(A) When the switch $S$ is closed,the capacitor begins to charge. Initially,the capacitor acts as a short circuit (zero resistance),allowing current t

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The bulb will light up for an instant when the capacitor starts charging.

(A) When the switch $S$ is closed,the capacitor begins to charge. Initially,the capacitor acts as a short circuit (zero resistance),allowing current to flow through the circuit,which causes the bulb to light up for an instant. As the capacitor charges,the potential difference across it increases,and the current in the circuit decreases. Once the capacitor is fully charged,it acts as an open circuit,blocking the flow of direct current $(DC)$. Therefore,the current becomes zero,and the bulb stops glowing.

Class 12 Physics Electric Potential and Capacitance Charging and Discharging of Capacitance and RC circuit (DC)

Medium

$A$ light bulb,a capacitor,and a battery are connected in a series circuit as shown in the figure,with switch $S$ initially open. When the switch $S$ is closed,which one of the following is true?

A
The bulb will light up for an instant when the capacitor starts charging.
B
The bulb will light up when the capacitor is fully charged.
C
The bulb will not light up at all.
D
The bulb will light up and go off at regular intervals.

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Class 12

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Chapter

Electric Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

In the given circuit, the potential difference across the plates of the capacitor $C$ in steady state is (in $V$)

MediumAP EAMCET 2025

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The potential difference between the plates of a parallel plate capacitor is changing at the rate of $10^6\, V/s$. If the capacitance is $2\,\mu F$,the displacement current in the dielectric of the capacitor will be......$A$

Medium

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An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

KVPY 2020

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The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

DifficultJEE MAIN 2024

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The time in seconds required to produce a potential difference of $20 \ V$ across a capacitor of $1000 \ \mu F$ when it is charged at a steady rate of $200 \ \mu C/s$ is

DifficultAP EAMCET 2002

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$A$ light bulb,a capacitor,and a battery are connected in a series circuit as shown in the figure,with switch $S$ initially open. When the switch $S$ is closed,which one of the following is true?

Similar Questions

In the given circuit, the potential difference across the plates of the capacitor $C$ in steady state is (in $V$)

The potential difference between the plates of a parallel plate capacitor is changing at the rate of $10^6\, V/s$. If the capacitance is $2\,\mu F$,the displacement current in the dielectric of the capacitor will be......$A$

An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

The time in seconds required to produce a potential difference of $20 \ V$ across a capacitor of $1000 \ \mu F$ when it is charged at a steady rate of $200 \ \mu C/s$ is

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