## Drop in current as a capacitor discharges

We can use the circuit in Figure 1 to investigate how the current flowing from a capacitor changes as it discharges.

(We will assume that the capacitor shown has a capacitance of 1000 mF and is discharging through a 10kW resistor and was charged so that the initial potential difference across it was 12V).

When it is fully charged the charge on its plates will be Q = CV = 1000x10-6x12 = 12mC. At the moment the switch is moved to position 2 the initial current flowing from it will be given by I = V/R = 12.0/10000 = 1.2 mA. Therefore after 1s 1.2 mC of charge will have flowed from the capacitor, leaving 10.8 mC on its plates and the potential difference across it will have dropped to V = 10.8 mC x1000 µF = 10.8V.

The current flowing will now be 1.08 mA (since I = V/R), and after another second the voltage will be V = 9.72V.

If you repeat the process you can obtain a series of values for the voltage across the capacitor at any given time.

## The time constant

The time that it takes the potential difference across the capacitor to fall to 1/e of its original value is called the time constant for the circuit. If a capacitor C is discharged through a resistance R then the time constant is equal to RC. You can see that the time constant is independent of the initial voltage and this makes it a very useful quantity when using capacitors in timing circuits.

Time constant (t) for a capacitor C connected to a resistor R = RC

It is important to realise that if the voltage falls to 1/e of the original value in RC seconds then it will fall to (1/e)/e = 1/e2 in 2RC s. You can find out more detail about the use of the time constant in the section showing the mathematical treatment of charge and discharge.

Example problems
1. What is the time constant for a 4700 mF capacitor discharged through a 1000kW resistance?
Time constant = RC = 4700x10-6x1000x103 = 4700 s

2. What will be the voltage across a 100 mF capacitor charged to 20V discharging through a 200kW resistor after (a) 20s  (b) 40s  (c) 2 minutes
For this capacitor – resistor arrangement RC = 200x103x100x10-6 = 20 s
(a) time = RC therefore voltage = 20/e = 7.76 V
(b) time = 2xRC therefore voltage = 20/e2 = 2.71 V
(c) time = 6xRC therefore voltage = 20/e6 = 0.05 V

## Measurement of the charge stored in a capacitor

The current from a capacitor that is discharging is changing all the time so it is difficult to find the charge on the capacitor at any given time.

Taking a small interval of time IDt the charge flowing in that time is ID t where I is the average current during that small time. Therefore if we add together all the little areas made up of the shapes IDt then we will get the total area under the curve, see Figure 2. This is of course also the total charge initially present on the charged capacitor.

However since charge = current x time:

The area under the I- t graph = current x time = charge stored

Student investigation
The following readings were obtained from a capacitor discharge experiment. Use them to find the value of the capacitor used. The capacitor was discharged through a resistance of 15 kW.

 Voltage across capacitor (V) Time (s) Voltage across capacitor (V) Time (s) 12 0 3.85 25 9.56 5 3.07 30 7.62 10 2.44 35 6.07 15 1.55 45 4.83 20

Problem
 Time (s) 0 2 4 6 8 10 12 Current (mA) 15 7.3 5.4 4.6 4.3 4.2 4.1

Plot a current - time graph from the data above and use it to estimate the charge lost by the capacitor from t = 0 to t = 10s.