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(A) The impedance of the $RC$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.When a dielect

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increases

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(A) The impedance of the $RC$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.When a dielectric is placed between the plates of the capacitor,its capacitance $C$ increases $(C' = KC)$.Since $X_C = \frac{1}{\omega C}$,an increase in $C$ leads to a decrease in capacitive reactance $(X_C \downarrow)$.As $X_C$ decreases,the total impedance $Z = \sqrt{R^2 + X_C^2}$ of the circuit decreases $(Z \downarrow)$.According to Ohm's law for $AC$ circuits,the current $I = \frac{V}{Z}$. Since $Z$ decreases,the current $I$ in the circuit increases.Consequently,the power dissipated in the bulb $(P = I^2 R)$ increases,and the glow of the bulb increases.

$A$ bulb and a capacitor are connected in series across an $AC$ supply. $A$ dielectric is then placed between the plates of the capacitor. The glow of the bulb:

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