Quantitative Treatment of Capacitor Discharge and Charge
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:
The graph is described by
the relationship:
[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:
·
·
Since t = RC, we can
rewrite the voltage equation as:
And do the same for the charge and current equations.
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:
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}
Therefore:
The halflife is 69 % of
the time constant.
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?

Answer
(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} Vo = 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 
