An electric bulb and a capacitor are connected in series across an...

The impedance of the series RC circuit is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \dfrac{1}{2\pi f C}$.

When the frequency $f$ of the AC source is increased, the capacitive reactance $X_C$ decreases.

As a result, the total impedance $Z$ of the circuit decreases, which leads to an increase in the rms current $I = \dfrac{V}{Z}$.

The brightness of the bulb depends on the power dissipated, $P = I^2 R$. Since the current $I$ increases, the power dissipated increases, causing the brightness of the bulb to increase.

The potential difference across the bulb is $V_R = I R$, which increases as $I$ increases.

The total rms voltage of the source is $V = \sqrt{V_R^2 + V_C^2}$. Since $V$ is constant and $V_R$ increases, the potential difference across the capacitor $V_C$ must decrease.

Answer: Brightness increases, and the potential difference across the capacitor decreases.