Experiment 8
DC RC Circuits
Preparation
Prepare for this week's experiment by reviewing past material and by reading about capacitors,
capacitors in series and parallel, RC circuits, and energy stored in capacitors.
Principles
Consider a circuit consisting of a capacitor and a resistor wired in series with a constant DC
dq
voltage source like a battery. At a time t = 0 the circuit is closed and a current, I =
will
dt
begin to flow. Current cannot pass through the capacitor so charge, q, will build up on one plate.
This will induce a charge on the other plate and a voltage will develop across the capacitor which
we can express as
VC =
q
.
C
Kirchhoff's loop rule holds at all times for this circuit. The battery voltage, Vo, will equal the
sum of the resistor voltage, VR, and the capacitor voltage, VC, that is
V0 = VR + VC .
When the circuit is first closed there is no charge and therefore no voltage across the capacitor.
All the battery voltage will drop across the resistor. Since
I=
VR
,
R
the current will have a maximum value, Io , at t = 0. As the current flows charge builds up on
the capacitor, VC increases and VR and I must decrease. Eventually the capacitor acquires the
maximum amount of charge, Qo, and VC will equal Vo. Everything comes to a stop at that point.
You can find the capacitor voltage as a function of time by differentiating Kirchhoff's equation to
give
d
d
d q
.
V0 =
IR+
dt
dt
dt C
Since R, C, and Vo are constants and I =
dq
, the equation becomes
dt
0=R
dI I
+ .
dt C
Separating the variables gives
dI
-1
=
dt .
I
RC
You can integrate this to show that the current as a function of time will be
ln
I
-t
=
I0
RC
therefore
I = I 0 e-t/RC .
You can use Ohm's and Kirchhoff's Laws to show that the voltage across a capacitor in an RC
circuit as a function of time is given by
VC = V0 (1-e-t/RC )
while the capacitor is charging.
Now consider removing the battery and connecting the capacitor to the resistor. Charge will
migrate from the positive to the negative plate causing a current to flow through the resistor. The
value of this current will not be constant. It will equal the rate of decrease in the charge so
-dq
.
I=
dt
Now the loop rule gives us
q
dq q
0 = IR- = -R - .
C
dt C
You can rearrange this, separate the variables and integrate to find the charge on the capacitor as
a function of time. You can differentiate this to show that the current as a function of time will
be
I = I 0 e-t/RC ,
and the voltage will be
VC = V0 e-t/RC .
If you plot the VC as a function of time on Cartesian paper you will get an exponential curve. If
you plot the same data on semilog paper you will get a straight line. You can then take the slope
of this line and use it to find RC, and if R is known, to find C. Taking a natural log of the
previous equation gives us
lnVC = lnV0 -
t
.
RC
If we differentiate this equation we get
d(lnVC ) = 0 -
1
dt
RC
rearrange this to get
d(lnVC )
-1
.
=
dt
RC
The quantity RC is known as the time constant of the circuit where R is the total resistance. The
units of RC are seconds and that the time, t, for a capacitor to discharge to some fraction of its
maximum value, VC/Vo, is
⎛V
t = RCln ⎜ 0
⎝ VC
⎞
⎟.
⎠
The equivalent capacitance of N capacitors wired in series is given by
N
1
1
=∑ .
Ceq
i=1 Ci
The equivalent capacitance of N capacitors in parallel is given by:
N
Ceq =
∑C .
i
i=1
Study Figure 1. When the switch is closed the capacitor will be charged almost instantly. When
the switch is opened the capacitor will take much longer to discharge than it did to charge.
In this circuit the voltmeter is wired in series. This can be done because its internal resistance is
close to the resistance of R1. The net resistance in the circuit is very high and the current is
therefore very small. The internal resistance of the meter, Rm, is very high and must be
calculated in order to find the total resistance of the circuit. Because of the resistor R1 in the
circuit, the voltage across the capacitor, Vc, and the voltage across the meter, Vm, will always
differ by a constant value. You can use Ohm's law to show that
R +R
VC = 1 m Vm .
Rm
Equipment
1
1
1 set
2
2
3
1
1
1
2
1
protoboard
multimeter
red and black leads and alligator clips
2' banana wires, one red, one black
18" banana wires, one red, one black
small wires
0.1 µF capacitor
0.5 µF capacitor
1 µF capacitor
22 MΩ or larger resistor
stopwatch
Procedure
1.
Use the impedance bridge to measure the capacitance of the capacitors. Use the DMM to
measure the resistance of the resistors.
2.
Record the power supply voltage. This will also be the voltage across the capacitor when it
is fully charged (Vc at t=0).
3.
Wire the circuit in Figure 1. Record the meter voltage. This is Vm at t =0.
4.
Disconnect the power
supply and record Vm
at 4 second intervals
until the voltage drops
to just below 0.5 VDC.
One person in the group
should watch the time
and another should
record the voltages. Do
a practice run without
writing data to get an
idea of how many
voltages you will need
to record. This number will vary for each set of capacitors.
5.
Do this three times, once for the 0.5 µF capacitor alone, once for the 0.5 µF and the 1 µF
capacitors in series, and once for the 0.1 µF and the 0.5 µF capacitors in parallel.
Data
Data for the experiment should consist of the measured values of the resistor and the capacitors,
the power supply voltage, the initial DMM voltage and the voltages as a function of time.
Analysis
1.
Use the values for Vm and Vc at t=0 to solve for Rm. Calculate the constant needed to
convert Vm to Vc. Find the equivalent capacitance for the series and parallel capacitor
combinations using the measured values.
2.
Multiply each Vm by the appropriate constant in order to find each Vc value. Then find the
natural log of each Vc. Find the slope of lnVC vs. t for each set of capacitors. Find the
intercept of the line. Take the anti-log (inverse of the natural log or marked e x on some
calculators) of this value and compare it to the power supply voltage.
3.
Use the slope of each line to calculate the capacitance for each circuit and find the percent
error from the measured values. Remember to use the total resistance of the circuit, Rm +
R1, for your calculations.
4.
Plot Vc vs. t on semilog paper for each circuit; plot all three lines on one piece of paper.
Questions
1.
Draw a neat schematic of the circuit used in this experiment.
2.
Explain how current flows in the circuit. Derive the equation you will use to find the
resistance of the voltmeter, Rm.
3.
When the switch is closed the capacitor will be charged almost instantly. Why? What will
the maximum voltage across the capacitor equal? When the switch is opened the capacitor
will take much longer to discharge than it did to charge. Explain why.
4.
Write a paragraph explaining what a capacitor is and how it stores energy. What do we call
the unit for capacitance and what basic units is it made up of?
5.
Calculate how much charge was stored on each of your capacitors when it was fully
charged to the power supply voltage. What average current would you get if each
capacitor was shorted out and completely discharged in 0.01 s? If you open a copier you
will see warnings not to touch the capacitors, even if the machine is turned off. Why do
you think doing such a thing could be dangerous?
If it applies to you, write "I have not cheated on this lab report" and sign your name.
Grading
4 pts
Data and Analysis.
2 pts
Each for questions 1, 2, and 3.
4 pts
Each for questions 4 and 5.