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(C) When the switch is turned on,the capacitor begins to charge. Initially,the charge on the capacitor is $0$,so the potential difference across it is

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Sharply rises initially and then gradually decreases

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(C) When the switch is turned on,the capacitor begins to charge. Initially,the charge on the capacitor is $0$,so the potential difference across it is $0 \ V$. Consequently,the entire battery voltage appears across the bulb,causing it to glow with maximum intensity.As the capacitor charges,the potential difference across it $(V_c = q/C)$ increases over time. According to Kirchhoff's voltage law,the voltage across the bulb $(V_b)$ is given by $V_b = V_{battery} - V_c$. As $V_c$ increases,$V_b$ decreases.Therefore,the intensity of the light,which depends on the power dissipated by the bulb $(P = V_b^2 / R)$,will be maximum at the start and will gradually decrease as the capacitor becomes fully charged.

An electric bulb,a capacitor,a battery,and a switch are all connected in series in a circuit. How does the intensity of light vary when the switch is turned on?

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