A 10 μ F capacitor is connected to a 210 V , 50 Hz source as shown in figure. The peak current in the circuit is nearly ( π=3.14 ):

Physics · Alternating Current

NEET (UG) 2024 — Question 37

A $10 \mu \mathrm{~F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly ( $\pi=3.14$ ):

Option A:
0.58 A
Option B:
1.20 A
Option C:
0.93 A
Correct
Option D:
0.35 A

Answer: C

Step-by-step solution

For a pure capacitor on AC, peak current $I_0 = V_0 \omega C$ Given $210~\text{V},~50~\text{Hz}.~$ Assuming $210~V$ is RMS, then $V_0 = \sqrt{2}\times 210 \approx 297~\text{V}$ Angular frequency: $\omega = 2\pi f = 2 \times 3.14 \times 50 \approx 314~\text{s}^{-1}$ Capacitance: $C = 10~\mu\text{F} = 10 \times 10^{-6}~\text{F}$ $I_0 = 297 \times 314 \times 10^{-5} \approx 0.932~\text{A}\ (\text{rms}) \implies I_0 \approx 1.20~\text{A}\ (\text{peak})$ Thus, the peak current is about $1.2~A.$

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Exam: NEET (UG) 2024
Paper: 2024 paper
Subject: Physics
Chapter: Alternating Current
Topic: Average, Peak and RMS value of Alternating Current and Voltage