Sanity check: does
RC
make sense as a time? Let’s check its units:
•
CGS:
R
×
C
has units (sec/cm)
×
(cm) = seconds.
•
SI:
R
×
C
is (ohms)
×
(farads) = (volts/amps)
×
(coulombs/volts) = coulombs/amps =
seconds.
So it does make sense.
Let’s look at the current
I
(
t
):
I
(
t
) =

dQ
dt
=

Q
0
d
dt
e

t/RC
=
Q
0
RC
e

t/RC
.
Just like the charge, the current decays exponentially. In particular, it starts out at some
big value, and then falls away — just as our physical intuition told us it would.
81
Subscribe to view the full document.
9.2
Variable currents 2: Charging a capacitor
Let’s consider a different kind of circuit. The capacitor begins, at
t
= 0, with
no
charge;
but, the circuit now contains a battery:
R
s
C
V
b
Now, what happens when the switch is closed?
Let’s again think about this physically before doing the math. When the switch is first
closed, the EMF of the battery can very easily drive a current, so we expect the current to
jump up, and charge to begin accumulating on the capacitor. As this charge accumulates,
an
~
E
field will build up in a direction that
opposes
the flow of current! Thus, we expect that,
as time goes on, the flow of current will decrease. Eventually it will stop, when there is just
enough charge on the capacitor to totally oppose the battery.
Let’s check this. For Kirchhoff, we have an EMF
V
b
from the battery, and two voltage
drops:

Q/C
from the capacitor, and

IR
across the resistor:
V
b

Q
C

IR
= 0
.
We again need to relate the current
I
and the capacitor’s charge
Q
. Thinking about it for a
second, we must have
I
= +
dQ
dt
.
Why a plus sign?
In this case we are
adding
charge to the capacitor (more accurately,
increasing
the charge separation). Large current means a large increase in the charge sepa
ration. When doing a problem involving a circuit like this, it is important to stop and think
carefully about how the capacitor’s charge relates to the charge flowing onto it or off of it.
We end up with the following differential equation:
V
b

Q
C

R
dQ
dt
= 0
.
Rearrange:
dQ
dt
=

Q

CV
b
RC
82
or
dQ
Q

CV
b
=

dt
RC
.
Integrate: we require
Q
(
t
= 0) = 0, so we find
ln
"
CV
b

Q
(
t
)
CV
b
#
=

t
RC
.
Exponentiating both sides and solving for
Q
(
t
), we find
Q
(
t
) =
CV
b
‡
1

e

t/RC
·
.
This charge starts off at zero, builds up quickly at first, then starts building up more and
more slowly. It asymptotically levels off at
Q
=
CV
b
as
t
→ ∞
. Let’s look at the current:
I
(
t
)
=
CV
b
d
dt
‡
1

e

t/RC
·
=
V
b
R
e

t/RC
.
As we intuitively guessed, the current starts large, but drops off as time passes and the
capacitor’s electric field impedes it.
Notice that the voltage across the capacitor
V
C
(
t
) =
Q
(
t
)
C
=
V
b
‡
1

e

t/RC
·
and the voltage across the resistor
V
R
(
t
) =
I
(
t
)
R
=
V
b
e

t/RC
sum to give a constant,
V
C
(
t
) +
V
R
(
t
) =
V
b
‡
1

e

t/RC
·
+
V
b
e

t/RC
=
V
b
,
the EMF of the battery. If you think about Kirchhoff’s rules for a moment, this should make
a lot of sense!
9.3
Charging and discharging revisited
Suppose you’re taking a test and you don’t want to slog through setting up a bunch of
integrals and running the risk of botching the details because of a silly mistake. Provided
you understand
physically
what is going on in the above discussions, you can write down
the rules for charge on a capacitor without doing any math. You only need to know three
Subscribe to view the full document.
 Spring '08
 Covault