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 determined by the formula: \[V_L = I X_L,\] where \( X_L = \omega L \) represents the inductive reactance and \( \omega = 2\pi f \) is the angular frequency of the AC supply. Step 1: Determine the angular frequency \( \omega \): \[\omega = 2\pi f = 2\pi \cdot 50 = 3.14 \times 100 = 314\, \text{rad/s}.\] Step 2: Compute 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 the relationship \( V_L = I X_L \): \[I = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10 \, \text{A}.\] Final Answer: \[\boxed{10 \, \text{A}}.\]