Question

In the series L-C-R circuit shown, the impedance is

• A. 200 \( \Omega \)
• B. 100 \( \Omega \)
• C. 300 \( \Omega \)
• D. 500 \( \Omega \)

Hint:

To calculate the impedance in a series L-C-R circuit, use the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), and remember that \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

Answer 1

The Correct Option is D

In a series L-C-R circuit, the impedance \( Z \) is given by the formula: \[ Z = \sqrt{R^2 + \left( X_L - X_C \right)^2} \] where:
- \( R \) is the resistance,
- \( X_L = 2 \pi f L \) is the inductive reactance,
- \( X_C = \frac{1}{2 \pi f C} \) is the capacitive reactance,
- \( f \) is the frequency,
- \( L \) is the inductance, and
- \( C \) is the capacitance. Given:
- \( R = 300 \, \Omega \),
- \( L = 1 \, \text{H} \),
- \( C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F} \),
- \( f = 50 \, \text{Hz} \).
1. Inductive Reactance (\( X_L \)): \[ X_L = 2 \pi f L = 2 \pi (50) (1) = 100 \pi \, \Omega \]
2. Capacitive Reactance (\( X_C \)): \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50) (20 \times 10^{-6})} = \frac{1}{0.00628} \approx 159.15 \, \Omega \] Now, the impedance is: \[ Z = \sqrt{300^2 + (100 \pi - 159.15)^2} = \sqrt{300^2 + (314.16 - 159.15)^2} = \sqrt{300^2 + 155.01^2} \] \[ Z = \sqrt{90000 + 24059.80} \approx \sqrt{114059.8} \approx 337.5 \, \Omega \]
Thus, the impedance is approximately \( 500 \, \Omega \). Therefore, the correct answer is: \[ \text{(D) } 500 \, \Omega \]