Question:
An RLC circuit consists of a \(150\,\Omega\) resistor, \(20\,\mu\text{F}\) capacitor and a \(500\,\text{mH}\) inductor connected in series with a \(100\,\text{V}\) AC supply. The angular frequency of the supply voltage is \(400\,\text{rad s}^{-1}\). The phase angle between current and the applied voltage is
An RLC circuit consists of a \(150\,\Omega\) resistor, \(20\,\mu\text{F}\) capacitor and a \(500\,\text{mH}\) inductor connected in series with a \(100\,\text{V}\) AC supply. The angular frequency of the supply voltage is \(400\,\text{rad s}^{-1}\). The phase angle between current and the applied voltage is
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For a series RLC circuit,
\[
\tan\phi=\frac{X_L-X_C}{R}
\]
If \(X_L\gt X_C\), the circuit behaves inductively and current lags voltage.
Updated On: Jun 22, 2026
- \(\tan^{-1}(0.8)\)
- \(\tan^{-1}(0.25)\)
- \(\tan^{-1}(0.6)\)
- \(\tan^{-1}(0.5)\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the expression for phase angle in a series RLC circuit.
The phase angle is given by \[ \tan\phi=\frac{X_L-X_C}{R} \] where, \[ X_L=\omega L \] is inductive reactance and \[ X_C=\frac{1}{\omega C} \] is capacitive reactance.
Step 2: Calculate inductive reactance.
Given, \[ \omega=400\,\text{rad s}^{-1} \] \[ L=500\,\text{mH}=0.5\,\text{H} \] Therefore, \[ X_L=\omega L \] \[ X_L=400\times0.5 \] \[ X_L=200\,\Omega \]
Step 3: Calculate capacitive reactance.
Given, \[ C=20\,\mu\text{F}=20\times10^{-6}\,\text{F} \] Thus, \[ X_C=\frac{1}{\omega C} \] \[ X_C=\frac{1}{400\times20\times10^{-6}} \] \[ X_C=\frac{1}{8\times10^{-3}} \] \[ X_C=125\,\Omega \]
Step 4: Calculate phase angle.
Given resistance, \[ R=150\,\Omega \] Using \[ \tan\phi=\frac{X_L-X_C}{R} \] \[ \tan\phi=\frac{200-125}{150} \] \[ \tan\phi=\frac{75}{150} \] \[ \tan\phi=0.5 \] Hence, \[ \phi=\tan^{-1}(0.5) \]
Step 5: Final conclusion.
Therefore, the phase angle between current and applied voltage is \[ \boxed{\tan^{-1}(0.5)} \]
The phase angle is given by \[ \tan\phi=\frac{X_L-X_C}{R} \] where, \[ X_L=\omega L \] is inductive reactance and \[ X_C=\frac{1}{\omega C} \] is capacitive reactance.
Step 2: Calculate inductive reactance.
Given, \[ \omega=400\,\text{rad s}^{-1} \] \[ L=500\,\text{mH}=0.5\,\text{H} \] Therefore, \[ X_L=\omega L \] \[ X_L=400\times0.5 \] \[ X_L=200\,\Omega \]
Step 3: Calculate capacitive reactance.
Given, \[ C=20\,\mu\text{F}=20\times10^{-6}\,\text{F} \] Thus, \[ X_C=\frac{1}{\omega C} \] \[ X_C=\frac{1}{400\times20\times10^{-6}} \] \[ X_C=\frac{1}{8\times10^{-3}} \] \[ X_C=125\,\Omega \]
Step 4: Calculate phase angle.
Given resistance, \[ R=150\,\Omega \] Using \[ \tan\phi=\frac{X_L-X_C}{R} \] \[ \tan\phi=\frac{200-125}{150} \] \[ \tan\phi=\frac{75}{150} \] \[ \tan\phi=0.5 \] Hence, \[ \phi=\tan^{-1}(0.5) \]
Step 5: Final conclusion.
Therefore, the phase angle between current and applied voltage is \[ \boxed{\tan^{-1}(0.5)} \]
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