An electric bulb, a capacitor, a battery, and | vedclass

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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

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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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