The RC Series Circuit
The voltage across the resistor and capacitor is an RC series circuit are examined using an
oscilloscope. By measuring the voltage across either the resistor or capacitor at various times,
the time constant of the circuit is determined.
Theory, Charging
A resistor and an uncharged
capacitor are connected in series to a battery
of emf Vo and a switch, S. (Refer to Figure
1.) When the switch is closed at time t = 0,
with no initial charge on the capacitor,
current will begin to flow in a clockwise
direction around the circuit. In this case the
charge on the capacitor is gradually
increasing. It can be shown that after a time
t has elapsed, the charge on the capacitor is
q  CVo (1  e t / RC ).
Figure 1. The RC series circuit.
(1)
The voltage across the capacitor, which is q/C, is then
Vc = Vo(1 - e-t/RC).
(2)
Equation (2) indicates that the voltage across the capacitor starts at zero and builds up
exponentially to a final value Vo. At a time t=RC, Vc reaches a value that is 63% of Vo. This
term is referred to as the time constant of the RC circuit. That is,
 = RC.
(3)
After a time equal to three time constants the voltage across the capacitor reaches a value that is
95% of Vo, and after five time constants the voltage is 99.3% of Vo, a value that is essentially Vo.
The current in the circuit starts at a maximum value but then decreases as the capacitor
becomes charged. The decrease is, in fact, an exponential decrease which can be written as
V 
I   0  e-t/RC.
R
(4)
The voltage across the resistor, which is the product of current and resistance, then becomes
VR = Vo e -t/RC.
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(5)
This expression shows that the voltage across the resistor decreases to 37% of it initial value, Vo
after a time equal to one time constant, where the time constant is again  =RC. After a long
enough time elapses, the capacitor will be fully charged and the voltage across the capacitor will
be Vo.
Theory, Discharging
Suppose now the capacitor has an
initial charge Qo (=CVo), the switch is
closed, and the battery is removed from the
circuit. (Refer to Figure 2.) The voltage
across the capacitor will decrease as the
positive and negative charges leave the
plates of the capacitor and neutralize each
other. This is referred to as the discharging
situation and the voltage across the capacitor
is
-t/RC
VC = Vo e
Figure 2. The RC series circuit.
.
(6)
(Time. t = 0 is assumed to at the time when the capacitor first begins to discharge.) Because
current is flowing in the opposite direction from when it was charging, the voltage across the
resistor starts out negative and gradually reaches zero as the current decreases. The voltage
across the resistor in the discharging situation is then
VR = Vo e-t/RC.
(7)
After a time equal to one time constant, Vc decreases from Vo to 37% of Vo, and VR increases
from -Vo to 37% of –Vo.
Graphs of the voltage delivered by the battery and the voltages across the capacitor and
resistor for the entire charging and discharging sequences of the RC series circuit are shown in
Figure 3.
In the experiment, an oscilloscope is used to measure and observe the voltages, and a
signal generator set in the square wave mode acts as a battery delivering a voltage Vo, in the
charging cycle and a short in the discharging cycle. Because the signal generator has an output
resistance, RG that is in series with the circuit, a more realistic description of the circuit is shown
in Figure 4.For this situation, the voltages Vc and VR corresponding to the charging situation are
VC = Vo [1- e-t/( RG +R )C],
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(8)
Figure 3. The voltage delivered by the battery (V), the voltage across the
capacitor (Vc), and the voltage across the resistor (VR) for the
RC series circuit.
 VoR  -t/( RG +R )C.
VR  
e
 RG  R ' 
(9)
In the discharging situation, the voltages are
VC = Vo e-t/( RG +R )C,
(10)
 VoR  -t/( RG +R )C.
VR  
e
 RG  R' 
(11)
and
Notice that the time constant is now
 = (RG+R) C
The uncertainty in this value can be estimated with the formula
 = (RG+R) C + C(RGR' ; but RG is negligible here.
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(12)
Figure 4. The circuit components connected to measure the voltage across
(a) the capacitor and (b) the resistor.
Apparatus
o oscilloscope with leads
o signal generator
o 3 leads
o decade resistance box, set at 1000 , ± 5 
o capacitance box, set at 0.150 F, ± 0. 015 F
Procedure
1) Connect the circuit elements as shown in Figure 4(a). Set the voltage output knob on the
signal generator to1 volt if possible. At this setting the output resistance of the signal
generator, RG is 52 ohms.
2) Turn on the oscilloscope and the signal generator. Adjust the horizontal sweep knob
(TIME/DIV) and the frequency of the signal generator until a series of charging curves
and discharging curves for the capacitor are seen on the screen. Be sure that the starting
point and the flat portion of the curves are seen. (Refer to Figure 5.)
3) Adjust the amplitude knob on the signal generator until the vertical distance between the
beginning of the charging curve and the flat portion of the curve is 6 divisions high. This
allows you to take advantage of the percent scale on the oscilloscope grid.
4) Adjust the horizontal sweep (TIME/DIV) until the charging curve is expanded so that the
starting point and the 63% point fill horizontally as much of the screen as possible. (A
factor of five expansions can be obtained by pulling out the horizontal position knob.
When this is done, the horizontal sweep speed-reading is divided by five.)
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Figure 5. The voltage across the capacitor when charging.
5) Determine the time constant of the circuit from the voltage curve for the capacitor that is
observed on the oscilloscope by counting divisions (± 0.1 div) and noting setting (± 3%)
6) Disconnect and reconnect the circuit according to the circuit diagram in Figure 4(b). The
voltage across the resistor will look similar to the curve in Figure 6.
Figure 6. The voltage across the resistor when the capacitor
charging, then discharging.
7) Adjust the oscilloscope and the signal generator until the flat portion of the voltage curve
is at zero volts. Expand the curve until the starting point and the 37% point fill
horizontally as much of the screen as possible.
8) Measure the time constant, .
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Analysis
Three values of the time constant are found. The first one corresponds to the
experimental value obtained from the voltage across the capacitor during the charging process,
the second to the experimental value obtained from the voltage across the resistor during the
charging of the resistor, and the third to the theoretical value obtained from (12).
Report the three values of the time constant and their uncertainties in a results table and
calculate the percentage error of the two experimental values. Include an error bar-graph.
Questions
1.
2.
3.
4.
5.
Explain why the resistor or the capacitor must be connected to “ground” at one end when
measuring its voltage with the oscilloscope.
Explain why the maximum voltage across the capacitor is larger than the maximum
voltage across the resistor.
Explain why the voltage changes polarity for the resistor when discharging is taking place
but not for the capacitor.
Starting with Kirchhoff’s loop rule you can show that the differential equation for the
capacitor during the charging process is: Vo –R(dq/dt) –q/C =0. Solving this equation
gives equation (1) and the (2). Write the differential equations for the other three cases:
the capacitor charge during the discharging process and the resistor current during the
charging and discharging processes.
Show that you can solve at least ONE of the differential equations in problem 4. That is,
derive one of the equations (9), (10), or (11).
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