CUET · PHYSICS · PYQ PAPER 2025
An inductor of 500 mH is in series with a resistance and a variable capacitor connected to a source of frequency 0.4 kHz. The value of capacitance of the capacitor to get a maximum current will be
- A \( 2.3 \mu F \)
- B \( 0.32 \mu F \)
- C \( 63 \mu F \)
- D \( 0.62 \mu F \)
Answer & Solution
Correct Answer
(B) \( 0.32 \mu F \)
Step-by-step Solution
Detailed explanation
For maximum current (resonance): \( X_L = X_C \) \( 2\pi f L = \frac{1}{2\pi f C} \) \( C = \frac{1}{(2\pi f)^2 L} \) \( C = \frac{1}{(2\pi \times 0.4 \times 10^3 \, \text{Hz})^2 \times 0.5 \, \text{H}} \) \( C = \frac{1}{(800\pi)^2 \times 0.5} \, \text{F} \)…
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