A 10 μF capacitor is connected to a 210 V, 50 Hz AC source. The peak current is nearly (π = 3.14):
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Question

A 10 μF capacitor is connected to a 210 V, 50 Hz AC source. The peak current is nearly (π = 3.14):

G1 Physics · Accepted Answer

Step 1: Capacitive reactance X(C) = 1/(2πfC). Step 2: Assuming 210 V is RMS, current RMS = V(RMS)(2πfC) = 210×2×3.14×50×10⁻⁵ ≈ 0.6594 A. Step 3: Peak I₀ = √2 I(RMS) ≈ 0.932 A. Correct A. Final answer: (A) 0.93 A.