Capacitors and RC Circuits

By: John Connor

Lab Partner: Adelina Lane
TA: Stefan Evans
PHYS182: Section 007
2/22/16 8:00 AM

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 Abstract
This experiment was aimed at developing a better understanding of capacitors and
capacitance, as well as how resistance affects the time it takes a capacitor to discharge. This was
achieved by creating different kinds of circuits and measuring the respective voltages or
discharge times in order to experimentally observe how capacitors behave. It was found that
capacitors hold charge after disconnected from a power source, and that capacitors behave
differently when in a parallel or series circuit. Also, it was found that the greater the resistance in
a circuit, the longer it takes a capacitor to discharge

Introduction
This experiment is centered around capacitors, and the idea of capacitance. The metal
plates of a capacitor are used to store and hold electrical energy. This is achieved by storing
electric charge on its plates, in which one of the plates becomes positively charged, and the other
becomes negatively charged. Because of this property, capacitors are used in many electrical
devices so as to extend the flow of energy in case of a short. The maximum amount of charge a
given capacitor can store per volt is known as its capacitance. The capacitance of a capacitor is
determined by the material between the plates, as well as the shape of its plates, and the distance
between the plates. In addition, the capacitance of capacitors in parallel circuits and series
circuits behave in a way that is different from resistors. The total capacitance in parallel circuits
is the sum of the individual capacitances of the capacitors. They can be thought of as one
capacitor with an increased surface area. Moreover, capacitors in series, act as if the plate area is
constant, but the distance between the plates increases, resulting in a value lower than the sum of
the individual capacitances. Also, when a capacitor discharges through a resistor, one is able to
observe the relationship between the magnitude of resistance in a circuit and how quickly the
voltage across a capacitor changes. This allows one to determine the half-life of a capacitor with
respect to a certain resistor.

Theory & Derivations

Description of the Procedure

The first part of this experiment consists of constructing a simple circuit, in which a
battery is connected to a capacitor. However, before setting up any circuit, one needs to connect a
wire between the two terminals of the capacitor in order to discharge it. So, to begin, one needs
to connect a battery to a capacitor and let it charge the capacitor for about 15 seconds. Then, the
voltage across the battery and the capacitor is measured with a voltmeter and recorded. Next, the
battery is disconnected from the capacitor, and the voltage is measured across the capacitor again
using the voltmeter.
The next part of the lab consists of constructing a circuit with two capacitors in parallel,
and a circuit with two capacitors in series. Firstly, the parallel circuit, contains two capacitors,
one with a capacitance of 1000µF and 470µF. Then, a similar procedure to the first part of the
experiment is followed, in which the voltages across the battery and the capacitors are measured
and recorded. In the series circuit, two capacitors, each with a capacitance of 470µF, are used.
Then, using the voltmeter, voltages across the battery and the capacitors are measured. Then,
another series circuit is constructed with the first capacitor having a capacitance of 1000µF and
the second having a capacitance of 470 µF. The voltages are again measured.

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 The final portion of the experiment uses a resistor to discharge a capacitor. One can
achieve this by charging a 470µF capacitor with a battery, then disconnecting the wires from the
battery and connecting them to resistors with varying resistances (10kΩ, 30kΩ, and 100kΩ).
Then, the time it takes for the capacitor to discharge is observed and recorded. Moreover, the
time it takes for the capacitor to discharge from 8V to 4V to 2V to 1V to 0.5V and to 0.25V is
measured and recorded. This enables one to determine the half-life time of the capacitor. This
data is then graphed in order to see the relationship between time and ln(V).

Results and Sample Calculations

In the first part of the experiment, it was found that the voltage across the battery was 8.7
V, and the voltage across the capacitor was also 8.7 V, which was to be expected. Since there is
only so much charge that can be placed on a capacitor. This results in the voltage across the
capacitor being equal to the voltage of the battery. After disconnecting the battery from the
capacitor, the voltage was measured at 8.69 V, which shows the capacitor maintained the charge
from the battery even after it was disconnected. Since a capacitor is designed to hold and store
charge, it makes sense that once the battery is disconnected, the capacitor would maintain a
charge with a value close to the original measured voltage of 8.7 V. However, after awhile, the
voltage across the capacitor will decrease back to zero once it is disconnected from the battery.
The second part of the experiment dealt with capacitors in a parallel circuit. One can see
that the voltage across the battery will be equal to the voltage across each capacitor. The value
for the voltage across the battery was measured at 8.70 V, which was the same for the voltage
across the 470 µF capacitor, and the 1000 µF capacitor.. This is because each capacitor is
connected to the battery terminals, which results in the capacitors receiving the same voltage but
having different charges).
q = CV
q1 = C1V1 q2 = C2V2
q1 = (0.001F)(8.7V) q2 = (0.00047F)(8.7 V)
q1 = 0.0087 FV q2 = 0.0041 FV
In a series circuit, with capacitors, it can be seen that the sum of the voltages across each
capacitor is equal to the voltage across the battery. In this case, the voltage across the battery was
measured at 8.46 V, whereas the voltage across the first 470 µF capacitor was measured at 4.36
V, and the voltage across the second 470 µF capacitor was measured at 4.12 V.
Vtot = Vbatt = V1+V2
8.46 V = 4.12 V + 4.36 V
8.46 V = 8.46 V
The same can be said with regards to a battery connected to two non-identical capacitors in
series. In this case, the voltage across the battery was 8.50 V, whereas the voltage across the first
1000 µF capacitor was 2.53 V, and the voltage across the second 470 µF capacitor was 6.01 V.
This was to be expected, the relationship can be seen by the equation which says V= q/C. So, the
capacitor with the greater capacitance will have a lower voltage since the charge is the same on
both capacitors in a series circuit.
The last part of the experiment dealt with discharging a capacitor through a resistor. It
was found that discharging the capacitor with the 100kΩ resistor took the longest, followed by
the 30kΩ and 10kΩ resistors, respectively. Since a greater resistance results in a smaller current,
it makes sense that it would take longer for a capacitor to discharge when there is a greater
resistance.

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 One can see in Table 1 the data from the final portion of the experiment, where times
were recorded in occurrence with the capacitor losing charge. In addition, in Graph 1, one can
see that the voltage from the capacitor decays exponentially, since graphing ln (V) vs. time
resulted in a straight line with a negative slope, in which the slope was -0.0207. This should be
equal to -1/RC. The theoretical value of -1/RC in this case is
-1/(100000kΩ*0.00047F) = -0.0212
So, the slope of Graph 1 is very close to the theoretical value for this particular circuit.
To calculate the half-life for this RC circuit, one needs to use the quation
T 1 /2=τ ln ( 2 )
and since τ =RC ,
T 1 /2=RC∗ln ( 2 )
T 1 /2=¿ (100000kΩ*0.00047F)*ln(2)
T 1 /2=32.578 s
One can see from Table 1 that the half-life observed experimentally was 32.5 s, which is very
close to the theoretical value.

Conclusion & Discussion

At the center of this experiment was the capacitor and its properties. The function of a
capacitor is to obtain, hold, and store charge, especially in the case of short. It was seen in this
experiment that a capacitor can be charged up to the point where its voltage is the same as the
voltage across the battery. It was found that as the plates of a capacitor are charged up to that
point, it becomes more difficult to add more charge onto the plates, which is due to the repulsive
forces that occur between the positive charges attempting to be added and the already very
positive plate of the capacitor. It was also seen that even after disconnecting the battery from the
capacitor, the capacitor maintained its voltage for a period of time, which shows that capacitors
do indeed hold and store charge.
When studying capacitors in parallel circuits and series circuits, it was found that there
were certain trends regarding the voltage across the capacitors. In parallel circuits, it was
determined that the voltage across the battery and the voltages across each individual capacitor
were the same, regardless of the difference in capacitance. However, the charge on each
individual capacitor is not the same. The opposite can be said of capacitors in series, since each
capacitor holds the same charge but the voltages across the battery and the two capacitors are all
different. Moreover, the larger the capacitor in a series circuit, the smaller its voltage.
The last part of the experiment dealt with discharging a capacitor through a resistor.
Specifically, how a change in resistance would affect the time it takes for a capacitor to
discharge. It was found that the greater resistance a resistor has, the longer it will take for a
capacitor to discharge. This is due to the fact that greater resistance slows current, so a greater
resistance would also prevent the electrons from discharging quickly from the negative plate of a
capacitor. Finally, one can use the values for resistance and capacitance to calculate the half-life
of any capacitor.

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 Data

Table 1: Voltage vs. Time

Voltage (V) Time (s) ln (V)
8 0 2.08
4 32.5 1.39
2 66 0.69
1 99 0
0.5 133 -0.69
0.25 168 -1.39
This table shows the time in seconds it took for a capacitor to lose certain amounts of voltage.

Graph 1

Time vs. ln (V)

2.5

2
f(x) = - 0.02x + 2.06
1.5 R² = 1

0.5

0
0 20 40 60 80 100 120 140 160 180
-0.5

-1

-1.5

-2
This graph shows how voltage decays exponentially over time

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