An a.c. voltage source of fixed amplitude \( \varepsilon_0 \) and variable angular frequency \( \omega \) is connected with an electric bulb of resistance R (inductance zero) and a capacitor of capacitance C in series. On increasing the angular frequency of the source, the total impedance of the circuit
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → decreases and the bulb glows brighter
In a series RC circuit the impedance combines the resistance and the capacitive reactance, and the reactance falls as the frequency rises.
\( \displaystyle Z = \sqrt{R^{2} + X_C^{2}}, \qquad X_C = \frac{1}{\omega C} \)
As \( \omega \) increases, \( X_C \) decreases, so \( Z \) decreases. With a fixed source amplitude the current increases:
\( \displaystyle I_0 = \frac{\varepsilon_0}{Z} \)
A larger current gives a larger power \( I_{\mathrm{rms}}^{2} R \) in the bulb, so it glows brighter.
NCERT (2026-27):
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.”