Scott Hughes
9 March 2004
Massachusetts Institute of Technology
Department of Physics
8.022 Spring 2004
Lecture 9:
Variable currents; Th´evenin equivalence
9.1
Variable currents: Discharging a capacitor
Up til now, everything we have done has assumed that things are in steady state: all fields
are constant; charges are either nailed in place or else are flowing uniformly. In reality, such
a situation is the exception rather than the rule. We now start to think about things that
vary in time.
Suppose we have a capacitor
C
charged up until the potential difference between its
plates is
V
0
; the charge separation is
Q
0
=
CV
0
. We put this capacitor into a circuit with a
resistor
R
and a switch
s
:
R
s
C
What happens when the switch is closed?
Before thinking about this with equations, let’s think about what happens here physically.
The instant that the switch is closed, there is a potential difference of
V
0
across the resistor.
This drives a current to flow. Now, this current can only come from the charge separation
on the plates of the capacitor: the excess charges on one plate flow off and neutralize the
deficit of charges on the other plate. The flow of current thus serves to reduce the amount
of charge on the capacitor; by
Q
=
CV
, this must reduce the voltage across the capacitor.
The potential difference which drives currents thus becomes smaller, and so the current flow
should reduce. We expect to see a flow of current that starts out big and gradually drops
off.
To substantiate this, turn to Kirchhoff’s laws: at any moment, the capacitor supplies an
EMF
V
=
Q/C
. As the current flows, there is a voltage drop

IR
across the resistor:
Q
C

IR
= 0
.
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This is true, but not very helpful — we need to connect the charge on the capacitor
Q
to
the flow of current
I
. Thinking about it a second, we see that we must have
I
=

dQ
dt
.
Why the minus sign? The capacitor is
losing
charge. A large, positive current reflects a large
loss in charge on the capacitor.
Putting this into Kirchhoff, we end up with a first order differential equation:
Q
C
+
R
dQ
dt
= 0
.
To solve it, rearrange this in a slightly funny way:
dQ
Q
=

dt
RC
.
Then integrate both sides. We use the integral to enforce the
boundary conditions
: initially
(
t
= 0), the charge separation is
Q
0
. At some later time
t
, it is a value
Q
(
t
). Our goal is to
find this
Q
(
t
):
Z
Q
=
Q
(
t
)
Q
=
Q
0
dQ
Q
=

Z
t
t
=0
dt
RC
→
ln
"
Q
(
t
)
Q
0
#
=

t
RC
.
Taking the exponential of both sides gives the solution:
Q
(
t
) =
Q
0
e

t/RC
.
The charge decays exponentially. After every time interval of
RC
, the charge has fallen by
a factor of 1
/e
relative to its value at the start of the interval.
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 Spring '08
 Hughes
 Current, Mass, Resistor, Electrical resistance, Voltage drop, Kirchhoff

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