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(B) The impedance of an $A.C.$ circuit containing a resistor (lamp) and a capacitor in series is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac

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is reduced

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(B) The impedance of an $A.C.$ circuit containing a resistor (lamp) and a capacitor in series is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance.As the frequency $f$ of the source is reduced, the capacitive reactance $X_C = \frac{1}{2\pi fC}$ increases.Since $Z = \sqrt{R^2 + X_C^2}$, an increase in $X_C$ leads to an increase in the total impedance $Z$ of the circuit.The current in the circuit is given by $I = \frac{V}{Z}$. Since $V$ is constant and $Z$ increases, the current $I$ flowing through the lamp decreases.The brightness of the lamp depends on the power dissipated, $P = I^2R$. As the current $I$ decreases, the power dissipated $P$ also decreases.Therefore, the brightness of the lamp is reduced.

An electric lamp connected in series with a capacitor and an a.c. source is glowing with certain brightness. On reducing the frequency of the source, the brightness of the lamp

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