A bulb and a capacitor are connected in series to an a.c. source. A dielectric slab is now introduced between the plates of the capacitor. The intensity of the bulb will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Increases
In an a.c. circuit a capacitor offers capacitive reactance, which falls as the capacitance rises.
\( \displaystyle X_C = \frac{1}{\omega C} \)
A dielectric slab of dielectric constant \( K \) raises the capacitance to \( KC \), so \( X_C \) falls to \( X_C/K \). With the bulb of resistance \( R \) in series,
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{\sqrt{R^{2} + X_C^{2}}} \)
A smaller \( X_C \) means a smaller impedance, so the current increases and the bulb glows brighter (power \( I^{2}R \) rises).
Why the other options are incorrect
Option (1) Zero — Incorrect. An a.c. current does pass through a capacitor circuit, so the bulb does not go out.
Option (3) Decreases — Incorrect. The bulb would get dimmer only if \( C \) decreased; the dielectric increases it.
Option (4) Remain same — Incorrect. The reactance changes, so the current and the brightness change too.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Comparing it to im = vm/R for a purely resistive circuit, we find that (1/\( \omega \)C) plays the role of resistance. It is called capacitive reactance and is denoted by Xc,
\( \displaystyle X_c = 1/\omega C \qquad (7.17) \)”
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.”