Given below are two statements:
Statement I: In a purely capacitive circuit, the current reaches its maximum value one-fourth of a period after the voltage.
Statement II: The capacitive reactance is directly proportional to the frequency of the source.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Both Statement I and Statement II are incorrect.
For a capacitor joined to \( v=v_m\sin\omega t \), the current is \( i=i_m\sin(\omega t+\pi/2) \). So the current leads the voltage by a quarter of a cycle. The capacitor limits the current through its capacitive reactance \( X_C=1/\omega C \).
Statement I In a purely capacitive circuit, the current reaches its maximum value one-fourth of a period after the voltage. — False. The current leads the voltage, so it reaches its maximum a quarter period before the voltage. A maximum after the voltage is the case of a pure inductor.
Statement II The capacitive reactance is directly proportional to the frequency of the source. — False. The frequency \( \nu \) is in the denominator of \( X_C=1/(2\pi\nu C) \). So a higher frequency gives a smaller reactance. The reactance is inversely proportional to the frequency.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.”
AC Voltage Applied to a Capacitor – “The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance.”