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 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 \( 10 \, \text{A} \).

Answer 2

Step 1: Given data.
Applied voltage, V = 200 V
Frequency, f = 50 Hz
Inductance, L = 10 mH = 10 × 10⁻³ H
Voltage across the inductor, V_L = 31.4 V

Step 2: Formula for inductive reactance (X_L).
The inductive reactance is given by:
\[ X_L = 2\pi f L \]
Substitute the given values:
\[ X_L = 2 \times 3.14 \times 50 \times 10 \times 10^{-3} = 2 \times 3.14 \times 0.5 = 3.14 \, \Omega \]

Step 3: Find the current in the circuit.
The voltage across the inductor is related to current by:
\[ V_L = I X_L \]
\[ I = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10 \, \text{A} \]

Step 4: Final Answer.
The current in the circuit is:
\[ \boxed{10 \, \text{A}} \]

Final Answer: 10 A