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(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance.As t

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Bulb will give more intense light

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(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance.As the frequency $f$ of the $ac$ source increases,the capacitive reactance $X_C$ decreases.Since $Z = \sqrt{R^2 + (\frac{1}{2\pi f C})^2}$,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 $I = \frac{V}{Z}$. Since $V$ is constant and $Z$ decreases,the current $I$ in the circuit increases.The intensity of light emitted by the bulb is proportional to the power dissipated,$P = I^2 R$. As the current $I$ increases,the power dissipated increases,and the bulb glows more intensely.

$A$ 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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