Capacitors Tutorial 2 - Exponential Discharge
Representation and Quantitative Treatment of Capacitor Discharge
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
seconds is 25 % of the original (50 % of 50 %).
This time interval is called the half-life
of the decay. The decay curve
against time is called an
voltage, current, and charge all decay exponentially during the capacitor
We can plot a graph using a circuit like this:
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:
shape is unaffected by the voltage.
half life of the decay is independent of the voltage.
current follows exactly the
same pattern as
I = V/R.
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.
graph is described by the relationship:
= Q0 e t/RC
at the start;
e exponential number
R resistance (W).]
voltage and current, the equation becomes:
V0 e t/RC
= I0 e t/RC
(capacitance × resistance) which we see in the formula is called the
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= V0 e RC/RC = V0 e 1
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:
can link the half-life to the capacitance.
At the half life:
Q = Q0/2
t = t1/2
Q0 ÷ 2 = Q0 e t1/2/RC
Þ ½ = e t1/2/RC
Þ 2-1 = e t1/2/RC
Þ e + t1/2/RC = 2
Þ loge (2) =
[In text books you may see the natural logarithm written as ln]
loge (2) × RC
half-life is 69 % of the time constant.
A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor.
What is the time constant?
What is the voltage after 13 s?
What is the half-life of the decay?
(d) How long would it take the capacitor to discharge to 2.0 V
Time constant =
= 2000 W
× 5000 × 10-6 F = 10 s.
(b) Use V = V0 e t/RC
Þ V = 12.0 V × e 13 s/10 s = 12.0 × e 1.3 = 12.0 × 0.273
V = 3.3 volts
0.693 × RC
= 0.693 × 10 = 6.93 s.
We need to rearrange the formula by taking natural
= V0 e t/RC
V / V0
V - loge
V0 = -t/RC [When you divide two
numbers, you subtract their
0.693 2.485 = - t/10
Þ t = 1.792 × 10 = 17.9 s
capacitor, charged up to 12.0 V is connected to a 100 kW
What is the time constant?
What is the voltage after 10 s?
(c) How long does it take for the voltage to drop to 2.0 V?