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Physics 102
RC Circuits
March 7, 2006
Elizabeth Silva and Maria Zavala
Abstract:
This experiment will enable the student to expand on their knowledge of Ohm’s Law and
capacitors. They will discover the charging and discharging rates of a circuit.
Equipment:
● four resistors (100k ohms, 10k ohms, 1k ohms, 100 ohms)
● capacitor (5µF)
● power supply
● computer
● Pasco Datastudio with voltage sensors
Procedure:
1. Connect the series RC circuit shown below.
2. Set the power supply to 6 V.
3. Set up the Datastudio interface.
a.
b.
c.
d.
4.
5.
6.
7.
Choose the Datastudio icon.
Create New Experiment.
Sensors, choose Voltage sensor.
Double click Voltage sensor icon.
1. Sensitivity
A. Choose : Low (1x)
2. Sample rate: 100 Hz
e. Double click on the options button under Experiment
1. Sampling Options
A. Delayed Start
a. Data measurement: Rise above; 1.000V
f. Repeat steps d and e, to add a second Voltage sensor.
g. Choose graphs in the displays box.
1. Select Voltage Ch. A.
h. Repeat step g for the second probe.
Discharge the capacitor, by flipping the switch from point A to point B. Make sure to
leave in that position for least 10 seconds to fully discharge the capacitor.
Click the Start button on the menu bar and turn the power supply on.
Stop when the graph flattens out and turn the power supply off.
Repeat steps 5d through 8 for the other three resistors.
a. Change the sample rate for each resistor.
1. 100K Ω = 100 Hz
2. 10K Ω = 100 Hz
3. 1K Ω = 1000Hz
4. 100 Ω = 10000Hz
Data:
See attached Excel spreadsheet.
Calculations:
Theoretical value for capacitor Voltage values at t = 0.01 s
Vcap = V0 ( 1 - e-t/RC )
Vcap = 6V ( 1 - e-0.01/100000Ohms * 0.000005F )
Vcap = 1.087615V
Theoretical value for resistor Vres= 6V - Vcap
Vres= 6V - 1.087615V
Vres= 4.912385V
Experimental value VKirchhoff’s Law exp. -
VKirchhoff’s Law exp = Vcap + Vres
VKirchhoff’s Law exp = 1.573V + 4.452V
VKirchhoff’s Law exp = 6.025V
Theoretical value VKirchhoff’s Law theo. –
VKirchhoff’s Law theo. = Vcap + Vres
VKirchhoff’s Law theo. = 1.087615 V + 4.912385 V
VKirchhoff’s Law theo. = 6V
Graphs:
See attached Excel spreadsheet
Error Analysis:
The standard deviation +/- 5% for the resistors was not considered for this experiment.
Nor was the standard deviation of the 6V from the power source. As well, as the loss of
potential energy from the wires used to connect the components. All these factors
contributed to the errors in this lab.
In addition, the graphs indicate some noise from the power source. This is more evident
in the 100 ohms run. The values fluctuated significantly at the start of the run.
Percent error:
│theoretical value – actual value │ x
Theoretical value
100 = % error
VC 100K Ω at t = 0.5 s
│ 0.57097 – 1.511
│ x 100 = 165 %
0.57097
VR 100K Ω at t = 0.5 s = 16%
VC 10K Ω at t = 0.05 s = 9.8%
VR 10K Ω at t = 0.05 s = 14%
VC 1K Ω at t = 0.005 s = 4.4%
VR 1K Ω at t = 0.005 s = 5.5%
VC 100 Ω at t = 0.0005 s = 14.4%
VR 100 Ω at t = 0.0005 s = 1.3%
Questions:
1. What is the average value of V/Vo for each graph at t=RC?
First find t for each graph:
For VC t = 100K Ω * 5μF = 0.5 seconds
For VR t = 100K Ω * 5μF = 0.5 seconds
For VC 10K Ω
For VR 10K Ω
For VC 1K Ω
For VR 1K Ω
For VC 100 Ω
For VR 100 Ω
t = 0.05 seconds
t = 0.05 seconds
t = 0.005 seconds
t = 0.005 seconds
t = 5.0E-4 seconds
t = 5.0E-4 seconds
Then at the time constant t = RC for each VC and VR:
Sample equation:
For 100KΩ:
VC = 1.511V ; Vo = 6
V / Vo = 1.511V / 6V = 0.25
For VR 100K Ω V / Vo
For VC 10K Ω V / Vo
For VR 10K Ω V / Vo
For VC 1K Ω V / Vo
For VR 1K Ω V / Vo
For VC 100 Ω V / Vo
For VR 100 Ω V / Vo
= 0.76
= 0.69
= 0.32
= 0.66
= 0.35
= 0.54
= 0.37
2. Was Kirchhoff’s Law verified? What might account for the variations?
No Kirchhoff’s Law was not verified, the sum of the voltages going in did not
equal exactly the voltages going out. For the most part they were off by 100th of a volt.
The variations may be from the standard deviation of the resistors, which is +/- 5%.
There was also the standard deviation of the power supply, which was not known. But the
number was not taken into account. The loss of potential energy from the wires that
connects everything together needs to be considered, when calculating the percent error.
3. Describe the charging and discharging voltage curves for the capacitor and resistor in a
parallel RC circuit.
In a parallel RC circuit you can have two scenarios, one with a power source connected
and two with no power source connected.
One with power source, charging capacitor: The flow of electrons will follow the easiest
path, the path with the least resistance. That path with the least resistance is the one with
the capacitor. The capacitors’ voltage curve at this point is increasing and the resistors’
voltage curve is decreasing. As the capacitor continues to charge the current meets
resistance. The capacitors’ voltage curve at this point begins to decrease and the
resistors’ voltage curve begins to increase.
Two no power source, discharging capacitor: There is only one path the electrons follow,
from the capacitor to the resistor. The voltage curve for the capacitor will decrease, while
the voltage curve for the resistor will increase. These events are all happening
simultaneously.
Conclusion:
This experiment concluded for t = RC:
100K ohms
10K ohms
1K ohms
100 ohms
t = 0.5 s
t = 0.05s
t = 0.005s
t = 0.0005s
Vc = 1.511V
Vc = 4.165V
Vc = 3.960V
Vc = 3.247V
You need more in the conclusion.
Grade: 98
VR = 4.557V
VR = 1.896V
VR = 2.085V
VR = 2.236V