Purely capacitive AC circuit: If the operating frequency is doubled while the RMS voltage is held constant, how does the total circuit current change?

Introduction / Context:
In a purely capacitive AC circuit, the reactance depends on frequency. This determines the current drawn for a fixed applied voltage. Understanding this relationship is fundamental for capacitor sizing, reactive power considerations, and filter behavior.

Given Data / Assumptions:

• Ideal capacitor C with negligible series resistance.
• RMS voltage V is constant.
• Frequency is changed from f to 2f.

Concept / Approach:
Capacitive reactance is X_C = 1/(2πfC). The current in a pure capacitor is I = V / X_C = V * 2πfC. Therefore I is directly proportional to f when V and C are constant. Doubling f doubles I.

Step-by-Step Solution:
Start: X_C = 1/(2πfC).Current: I = V / X_C = V * 2πfC.If f → 2f, then I → 2I.Conclusion: current doubles.

Verification / Alternative check:
Example numbers: Let V = 10 V, C = 100 μF, f = 50 Hz. I1 = 10 * 2π * 50 * 100e−6 ≈ 0.314 A. At 100 Hz, I2 ≈ 0.628 A, exactly double.

Why Other Options Are Wrong:
(a) ignores the frequency dependence. (c) and (d) would require inverse or triple proportionality, which is not the case. (e) is unnecessary because a precise rule exists.

Common Pitfalls:
Using X_C ∝ 1/f but then forgetting that I = V / X_C ∝ f; mixing peak and RMS values does not change the proportionality.

Final Answer:
Current doubles