If you double the frequency in a circuit containing only a capacitor, how does Xc change?

If you double the frequency in a circuit containing only a capacitor, how does Xc change?

• A. Xc doubles
• B. Xc halves
• C. Xc increases
• D. Xc decreases by a factor of two

Answer: (C)

Capacitive reactance is inversely related to frequency: Xc = 1/(2π f C). So when frequency doubles, Xc becomes half of what it was before, because Xc(new) = 1/(2π (2f) C) = (1/2) × [1/(2π f C)] = Xc/2. In a circuit containing only a capacitor, the impedance magnitude follows this inverse rule, while the sign is just a phase factor. For a quick check, if a capacitor has Xc ≈ 1.59 kΩ at 1 kHz, increasing the frequency to 2 kHz makes Xc ≈ 0.795 kΩ, exactly half. Therefore the correct behavior is that Xc decreases by a factor of two (it halves) when the frequency is doubled.

Capacitive reactance is inversely related to frequency: Xc = 1/(2π f C). So when frequency doubles, Xc becomes half of what it was before, because Xc(new) = 1/(2π (2f) C) = (1/2) × [1/(2π f C)] = Xc/2. In a circuit containing only a capacitor, the impedance magnitude follows this inverse rule, while the sign is just a phase factor. For a quick check, if a capacitor has Xc ≈ 1.59 kΩ at 1 kHz, increasing the frequency to 2 kHz makes Xc ≈ 0.795 kΩ, exactly half. Therefore the correct behavior is that Xc decreases by a factor of two (it halves) when the frequency is doubled.