Question:
A \(70\,\text{mH}\) inductor is connected to \(220\,\text{V},\ 50\,\text{Hz}\) AC supply. The rms value of the current in the circuit is:
A \(70\,\text{mH}\) inductor is connected to \(220\,\text{V},\ 50\,\text{Hz}\) AC supply. The rms value of the current in the circuit is:
Show Hint
For a pure inductor in AC circuit:
\[
X_L=2\pi fL
\]
and
\[
I_{\text{rms}}=\frac{V_{\text{rms}}}{X_L}.
\]
Updated On: Jun 24, 2026
- \(\dfrac{100}{\sqrt{2}\pi}\,\text{A}\)
- \(10\,\text{A}\)
- \(\dfrac{50}{\pi}\,\text{A}\)
- \(\dfrac{10\sqrt{2}}{\pi}\,\text{A}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the formula for inductive reactance.
Inductive reactance is \[ X_L=\omega L \] where \[ \omega=2\pi f \] Given, \[ f=50\,\text{Hz} \] and \[ L=70\,\text{mH}=70\times10^{-3}\,\text{H} \]
Step 2: Calculate angular frequency.
\[ \omega=2\pi(50) \] \[ \omega=100\pi\,\text{rad/s} \]
Step 3: Calculate inductive reactance.
\[ X_L=100\pi\times70\times10^{-3} \] \[ X_L=7\pi\,\Omega \] Approximating, \[ X_L\approx 22\,\Omega \]
Step 4: Use Ohm’s law for AC circuit.
For a pure inductor, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{X_L} \] Given, \[ V_{\text{rms}}=220\,\text{V} \] Thus, \[ I_{\text{rms}}=\frac{220}{7\pi} \] Using \[ \pi\approx \frac{22}{7}, \] we get \[ I_{\text{rms}}=\frac{220}{22} \] \[ I_{\text{rms}}=10\,\text{A} \]
Step 5: Final conclusion.
Hence, the rms current in the circuit is \[ \boxed{10\,\text{A}} \]
Inductive reactance is \[ X_L=\omega L \] where \[ \omega=2\pi f \] Given, \[ f=50\,\text{Hz} \] and \[ L=70\,\text{mH}=70\times10^{-3}\,\text{H} \]
Step 2: Calculate angular frequency.
\[ \omega=2\pi(50) \] \[ \omega=100\pi\,\text{rad/s} \]
Step 3: Calculate inductive reactance.
\[ X_L=100\pi\times70\times10^{-3} \] \[ X_L=7\pi\,\Omega \] Approximating, \[ X_L\approx 22\,\Omega \]
Step 4: Use Ohm’s law for AC circuit.
For a pure inductor, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{X_L} \] Given, \[ V_{\text{rms}}=220\,\text{V} \] Thus, \[ I_{\text{rms}}=\frac{220}{7\pi} \] Using \[ \pi\approx \frac{22}{7}, \] we get \[ I_{\text{rms}}=\frac{220}{22} \] \[ I_{\text{rms}}=10\,\text{A} \]
Step 5: Final conclusion.
Hence, the rms current in the circuit is \[ \boxed{10\,\text{A}} \]
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