Question

A series LCR circuit is subjected to an AC signal of \( 200 \, \text{V}, 50 \, \text{Hz} \). If the voltage across the inductor (\( L = 10 \, \text{mH} \)) is \( 31.4 \, \text{V} \), then the current in this circuit is:

• A. \( 68 \, \text{A} \)
• B. \( 63 \, \text{A} \)
• C. \( 10 \, \text{A} \)
• D. \( 10 \, \text{mA} \)

Hint:

In an LCR circuit, the current through the inductor can be calculated by dividing the voltage across the inductor by its inductive reactance, \( I = \frac{V_L}{X_L} \), where \( X_L = \omega L \).

Answer 1

The Correct Option is C

The voltage across an inductor in an AC circuit is given by: \[ V_L = I X_L, \] where \( X_L = \omega L \) is the inductive reactance and \( \omega = 2\pi f \) is the angular frequency of the AC supply. Step 1: Calculate the angular frequency \( \omega \): \[ \omega = 2\pi f = 2\pi \cdot 50 = 3.14 \times 100 = 314\, \text{rad/s}. \] Step 2: Calculate the inductive reactance \( X_L \): \[ X_L = \omega L = 314 \cdot 10 \times 10^{-3} = 3.14 \, \Omega. \] Step 3: Calculate the current \( I \) using \( V_L = I X_L \): \[ I = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10 \, \text{A}. \] Final Answer: \[ \boxed{10 \, \text{A}}. \]