Question

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance $R = 3 \text{ k} \Omega$, an inductor of inductive reactance $X_L = 250 \pi \text{ } \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (take $\pi^2 = 10$)

• A. 400 $\mu$F
• B. 40 $\mu$F
• C. 25 $\mu$F
• D. 4 $\mu$F

Hint:

Maximum power always implies resonance in LCR circuits, meaning the reactive components cancel each other out ($Z = R$).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Average power in a series LCR circuit is maximized during resonance. At resonance, the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).
Step 2: Key Formula or Approach:
Resonance condition: \(X_L = X_C = \frac{1}{2\pi f C}\).
Step 3: Detailed Explanation:
Given: \(X_L = 250\pi \text{ } \Omega\), \(f = 50 \text{ Hz}\).
At resonance:
\[ 250\pi = \frac{1}{2\pi (50) C} \]
\[ 250\pi = \frac{1}{100\pi C} \]
\[ C = \frac{1}{25000\pi^2} \]
Using \(\pi^2 = 10\):
\[ C = \frac{1}{250000} \text{ F} = \frac{1}{2.5 \times 10^5} \text{ F} \]
\[ C = 0.4 \times 10^{-5} \text{ F} = 4 \times 10^{-6} \text{ F} = 4 \text{ }\mu\text{F} \]
Step 4: Final Answer:
The capacitance value should be 4 $\mu$F.