PhysicsNEET 2024AC through R, L, C and LCR circuits

Alternating Current

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

A.0.93 A✓
B.1.20 A
C.0.35 A
D.0.58 A

Solution

$X_{C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 1 0 ^{- 5}} \approx 318$ $Ω$ . Peak current $= \frac{V _{0}}{X _{C}} = \frac{210 2}{318} \approx 0.93$ A.

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From Physics › Alternating Current › AC through R, L, C and LCR circuits — lectures, notes and every past question on it.

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