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(A) $1$. The brightness of the bulb depends on the power dissipated in the circuit,which is given by $P = I_{rms}^2 R_{bulb}$.$2$. The current in the

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(A) $1$. The brightness of the bulb depends on the power dissipated in the circuit,which is given by $P = I_{rms}^2 R_{bulb}$.$2$. The current in the $RC$ circuit is given by $I_{rms} = \frac{V_{rms}}{Z}$,where $Z = \sqrt{R_{total}^2 + X_C^2}$ is the impedance of the circuit.$3$. Here,$X_C = \frac{1}{\omega C}$ is the capacitive reactance.$4$. When a dielectric slab is inserted into the capacitor,the capacitance $C$ increases ($C' = KC$,where $K > 1$).$5$. As $C$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.$6$. Since $Z = \sqrt{R_{total}^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$.$7$. $A$ decrease in impedance $Z$ results in an increase in the current $I_{rms}$ flowing through the circuit.$8$. Since $P = I_{rms}^2 R_{bulb}$,an increase in $I_{rms}$ leads to an increase in the power dissipated by the bulb,hence the brightness of the bulb will increase.

If a dielectric slab is inserted in the capacitor of the given $RC$ circuit,then the brightness of the bulb will:

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