A light bulb connected in series with a capac | vedclass

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(A) The brightness of the bulb depends on the current $I$ flowing through the circuit. The current is given by $I = V / Z$, where $Z$ is the impedance

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is reduced, is reduced

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(A) The brightness of the bulb depends on the current $I$ flowing through the circuit. The current is given by $I = V / Z$, where $Z$ is the impedance of the circuit. For an $R-C$ series circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = 1 / (2 \pi f C)$ is the capacitive reactance.$1$. When the capacitance $C$ is reduced, $X_C = 1 / (2 \pi f C)$ increases. Consequently, the total impedance $Z = \sqrt{R^2 + X_C^2}$ increases. Since $I = V / Z$, the current $I$ decreases, and the brightness of the bulb is reduced.$2$. When the frequency $f$ is reduced, $X_C = 1 / (2 \pi f C)$ increases. Consequently, the total impedance $Z = \sqrt{R^2 + X_C^2}$ increases. Since $I = V / Z$, the current $I$ decreases, and the brightness of the bulb is reduced.Therefore, in both cases, the brightness of the bulb is reduced.

$A$ light bulb connected in series with a capacitor and an $a.c.$ source is glowing with certain brightness. On reducing the value of capacitance and frequency respectively, the brightness of the bulb

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