An AC voltage of angular frequency $\omega $ and fixed amplitude ${V_0}$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega $ is increased:
(A) The bulb glows dimmer
(B) The bulb glows brighter.
(C) Total impedance of the circuit is unchanged.
(D) Total impedance of the circuit increases.

Hint: A bulb glows brighter when the power supplied to it is increased. The impedance of a circuit is defined as the overall opposition of the circuit to the current flow. It is a combined effect of resistance and capacitance.
Formula used
$Z = \sqrt {{R^2} + {X_C}^2} $
$P = {V_{rms}}{I_{rms}}\cos \phi $
Here, Z is the impedance of the circuit.
R is the resistance of the circuit.
${X_C}$ is the capacitive reactance in the circuit.
P is the power supplied to the bulb.
${V_{rms}}$ is the root mean square voltage.
${I_{rms}}$ is the root mean square current.
And $\phi $ is the phase angle.

Complete step by step solution:
In the given circuit, the total resistance R and the total capacitive reactance ${X_C}$ constitute the impedance of the circuit.
It is given by-
$I = \sqrt {{R^2} + {X_C}^2} $
The capacitive reactance is given by-
${X_C} = \dfrac{1}{{{\omega _C}}}$
Keeping this value in the formula,
$Z = \sqrt {{R^2} + {{\left( {\dfrac{1}{{{\omega _C}}}} \right)}^2}} $
$Z \propto \dfrac{1}{{{\omega _C}}}$
If the value of $\omega $ is increased, the impedance in the circuit decreases, thus option (C) and (D) are incorrect.
The brightness of a bulb is determined by the magnitude of power supplied to it,
It is given by-
$P = {V_{rms}}{I_{rms}}\cos \phi $
The cosine of phase angle can be given by-
$\cos \phi = \dfrac{R}{Z}$
The formula for power can be rewritten by substituting this value as,
$P = {V_{rms}}{I_{rms}}\dfrac{R}{Z}$
It is given that the Voltage remains constant in the circuit, R of a circuit is a constant value and thus the current supplied to it is also constant.
The only value that changes with the change in $\omega $ is the impedance of the circuit.
$P \propto \dfrac{1}{Z}$
 Impedance decreases with the increase in $\omega $.
Thus, the power supplied to the bulb increases and the bulb glows brighter.

Therefore, option (B) is correct.

Note: The impedance of the circuit has two components, real and imaginary. The resistance of the circuit constitutes the real part of the impedance, while the imaginary part contains the reactance values of the inductive and capacitive behaviour of the circuit.