A 10 μF capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly ( π = 3.14 ): ![image](https://pub-7faa453bd1ec4bf0 - Physics MCQ NEET (UG) 2024 2026 | ExamTest.live

mediumMCQNEET (UG) 20242026Physics

1 mark

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

Solution & Step-by-step Explanation

Given: C = 10 \times 1 0^{- 6} F V_{r m s} = 210 V f = 50 Hz Capacitive reactance X_{C} : X_{C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 1 0 ^{- 6}} X_{C} = \frac{1 0 ^{6}}{314 \times 10} \approx \frac{1 0 ^{5}}{314} \approx 318.5 Ω RMS current I_{r m s} : I_{r m s} = \frac{V _{r m s}}{X _{C}} = \frac{210}{318.5} \approx 0.659 A Peak current I_{0} : I_{0} = 2 I_{r m s} = 1.414 \times 0.659 \approx 0.932 A

Try it yourself before checking the explanation above.

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

0.58 A

0.93 A

1.20 A

0.35 A

Practice with timed mock tests and track your performance across Physics.