
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 k. The
voltmeter is an "analogue" type using the 7·5v
calibration (on this calibration, it has a resistance of 75 k). 
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 7·5v calibration of the voltmeter, its
resistance is 75 k.) 

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, 4·5 volts 

repeat for other voltages. 


Repeat
one or two of the readings with the 75 k
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 = 3·5
V and v = 1·5 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? 

