In an AC circuit, a sine-wave voltage is applied across a capacitor. When the frequency of the applied voltage is decreased, what happens to the capacitor current?

Introduction / Context:
Capacitors in alternating-current (AC) circuits pass current that depends on both the capacitance and the frequency of the applied sine-wave voltage. This question tests understanding of how frequency affects capacitive reactance and, in turn, the current through a capacitor.

Given Data / Assumptions:

• Sine-wave voltage source across a capacitor.
• Frequency is decreased from some initial value.
• Capacitance is constant; voltage amplitude is unchanged.

Concept / Approach:
Capacitive reactance is Xc = 1 / (2 * pi * f * C). Current magnitude in a pure capacitive branch is I = V / Xc. Using reactance, I can also be written I = 2 * pi * f * C * V. Thus current is directly proportional to frequency when V and C are constant.

Step-by-Step Solution:
Xc = 1 / (2 * pi * f * C)If f decreases, then Xc increases.I = V / XcAs Xc increases, I must decrease for fixed V.Equivalently, I = 2 * pi * f * C * V shows I decreases when f decreases.

Verification / Alternative check:
Consider two frequencies f1 and f2 with f2 = f1 / 2. Then Xc doubles, so I halves. This confirms the inverse relationship between current and reactance (and direct proportionality to frequency).

Why Other Options Are Wrong:

• increases: False, because decreasing f reduces I.
• remains constant: False; I depends on f.
• ceases: False unless f approaches 0 Hz (true DC), where steady-state current is zero, but not for a mere decrease.

Common Pitfalls:
Confusing capacitors with resistors (whose current does not depend on frequency) or inductors (which have opposite frequency behavior for reactance). Another mistake is thinking voltage alone sets current without considering reactance variation with frequency.

Final Answer:
decreases