A bulb and a capacitor are connected in serie | vedclass

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(A) The circuit consists of a bulb (resistor $R$) and a capacitor $(C)$ connected in series,forming an $R-C$ circuit.The impedance of an $R-C$ circuit

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Bulb will give more intense light

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(A) The circuit consists of a bulb (resistor $R$) and a capacitor $(C)$ connected in series,forming an $R-C$ circuit.The impedance of an $R-C$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi fC}$ is the capacitive reactance.When the frequency $f$ of the $AC$ source is increased,the capacitive reactance $X_C = \frac{1}{2\pi fC}$ decreases.Since $Z = \sqrt{R^2 + X_C^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.According to Ohm's law for $AC$ circuits,the current $i$ is given by $i = \frac{V}{Z}$. Since the source voltage $V$ is constant and $Z$ decreases,the current $i$ flowing through the circuit increases.The brightness of the bulb is proportional to the power dissipated,$P = i^2R$. As the current $i$ increases,the power dissipated by the bulb increases,and thus the bulb will give more intense light.

$A$ bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased,while keeping the voltage of the source constant,then

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