IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions

# Charging and Discharging of Capacitors

## Introduction

Capacitor is a very important component of many devices. When connected to a battery, the capacitor stores electrostatic energy. This energy is in the form of charge on its plates which raises the potential difference between the plates. When required, this capacitor can release this stored energy and gets discharged.

## Apparatus

Two capacitors of high capacitance say $$1000\, \mathrm{\mu{F}}$$, a high value resistor say $$30\, \mathrm{k\Omega}$$, a LED, a 9 V battery.

## Procedure

1. Connect the capacitor to the battery through the resistor. Since the capacitor is electrolytic capacitor, see that the positive of the capacitor is connected to the positive of the battery. Allow it to charge for more than a minute.
2. Now remove the battery and connect the capacitor to an LED through the resistor. Again remember to put the positive of the capacitor to the longer lead of the LED.
3. See that the LED glows very brightly, but gradually the brightness decreases. Note the time for which the LED glows quite brightly. This gives a rough estimate of the time constant. See how much it differs from the product $$RC$$.
4. Now repeat the whole process with two capacitors in series. Note the time constant for this. See that the measured time constant is less.
5. Again repeat the process with two capacitors in parallel. Note the time constant. See that the time constant has increased.

## Discussion

When a capacitor in series with a resistor is connected to a DC source, opposite charges get accumulated on the two plates of the capacitor. We say the capacitor gets charged. The time taken to charge it to 63\% of the maximum charge is called the time constant of the capacitor. It is equal to the product of capacitance and resistance. If the value of the capacitance and resistance is large, the time constant is large enough to be measurable easily without the use of sophisticated instruments.

If this capacitor is now disconnected from the power supply and its plates are connected to a LED through the resistor, the capacitor will get discharged. In this process a current flows through the LED and it glows. In one time constant $$\tau=RC$$, 63\% of the total charge of the capacitor is neutralized and the current drops to 37\% of the maximum value. The intensity of the glow of the LED is maximum in the beginning and then gradually decreases. In one time constant the glow decreases significantly. This time can be roughly estimated by us and it gives a fair idea of the time constant.

When two capacitors are put in series, equivalent capacitance of the circuit decreases so the time constant decreases. When two capacitors are put in parallel, equivalent capacitance of the circuit increases so the time constant increases.