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Chapter 13
RC Circuits
99
13
RC Circuits
Suppose you connect a capacitor across a battery, and wait until the capacitor is charged to the
extent that the voltage across the capacitor is equal to the EMF
V
0
of the battery.
Further
suppose that you remove the capacitor from the battery.
You now have a capacitor with voltage
V
0
and charge
q
0
, where
q
0
=
C V
0
.
The capacitor is said to be charged.
Now suppose that you connect the capacitor in series with
an open switch and a resistor as depicted below.
The capacitor remains charged as long as the switch remains open.
Now suppose that, at a clock
reading we shall call time zero, you close the switch.
From time 0 on, the circuit is:
The potential across the resistor is now the same as the potential across the capacitor.
This
results in current through the resistor:
+
+
−
−
C
q
0
,
V
0
C
R
R
C
C
R
+
+
−
−
q
0
,
V
0
+
+
−
−
q
0
,
V
0
+
+
−
−
q
,
V
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RC Circuits
100
Positive charge flows from the upper plate of the capacitor, down through the resistor to the
lower plate of the capacitor.
The capacitor is said to be discharging.
As the charge on the
capacitor decreases; according to
q
=
CV
, which can be written
V
=
q
/
C
, the voltage across the
capacitor decreases.
But, as is clear from the diagram, the voltage across the capacitor is the
voltage across the resistor.
What we are saying is that the voltage across the resistor decreases.
According to
V
=
I
R
, which can be written as
I
=
V
/
R
, this means that the current through the
resistor decreases.
So, the capacitor continues to discharge at an everdecreasing rate.
Eventually, the charge on the capacitor decreases to a negligible value, essentially zero, and the
current dies down to a negligible value, essentially zero.
Of interest is how the various
quantities, the voltage across both circuit elements, the charge on the capacitor, and the current
through the resistor depend on the time
t
.
Let’s apply the loop rule to the circuit while the
capacitor is discharging:
+
V
–
V
R
= 0
Using
q
=
CV
expressed as
C
q
V
=
and
V
R
=
I
R
, we obtain
0
=
−
IR
C
q
.
I
is the charge flow rate through the resistor, which is equivalent to the rate at which charge is
being depleted from the capacitor (since the charge flowing through the resistor comes from the
capacitor).
Thus
I
is the negative of the rate of change of the charge on the capacitor:
dt
dq
I
−
=
C
R
+
+
−
−
V
I
+
V
R
−
1
KVL
1
I
C
R
+
+
−
−
q
,
V
+
−
Chapter 13
RC Circuits
101
Substituting this (
dt
dq
I
−
=
) into our loop rule equation (
0
=
−
IR
C
q
) yields:
0
=
+
R
dt
dq
C
q
q
RC
dt
dq
1
−
=
Thus
q
(
t
) is a function whose derivative with respect to time is itself, times the constant
RC
1
−
.
The function is essentially its own derivative.
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This note was uploaded on 02/29/2012 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.
 Fall '08
 RABE
 Physics, Charge, RC Circuits

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