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Discharging a Capacitor

A capacitor is a device used to store electric charge. The capacitance of a capacitor is a measure of the quantity of charge, Q, it can store for a given potential difference, V. Capacitance is defined by the following equation:

C = Q/V

and so the units of capacitance are CV-1. 1 CV-1 is called 1Farad (1F)

The capacitor is being studied here as it gives us another example of an exponential variation.

1. Preparation:

a) Remind yourself how to measure the slope of a curved graph at a given point.

b) See part 3 below.

2. The aim of the experiment is to plot a graph which shows how the voltage across a capacitor varies as it is discharging through a resistor.

R = 75 kBLACKOHM. The voltmeter is an "analogue" type using the 75v calibration (on this calibration, it has a resistance of 75 kBLACKOHM).

Do the experiment first without the resistor R in the circuit.

When the switch is closed, the capacitor charges (almost immediately) to the same voltage as the supply. As soon as the switch is opened, the capacitor starts to discharge through the voltmeter. (When using the 75v calibration of the voltmeter, its resistance is 75 kBLACKOHM.)
- charge the capacitor, read the voltmeter with the switch closed; this is the voltage at t = zero
- open the switch and start a watch simultaneously
- measure the time taken for the voltage to fall to, for example, 5 volts
- recharge C and measure the time taken for the voltage to fall to some lower value, for example, 45 volts
- repeat for other voltages.
Repeat one or two of the readings with the 75 kBLACKOHM resistor connected in parallel with the voltmeter, as shown.
Plot a graph of voltage against time.
3. If the graph is exponential, it will be found that the rate of fall of voltage is directly proportional to voltage.
Or, fall in voltage per second = (a constant) voltage
but fall in voltage per second is the slope of the graph
so, if we measure the slope at various voltages v we should find that

gradient / v = a constant

4. a) Prove that your graphs are exponential. To do this, measure the slope at three points on the curve, for example, at v = 5 V, v = 35 V and v = 15 V.
b) Another way to prove that the results show an exponential fall in voltage is to find how long it takes for the voltage to fall to half of its starting value. This "halving time" should be constant no matter what time you consider as the start. (You could, of course, consider the time taken for the voltage to fall to some other fraction of its initial value.)
c) In theory, how long would it take to completely discharge a capacitor? In practice, how long (approximately) did it take? Why is there this difference between theory and practice?

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