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Transcript
```RC-1
RC Circuits
An RC circuit is a circuit with a resistor (R) and a capacitor (C). RC circuits are used to
construct timers and filters.
Example 1. Very simple RC circuit: a capacitor C, charged to an initial voltage V0 = Q0/C,
attached to a resistor R with a switch.
switch
+Q0
C
C
R
–Q0
Q
V
 V0 
Q0
C
Close the switch at time t = 0 , so current I starts to flow. The charged capacitor is acting like a
battery: it produces a voltage difference across the resistor which drives the current through the
resistor:
C
At t = 0+,
+Q
I
–Q
R
I
Vacross C = Vacross R , VC = VR ,
dQ
dt
I0 
V0
.
R
(– sign because Q is decreasing)
Q
Q
dQ
 IR ,
 
R ,
C
C
dt
dQ
1
 
Q
dt
RC
RC = "time constant" =  , has units of time
is a differential equation of the form
dx
 a  x , where a is a constant.
dt
This equation says: (rate of charge of x)  x  exponential solution : x  x0 exp  a t 
d  x 0  ea t 
dx

 a  x 0  ea t  a  x .
Check:
dt
dt
a > 0  exponential growth ,
The solution to
dQ
1
 
Q
dt
RC
It works!
a < 0  exponential decay
is
 t 
 t
Q(t)  Q0 exp  
  C V0 exp   
 RC 
 
Notice that at t = 0, the formula gives Q = Q0 .
Last update: 4/30/2017
Dubson Phys1120 Notes, University of Colorado
RC-2
In time  , Q falls by a factor of exp(–1) = 1/e  0.37..
In time 2, Q falls by a factor of exp(–2) = (1/e)(1/e)  0.14..
 Q approaches zero asymptotically, and so does V and I
Q
 t 
 t
Q(t)  Q 0 exp  
  Q 0 exp   
 RC 
 
Q0
Q0/e
t

I 
V=Q/C
V
dQ
  0 exp( t / )
dt
R
t
t
Example 2: More complex RC circuit: Charging a capacitor with a battery.
switch
E
C
VC = Q / C
Let's use symbol E for battery voltage (E
short for emf) because there are so many
other V's in this example.
Before switch is closed, I = 0, Q = 0.
R
VR = I R
Close switch at t = 0.
Always true that E = VC + VR ,
by Kirchhoff's Voltage Law (Loop Law)
The charge Q on the capacitor and the voltage VC = Q / C across the capacitor cannot change
instantly, since it takes time for Q to build up, so ..
At t = 0+ , Q = 0 , VC = 0, E = VC + VR = VR = I R  I0 = E / R
Although Q on the capacitor cannot change instantly, the current I = dQ/dt can change instantly.
Last update: 4/30/2017
Dubson Phys1120 Notes, University of Colorado
R2
RC-3
"Current through a capacitor" means dQ/dt . Even though there is no charge ever passing
between the plates of the capacitor, there is a current
+
I
going into one plate and the same current is coming out
- I
+
of the other plate, so it is as if there is a current passing
+
through the capacitor.
+
C
E
E = VR  VC
Q
C
E = IR 
I
R
E =
,
I
dQ
dt
dQ
Q
R 
dt
C
Qualitatively, as t  , Q  , VC = Q/C  , VR  , I = VR / R . After a long time , t >> = RC ,
the current decreases to zero: I = 0, VC = E , Q = C E
Vc = Q / C
Analytic solution:
E
VC (t)  E [1  exp(t / RC) ]
Q(t)  E  C [1  exp( t / RC) ]
t
Things to remember:

Uncharged capacitor acts like a "short" ( a wire ) since VC = Q / C = 0.

After a long time, when the capacitor is fully charged, it acts like an "open-circuit" ( a
break the wire). We must have IC = 0 eventually, otherwise Q   , VC   .
Last update: 4/30/2017
Dubson Phys1120 Notes, University of Colorado
```
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