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

Answer: (C)

10 A

Answer 2

The voltage across the inductor is given by:

$V_L = I X_L,$

where:
- $I$ is the current in the circuit,
- $X_L$ is the inductive reactance given by:

$X_L = \omega L = 2\pi f L.$

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

Step 1: Calculate the Inductive Reactance

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

Step 2: Calculate the Current Using the formula $V_L = I X_L$:

$31.4 = I \times 3.14.$

Solving for $I$:

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

Therefore, the current in the circuit is $10 \, \text{A}$.