A 2 μF capacitor is connected directly across a 150 V$_{rms}$, 60 Hz A.C source. Calculate the r m s value of the current.

Analyze the question: A capacitor in an A.C. circuit opposes current through capacitive reactance (Xc). We need the rms current when the capacitor is connected directly across the source.

Concept:

• Capacitive reactance: Xc = 1 / (2πfC)
• Ohm's law for the A.C. circuit: I_rms = V_rms / Xc = 2πfC·V_rms

Solution:

Given: C = 2 μF = 2 × 10⁻⁶ F, f = 60 Hz, V_rms = 150 V

1. Reactance: Xc = 1 / (2π × 60 × 2 × 10⁻⁶) Xc = 1 / (7.54 × 10⁻⁴) Xc ≈ 1326 Ω

2. Current: I = V / Xc = 150 / 1326 ≈ 0.113 A

Check: 2π × 60 = 376.99; 376.99 × 2 × 10⁻⁶ = 7.54 × 10⁻⁴; × 150 = 0.1131 A.

Answer: The rms current is about 0.113 A, which is option A.

Note that option D (1324 A) is the reactance value (about 1326 Ω, with the unit and rounding wrong), a trap for those who stop after finding Xc.

Exam tip: Always convert μF to farads (×10⁻⁶) first. Remember that for a capacitor, I = 2πfCV. A quick sanity check: with a small capacitor like 2 μF, the current should be small (fractions of an amp), so options C and D are clearly unrealistic.