Lab 6: R-C Circuits
Only 6 more labs to go!!
A capacitor is a device used to store energy. In this lab we will measure how a capacitor “stores” up
or discharges energy.
If we use Kirkchoff’s loop rule to the circuit below we get the following equation:
R
Vapplied  VR  VC  0
C
Q
Vapplied  IR   0
C
Remember the definition of current is:
So the equation becomes:
Q
I
t
Q
Q
Vapplied 
R 0
t
C
Switch
Vapplied
This equation is called a differential
equation.
t


RC
The solution to this differential equation is:
Q  CVapplied 1  e 


This tells us that the charge on the capacitor is exponential in time.
t


RC
Q  CVapplied 1  e  t  


Q  CVapplied 1  e   CVapplied
Charge, Q
Q = CVapplied

t = RC = 

time (s)
Since we now know how the capacitor charges up we can write an equation that describes the voltage
as a function of time:
t
Voltage, V
Vcap


Q
RC
  Vapplied 1  e 
C


t = RC = 
time (s)
C is called the RC-time constant. It represents the time which the voltage drop across the capacitor
reaches 63% of it’s max value.
Let’s look at this example:
R = 105 
C = 5 F
1. Find the time constant, C.
 C  RC  105   5F  0.5 sec
Switch
Vapplied = 12 V
2. Find the charge on the capacitor at t = 0.2 sec
t
0.2 sec




RC
0
.
5
sec
  1.9 105 C
Q  CVapplied 1  e   5F 12V 1  e





3. Find the voltage on the capacitor at t = 0.3 sec
Vcap
t
0.3sec




RC
0
.
5
sec
  5.42V
 Vapplied 1  e   12V 1  e





4. How long will it take the capacitor to charge to 9V?
Vcap
t
t




RC
0
.
5
sec
  t  0.69 sec
 Vapplied 1  e   9V  12V 1  e





Now we will look at how the capacitor discharges its energy.
Let’s look at this circuit:
R
C
Use Kirchoff’s loop:
VR  VC  0
Switch
The solution to this
differential equation is:
Q  Q0 e
t
RC
 V0Ce
t
RC
If we take the derivative of this equation
with respect to time we can get the current:
Charge, Q
Q
IR   0
C
Q
Q
R 0
t
C
time (s)
 Q0
I
e
RC
t
RC
1.
R=105 
Find the voltage across the capacitor at time,
t = 0.3 s.
Vcap  V0 e
t
RC
 9Ve
0.3 sec
0.5 sec
C=5F
 4.94V
Switch
2. How long will it take the capacitor to reach 3 v?
Vcap  V0e
t
RC

Vcap
V0
e
t
RC
Vapplied = 9V
 Vcap 
t


 ln 


RC
 V0 
 Vcap 
3V 

  0.5 sec ln    0.55s
t   RC ln 
 9V 
 V0 