Question

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

(1) The bulb will light up for some time interval after the capacitor starts charging
(2) The bulb will light up when the capacitor is fully charged
(3) The bulb will not light up at all
(4) The bulb will light up and go off at regular intervals

Answer 1

Hint:Capacitor is a device that is used to store electric charge, and it supplies back the charge to the circuit whenever required. The capacitor does not allow current to pass through it, which means the current flow at the fully charged condition of a capacitor is zero.

Complete step by step answer:
It is given that initially, switch S is open, which means the circuit is not complete as a result of which there will be no current flow.
When switch S is closed, the given circuit is completed, the current starts flowing by the flow of charge from the negative terminal to the positive terminal. Initially, the current flowing through the capacitor is charging the capacitor due to current being passed through it, and the bulb will start glowing. But after some time, the capacitor will be fully charged, and we know that flow of current through a fully charged capacitor is zero. This means that now the capacitor will not allow current to pass through it; due to this, no current will reach the bulb, and hence bulb will not glow.
Therefore, based on the above explanation, we can say that the bulb will light up for some time interval after the capacitor starts charging when switch S is closed, and option (1) is correct.

Note: The capacity of a capacitor to store charge is termed as its capacitance and measured in Farad. The capacitor has an insulating layer that does not allow direct current to pass through it, allowing alternating current to pass through it with no resistance.