Practice question

Given: A 50 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the peak current? Formula: X_C = 1/omega C, omega = 2π × 50 = 314 rad/s. Substitution & Calculation: C = 50 × 10⁻⁶F . X_C = frac1314 × 50 × 10⁻⁶approx 63.7 Ω . RMS current: I = V/X_C = 220/63.7 approx 3.45 A . Peak current: i_m = √2 I = 1.414 × 3.45 approx 4.88 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.