In a purely capacitive AC circuit, what happens to the capacitive reactance Xc if the operating voltage is doubled while frequency and capacitance remain unchanged?

Introduction / Context:
Capacitive reactance determines how strongly a capacitor opposes AC. It is important to know which variables affect Xc to avoid incorrect assumptions when voltage levels change in a system.

Given Data / Assumptions:

• Purely capacitive circuit.
• Frequency f and capacitance C are constant.
• Only the applied voltage magnitude is doubled.

Concept / Approach:
The formula for capacitive reactance is Xc = 1 / (2 * pi * f * C). There is no voltage term in this equation. Therefore, changing the applied voltage does not change Xc; it only changes the resulting current I = V / Xc proportionally to V.

Step-by-Step Solution:
Step 1: Write Xc = 1 / (2 * pi * f * C).Step 2: Observe that voltage does not appear in this expression.Step 3: Conclude that doubling voltage leaves Xc unchanged, while current would double.

Verification / Alternative check:
Compute a numeric example: if Xc = 100 Ω at given f and C, then V doubles from 10 V to 20 V, but Xc remains 100 Ω; current increases from 0.1 A to 0.2 A accordingly.

Why Other Options Are Wrong:

• Doubles or halves: These imply voltage dependence which does not exist for Xc.
• Multiplies by 7: Arbitrary and unsupported by the formula.
• None of the above: Incorrect because no effect is correct.

Common Pitfalls:
Confusing reactance with current. Current changes with voltage because I = V / Xc, but Xc itself depends only on frequency and capacitance.

Final Answer:
has no effect on the capacitive reactance.