Capacitors Tutorial 2  Exponential Discharge
Graphical
Representation and Quantitative Treatment of Capacitor Discharge
Many
electronic circuits use the charge and discharge of a capacitor.
If we discharge a capacitor, we find that the charge decreases by the
same fraction for each time interval. If
it takes time t for the charge to
decay to 50 % of its original level, we find that the charge after another
t
seconds is 25 % of the original (50 % of 50 %).
This time interval is called the halflife
of the decay. The decay curve
against time is called an
exponential
decay.
The
voltage, current, and charge all decay exponentially during the capacitor
discharge.
We can plot a graph using a circuit like this:
We
can note the voltage and current at time intervals and plot the data, which
gives us the exponential graph shown below.
We should note the following about the graph:
Its
shape is unaffected by the voltage.
The
half life of the decay is independent of the voltage.
The
current follows exactly the
same pattern as
I = V/R.
The
charge is represented by the voltage, as
Q = CV.
The graph is asymptotic, i.e. in theory the capacitor does not completely discharge. In practice, it does.
The
graph is described by the relationship:
Q
= Q_{0} e^{ –t/RC}
[Q
– charge (C);
Q_{0} – charge
at the start;
e – exponential number
(2.718…);
t
– time (s);
C
– capacitance (F);
R – resistance (W).]
For
voltage and current, the equation becomes:
V=
V_{0} e^{ –t/RC}
I
= I_{0} e^{ –t/RC}
The
product RC
(capacitance × resistance) which we see in the formula is called the
time
constant. The units for the
time constant are seconds. We
can go back to base units to show that ohms × farads are seconds.
So if we discharge the capacitor for RC seconds, we can easily find out the fraction of charge left:
V= V_{0} e^{ –RC/RC} = V_{0} e^{ –1}
= 0.37
V_{0}
So
after RC
seconds the voltage is 37 % of the original.
This is used widely by electronic engineers.
To increase the time taken for a discharge we can:
Increase
the resistance.
Increase
the capacitance.
We
can link the halflife to the capacitance.
At the half life:
Q = Q_{0}/2
t = t_{1/2}
Q_{0} ÷ 2 = Q_{0} e ^{– t}_{1/2}^{/RC}
Þ ½ = e ^{– t}_{1/2}^{/RC}
Þ 2^{1} = e ^{– t}_{1/2}^{/RC}
Þ e ^{+ t}_{1/2}^{/RC} = 2
Þ log_{e} (2) =
t_{1/2}/RC
[In text books you may see the natural logarithm written as ‘ln’]
Þ
t_{1/2 } =
log_{e} (2) × RC
=
0.693 ×
RC
The
halflife is 69 % of the time constant.
Worked Example
A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor.
(a)
What is the time constant?
(b)
What is the voltage after 13 s?
(c)
What is the halflife of the decay? (d) How long would it take the capacitor to discharge to 2.0 V 
(a)
Time constant =
RC
= 2000 W
× 5000 × 10^{6} F = 10 s. 
(b) Use V = V_{0} e ^{–t/RC} Þ V = 12.0 V × e ^{– 13 s/10 s} = 12.0 × e ^{– 1.3} = 12.0 × 0.273
Þ
V = 3.3 volts

(c)
t_{1/2
} =
0.693 × RC
= 0.693 × 10 = 6.93 s. 
(d)
We need to rearrange the formula by taking natural
logarithms.
V
= V_{0} e ^{–t/RC}
Þ
V / V_{0 }
=
e ^{–t/RC}
Þ
log_{e}
V  log_{e}
V_{0} = t/RC [When you divide two
numbers, you subtract their
logs]
Þ
0.693 – 2.485 =  t/10
Þ
t/10
= 1.792
Þ
+t/10
= +1.792 Þ t = 1.792 × 10 = 17.9 s 
Question 1 
A
470
mF
capacitor, charged up to 12.0 V is connected to a 100 kW
resistor.
(a)
What is the time constant?
(b)
What is the voltage after 10 s? (c) How long does it take for the voltage to drop to 2.0 V? 