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A pure inductor of inductance \( 0.1 \, \text{H} \) is connected to an AC source (of rms voltage 220 V and angular frequency 300 Hz). The rms current is

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For an AC circuit with a pure inductor, the rms current depends on the voltage and the inductive reactance \( X_L \), which is related to the frequency and inductance of the circuit.
Updated On: Apr 19, 2026
  • \( \frac{3}{22} \, \text{A} \)
  • \( \frac{22}{3} \, \text{A} \)
  • \( \frac{11}{150} \, \text{A} \)
  • \( \frac{150}{11} \, \text{A} \)
  • \( \frac{22}{6\pi} \, \text{A} \)
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The Correct Option is

Solution and Explanation

Step 1: Understanding the Concept:
In an AC circuit with a pure inductor, the opposition to the current flow is called inductive reactance (\(X_{L}\)).
The root mean square (rms) current can be calculated using Ohm's law applied to AC circuits.
Step 2: Key Formula or Approach:
The inductive reactance is found using the formula \( X_{L} = \omega L \), where \(\omega\) is the angular frequency.
The rms current is then calculated using the relation \( I_{\text{rms}} = \frac{V_{\text{rms}}}{X_{L}} \).
Step 3: Detailed Explanation:
We are given the inductance \( L = 0.1 \text{ H} \), and the rms voltage \( V_{\text{rms}} = 220 \text{ V} \).
The problem states the angular frequency is \(300 \text{ Hz}\), which indicates \(\omega = 300 \text{ rad/s}\) based on standard problem conventions despite the unit.
First, we find the inductive reactance:
\[ X_{L} = \omega L = 300 \times 0.1 = 30 \, \Omega \] Next, we substitute this into the current formula:
\[ I_{\text{rms}} = \frac{220}{30} \] \[ I_{\text{rms}} = \frac{22}{3} \text{ A} \] Step 4: Final Answer:
The rms current flowing through the circuit is \(\frac{22}{3} \text{ A}\).
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