the rules for charge on a capacitor without doing any math. You only need to know three
things:
1. The time variation has to look like
e

t/RC
.
2. You need the initial (
t
= 0) charge on the capacitor, and
3. You need the final (
t
→ ∞
) charge on the capacitor.
83
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Then, you just need to figure out to hook everything together so that you get a formula
that obeys the correct boundary conditions. For example, if a problem involves discharging
a capacitor, you know that you start out with some initial value
Q
0
, and that it must fall
towards zero as time passes.
The only formula that obeys these conditions and has the
correct time variation is
Q
(
t
) =
Q
0
e

t/RC
,
just what we derived carefully before.
If it involves charging up a capacitor, you want a
formula that has
Q
= 0 at
t
= 0, and that levels off at
CV
b
as
t
→ ∞
. Using 1

e

t/RC
as
our “time variation” piece of the solution gets the
t
= 0 behavior right; multiplying by
CV
b
ensures that we level off at the right value at large
t
. We end up with
Q
(
t
) =
CV
b
‡
1

e

t/RC
·
.
Again, just what we derived carefully before.
If you can remember these physical reasons why the charge behaves as it does, you can
save yourself some pain later on.
9.4
Th´
evenin equivalence
The
RC
circuits we’ve looked at so far are quite simple — there’s only one resistor, and it’s
hooked up to the capacitor and/or battery in series. How do we handle a more complicated
circuit? Consider the following:
C
V
b
R
R
2
1
How does the charge on the capacitor evolve with time?
This problem can be tackled quite simply by using Kirchhoff’s laws: we have multiple
loops, so we go around them and write down equations balancing the voltage drops and
EMFs; we make sure the currents balance at the junctions; math happens; we find
Q
(
t
). As
an exercise, this is straightforward, but slightly tedious.
84
There is in fact a very cute way to approach this, based on a theorem proved by L´
eon
Charles Th´
evenin
1
:
Th´
evenin’s Theorem:
Any combination of batteries and resistances with two
terminals can be replaced by a single voltage source
V
OC
and a single series
resistor
R
T
.
For our problem, this means that we can redraw the circuit as
V
OC
R
T
s
C
The answer is now obvious! We just steal results from before, and write down
Q
(
t
) =
CV
OC
‡
1

e

t/R
T
C
·
.
We just need to know how to work out
V
OC
and
R
T
.
1
I
tried
to
find
biographical
information
about
this
guy
and
totally
failed.
Fortunately,
an
anonymous
Spring
’04
student
was
more
persistant
than
me,
and
found
a
web
bio:
He was a French telegraph engineer, and hence very well versed in practical aspects of electrical circuits.
Interestingly, “his” theorem was actually first derived by Hermann Von Helmholtz.
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To do this, we take the capacitor out of the circuit and consider what remains as a pair
of terminals:
V
b
R
R
2
1
If we short circuit these terminals with a wire, a current
I
SC
— the “short circuit” current
— will flow. On the other hand, if we leave it open, there will be a potential difference —
the “open circuit” voltage,
V
OC
 Spring '08
 Covault