An ac source of variable angular frequency and constant amplitude is connected in series with a capacitance C and an electric bulb of resistance R. When angular frequency is increased then:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → the bulb glow brighter
The capacitive reactance falls as the frequency rises, so the impedance of the series R-C circuit falls and the current rises.
\( \displaystyle X_C = \frac{1}{\omega C}\ \downarrow \quad\Rightarrow\quad Z = \sqrt{R^{2}+X_C^{2}}\ \downarrow \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z}\ \uparrow \)
The power in the bulb is \( I^{2}R \), which increases with the current, so the bulb glows brighter.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Example 7.3 A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. … With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.”