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(B) The capacitive reactance of a capacitor is given by the formula $X_{C} = \frac{1}{2 \pi f C}$.Here,$f$ is the frequency of the $A.C.$ source and $

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is reduced,is reduced

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(B) The capacitive reactance of a capacitor is given by the formula $X_{C} = \frac{1}{2 \pi f C}$.Here,$f$ is the frequency of the $A.C.$ source and $C$ is the capacitance.When the capacitance $C$ is reduced,the capacitive reactance $X_{C}$ increases.Similarly,when the frequency $f$ is reduced,the capacitive reactance $X_{C}$ increases.Since the bulb is in series with the capacitor,the total impedance $Z$ of the circuit is given by $Z = \sqrt{R^2 + X_{C}^2}$.As $X_{C}$ increases in both cases,the total impedance $Z$ of the circuit increases.According to Ohm's law for $A.C.$ circuits,the current $I = \frac{V}{Z}$.Since the impedance $Z$ increases,the current $I$ flowing through the bulb decreases.The brightness of the bulb depends on the power dissipated,$P = I^2 R$.As the current $I$ decreases,the power dissipated and the brightness of the lamp decrease in both cases.

$A$ light bulb connected in series with a capacitor and an $A.C.$ source is glowing with a certain brightness. On reducing the capacitance of the capacitor and the frequency of the source,the brightness of the lamp (respectively):

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