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(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + (\frac{1}{\omega C})^2}$.The current $I$ in the circ

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(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + (\frac{1}{\omega C})^2}$.The current $I$ in the circuit is given by $I = \frac{V}{Z}$.The brightness of the bulb depends on the power dissipated,which is $P = I^2 R$.As the frequency $f$ of the $AC$ source increases,the angular frequency $\omega = 2\pi f$ also increases.Since $X_C = \frac{1}{\omega C}$,an increase in $\omega$ leads to a decrease in the capacitive reactance $X_C$.As $X_C$ decreases,the total impedance $Z = \sqrt{R^2 + X_C^2}$ decreases.Since $I = \frac{V}{Z}$,a decrease in $Z$ leads to an increase in the current $I$ flowing through the circuit.As the current $I$ increases,the power dissipated $P = I^2 R$ increases,and therefore the brightness of the bulb increases.

An electric bulb and a capacitor are connected in series with an $AC$ source. On increasing the frequency of the source,the brightness of the bulb

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