\)
- \(30\,\text{W}\)
- \(31.6\,\text{W}\)
- \(20\,\text{W}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the given values.
Resistance is \[ R=3\,\Omega \] Inductive reactance is \[ X_L=1\,\Omega \] RMS voltage is \[ V_{\text{rms}}=10\,\text{V} \]
Step 2: Find the impedance of the series \(RL\) circuit.
For a series \(RL\) circuit, \[ Z=\sqrt{R^2+X_L^2} \] \[ Z=\sqrt{3^2+1^2} \] \[ Z=\sqrt{10}\,\Omega \]
Step 3: Find the RMS current.
\[ I_{\text{rms}}=\frac{V_{\text{rms}}}{Z} \] \[ I_{\text{rms}}=\frac{10}{\sqrt{10}} \] \[ I_{\text{rms}}=\sqrt{10}\,\text{A} \]
Step 4: Calculate power dissipated.
Power is dissipated only in the resistor.
Therefore, \[ P=I_{\text{rms}}^2R \] \[ P=(\sqrt{10})^2\times 3 \] \[ P=10\times 3 \] \[ P=30\,\text{W} \]
Step 5: Final conclusion.
Therefore, the power dissipated in the circuit is \[ \boxed{30\,\text{W}} \]
Resistance is \[ R=3\,\Omega \] Inductive reactance is \[ X_L=1\,\Omega \] RMS voltage is \[ V_{\text{rms}}=10\,\text{V} \]
Step 2: Find the impedance of the series \(RL\) circuit.
For a series \(RL\) circuit, \[ Z=\sqrt{R^2+X_L^2} \] \[ Z=\sqrt{3^2+1^2} \] \[ Z=\sqrt{10}\,\Omega \]
Step 3: Find the RMS current.
\[ I_{\text{rms}}=\frac{V_{\text{rms}}}{Z} \] \[ I_{\text{rms}}=\frac{10}{\sqrt{10}} \] \[ I_{\text{rms}}=\sqrt{10}\,\text{A} \]
Step 4: Calculate power dissipated.
Power is dissipated only in the resistor.
Therefore, \[ P=I_{\text{rms}}^2R \] \[ P=(\sqrt{10})^2\times 3 \] \[ P=10\times 3 \] \[ P=30\,\text{W} \]
Step 5: Final conclusion.
Therefore, the power dissipated in the circuit is \[ \boxed{30\,\text{W}} \]
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