Question:
A pure inductor of inductance 0.1 H is connected to an AC source (of rms voltage) 220 V and angular frequency of 300 Hz. The rms current is
A pure inductor of inductance 0.1 H is connected to an AC source (of rms voltage) 220 V and angular frequency of 300 Hz. The rms current is
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If frequency was given in Hz (f), we would use \( \omega = 2\pi f \). Here, "angular frequency" refers to \( \omega \).
Updated On: May 1, 2026
- \( \frac{3}{22} \text{ A} \)
- \( \frac{22}{3} \text{ A} \)
- \( \frac{11}{150} \text{ A} \)
- \( \frac{150}{11} \text{ A} \)
- \( \frac{11}{3\pi} \text{ A} \)
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The Correct Option is B
Solution and Explanation
Concept:
For a pure inductor, \( V_{rms} = I_{rms} X_L \), where \( X_L = \omega L \).
Step 1: {Extract values from Screenshot 2026-04-29 225540.jpg.}
\( L = 0.1 \text{ H} \), \( V_{rms} = 220 \text{ V} \), \( \omega = 300 \text{ rad/s} \).
Step 2: {Calculate inductive reactance.}
\[ X_L = 300 \times 0.1 = 30 \, \Omega \]
Step 3: {Solve for RMS current.}
\[ I_{rms} = \frac{220}{30} = \frac{22}{3} \text{ A} \]
Step 1: {Extract values from Screenshot 2026-04-29 225540.jpg.}
\( L = 0.1 \text{ H} \), \( V_{rms} = 220 \text{ V} \), \( \omega = 300 \text{ rad/s} \).
Step 2: {Calculate inductive reactance.}
\[ X_L = 300 \times 0.1 = 30 \, \Omega \]
Step 3: {Solve for RMS current.}
\[ I_{rms} = \frac{220}{30} = \frac{22}{3} \text{ A} \]
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