Capacitive reactance and AC magnitude: A 12 kHz sinusoidal voltage is applied to a 0.33 µF capacitor, and the measured rms current is 200 mA. Determine the rms voltage magnitude across the capacitor.

Introduction / Context:
In AC analysis, the magnitude of a capacitor’s impedance is given by its reactance Xc, which depends on frequency and capacitance. With known current and reactance, the voltage magnitude follows Ohm’s law for AC magnitudes: V = I * Xc.

Given Data / Assumptions:

• f = 12 kHz.
• C = 0.33 µF.
• Irms = 200 mA = 0.200 A.
• Pure capacitive branch; ignore series resistance.

Concept / Approach:
Capacitive reactance: Xc = 1 / (2 * pi * f * C). Then Vrms = Irms * Xc. Keep units consistent (Hz, farads, amperes).

Step-by-Step Solution:
C = 0.33 µF = 0.33 * 10^-6 FXc = 1 / (2 * pi * 12,000 * 0.33 * 10^-6)Xc ≈ 40.19 Ω (approximate)Vrms = Irms * Xc = 0.200 * 40.19 ≈ 8.04 VRounded to the nearest option: 8 V

Verification / Alternative check:
Proportional reasoning: A larger capacitance or frequency would reduce Xc and therefore reduce Vrms for a fixed current. The computed value is consistent with typical magnitudes for these parameters.

Why Other Options Are Wrong:

• 80 V: Ten times too large; would require Xc ≈ 400 Ω at 0.2 A.
• 800 mV and 80 mV: Too small by factors of 10 and 100 relative to the computed 8 V.

Common Pitfalls:
Forgetting to convert microfarads to farads or miscalculating 2 * pi * f * C. Also, do not confuse peak and rms values; the problem is clearly rms-based.

Final Answer:
8 V