CUET · PHYSICS · PYQ PAPER 2025
A \( 14.0 \, \mu\text{F} \) capacitor is connected to a 220 V, 50 Hz source. The value of peak current in the circuit is (Given: \( \pi = \frac{22}{7} \) and \( \sqrt{2} = 1.414 \)):
- A 1.57 A
- B 1.47 A
- C 1.37 A
- D 1.27 A
Answer & Solution
Correct Answer
(C) 1.37 A
Step-by-step Solution
Detailed explanation
\( I_P = 2 \pi f C V_{\text{rms}} \sqrt{2} \) \( I_P = 2 \times \frac{22}{7} \times 50 \, \text{Hz} \times 14.0 \times 10^{-6} \, \text{F} \times 220 \, \text{V} \times 1.414 \) \( I_P \approx 1.37 \, \text{A} \)
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