Q.An electric lamp is connected in series with a capacitor and an AC source, and is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance?

[!TLDR] Increasing C decreases the capacitive reactance $X_C=1/(\omega C)$, which lowers the total impedance $Z=\sqrt{R^2+X_C^2}$ (R here being the lamp's resistance), so the current -- and hence the lamp's brightness -- increases. [!ANSWER] The lamp glows brighter as the capacitance is increased.

The lamp (treated as a resistor of resistance R) is in series with the capacitor C, so the circuit's impedance is $Z=\sqrt{R^2+X_C^2}$, where $X_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC}$. Since $X_C$ is inversely proportional to C, INCREASING the capacitance C makes $X_C$ SMALLER, which in turn makes the overall impedance Z smaller (R itself is unchanged). A smaller impedance, for the same applied rms voltage, means a LARGER rms current flows through the series combination: $i_{rms}=e_{rms}/Z$. Since the power dissipated in (and hence the brightness of) the lamp depends on $i_{rms}^2R$, a larger current makes the lamp glow BRIGHTER. In the extreme limit of a very large capacitance, $X_C\to0$ and the capacitor offers almost no opposition at all, so the lamp approaches the brightness it would have if connected directly to the source with no capacitor in series. [!ANSWER] The lamp glows brighter as the capacitance is increased, because a larger C lowers $X_C=1/(\omega C)$, which lowers the circuit's impedance and so increases the current through the lamp.

Express the series R-C circuit's impedance as $Z=\sqrt{R^2+X_C^2}$ with $X_C=1/(\omega C)$, and track how increasing C changes $X_C$, then Z, then the current, then the lamp's power/brightness.

Assuming capacitance behaves like an ordinary resistance that reduces current when increased, when in fact capacitive reactance DECREASES as capacitance increases, having the opposite effect on the current.