Question

A series LCR circuit has \( R = 200 \, \Omega \), \( L = 663 \, \text{mH} \), and \( C = 265 \, \mu\text{F} \). The applied alternating voltage has an amplitude of 50 V and a frequency of 60 Hz so that \( X_L = 250 \, \Omega \) and \( X_C = 100 \, \Omega \). The peak current is

• A. 0.33 A
• B. 0.20 A
• C. 0.50 A
• D. 0.25 A

Hint:

In LCR circuits, the impedance depends on the values of resistance, inductive reactance, and capacitive reactance. Always use the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \) to calculate total impedance.

Answer 1

The Correct Option is B

Step 1: Understanding the LCR circuit.
In a series LCR circuit, the total impedance \( Z \) is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Where: - \( R = 200 \, \Omega \) - \( X_L = 250 \, \Omega \) - \( X_C = 100 \, \Omega \) Substituting the values into the equation: \[ Z = \sqrt{200^2 + (250 - 100)^2} = \sqrt{200^2 + 150^2} \] Step 2: Calculating the impedance.
Now, calculate the total impedance \( Z \): \[ Z = \sqrt{40000 + 22500} = \sqrt{62500} = 250 \, \Omega \] Step 3: Calculating the peak current.
The peak current \( I_0 \) is given by: \[ I_0 = \frac{V_0}{Z} \] Where \( V_0 = 50 \, \text{V} \). Substituting the values: \[ I_0 = \frac{50}{250} = 0.20 \, \text{A} \] Thus, the correct answer is (B) 0.20 A.