Question

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

• A. 68 A
• B. 63 A
• C. 10 A
• D. 10 mA

Answer 1

The Correct Option is C

The voltage across the inductor is expressed as:

\[ V_L = I X_L, \]

where:
- \( I \) represents the circuit current.
- \( X_L \) denotes the inductive reactance, calculated as:

\[ X_L = \omega L = 2\pi f L. \]

Given values:
- \( V_L = 31.4 \, \text{V} \),
- \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \),
- Frequency \( f = 50 \, \text{Hz} \).

Step 1: Inductive Reactance Calculation

\[ X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 10 \times 10^{-3} = 3.14 \, \Omega. \]

Step 2: Current Calculation using \( V_L = I X_L \):

\[ 31.4 = I \times 3.14. \]

Solving for \( I \):

\[ I = \frac{31.4}{3.14} = 10 \, \text{A}. \]

Consequently, the circuit current is \( 10 \, \text{A} \).