What happens to the capacitive reactance if the frequency of the ac source is doubled?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → It is halved.
The capacitive reactance is \( X_C=1/\omega C \), with \( \omega=2\pi\nu \), so it is inversely proportional to the frequency of the source.
Option (1) It is halved. — Correct. When the frequency is doubled, the new reactance is \( 1/(2\omega C) \), which is \( X_C/2 \). At the higher frequency the capacitor charges and discharges faster, so a larger current flows.
Option (2) It is doubled. — Incorrect. The inductive reactance \( X_L=\omega L \) doubles when the frequency doubles, because it is directly proportional to the frequency. \( X_C \) varies the opposite way.
Option (3) It is squared. — Incorrect. \( X_C \) depends on the first power of \( 1/\omega \). So doubling the frequency changes it by a factor of 2. Squaring a reactance would not even give a quantity in ohms.
Option (4) No effect, as it does not depend on the frequency. — Incorrect. Only a resistance does not depend on the frequency. \( X_C \) contains \( \omega \), so it changes whenever the source frequency changes.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Comparing it to im = vm/R for a purely resistive circuit, we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and is denoted by Xc,
\( X_c=1/\omega C \qquad (7.17) \)
… But it is inversely proportional to the frequency and the capacitance.”
Example 7.4 – “… If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.”