C
2
PH106. Experiment 3: Capacitors Lab
Background: The capacitance is a measure of a device’s ability to store charge. Capacitors are passive electronic devices which have fixed values of capacitance and negligible resistance. The capacitance, C, is the charge stored in the device, Q, divided by the voltage difference across the device,

V:
C = Q/

V. (1)
The SI unit of capacitance is the farad, 1 F = 1 C/V, In general, the capacitance can be calculated knowing the geometry of the device. For most practical devices, the capacitor consists of capacitor plates which are thin sheets of metal separated by a dielectric, insulating material. For this reason, the schematic symbol of a capacitor is has two vertical lines a small distance apart (representing the capacitor plates) connected to two lines representing the connecting wires (or leads).
Sometimes the sheets are rolled up in a spiral to increase the overall area while keeping the device compact. The capacitance of such a device is:
C =

 A/d. (2)
Where

is the dielectric constant of the space material,

 is the permittivity of free space, A is the surface area of the metal sheets, and d is the thickness of the dielectric. The capacitors used in this lab use mylar sheet as a dielectric with

= 3.2 and a thickness d = 10
 m.
Hypothesis: Combining capacitors in series or parallel will result in an equivalent capacitance given by the formulae below. To prove this, we will measure voltages across capacitive circuits and from the definition of capacitance determine the equivalent capacitance of various combinations.
Experiment: There are two ways to connect two passive (no polarity) components in an electronic circuit—series or parallel connection. In a series connection the components are connected at a single point, end to end as shown below:
C
1
C
2
For a series connection, the charge on each capacitor will be the same and the voltage drops will add.
We can find the equivalent capacitance, C eq
, from
Q/C eq
=

V =

V
1
+

V
2
= Q/C
1
+ Q/C
2
(3)
So
1/C eq
= 1/C
1
+ 1/C
2
(4)
In the parallel connection, the components are connected together at both ends as shown below:
C
1
C
2
For a parallel connection, the voltage drops will be the same, but the charges will add. Then the equivalent capacitance can be calculated by adding the charges:
C eq

V = Q = Q
1
+ Q
2
= C
1

V + C
2

V
So
(5)
C eq

C
1
+ C
2
(6)

Equipment: The experiment requires an AC power supply, a capacitor assortment, banana wires, a set of meter probes and a handheld multimeter.
Testing the Hypothesis:
Turn on the power supply and set the AC voltage between 5 and 10 V.
Measure this voltage with the multimeter and record it as V
A
.
Two capacitors in series
Connect two 0.1

F capacitors in series, with the connecting point at the meter, V
2
, and the second capacitor connected to V
A
. Measure V
2
with the meter and record it below.
Compute the expected value of V
2
using the measured value of the power supply voltage, the nominal values of the capacitors C
1
and C
2
with equations 3 and 4 (the meter is measuring the voltage across C
2
).
C eq
as a parallel combination of C
2
and C
3
Case 1: C
2
= C
3
Connect a third 0.1

F capacitor in parallel with C
2
. What is the equivalent capacitance, C eq
, of the parallel combination of C
2
and C
3
?
Measure and compute the voltage across the equivalent capacitance.
Case 2: C
2
= 10C
3
Now remove the third capacitor and replace it with a 0.01

F capacitor and repeat the measurement.
C eq
as a series combination of C
2
and C
3
Case 1: C
2
= C
3
Connect two 0.1

F capacitor in series as C
2
and C
3
. What is the equivalent capacitance now?
Measure and compute the voltage across the equivalent capacitance.
Case 2: C
2
= 10C
3
Now remove the third capacitor and replace it with a 0.01

F capacitor and repeat the measurement.
NOTE: Since we used AC, the measured voltage is actually the rms voltage,

V = I/

C, where I is the rms current and

is the angular frequency. We don’t actually measure I or

here, and the analysis is the same. This is covered in more detail in the chapter on AC circuits.
Fig. 1. Determining the power supply voltage, V
A
.
Fig. 2. Measuring V
2
.
Fig. 3. C
2
and C
3
in Parallel.
Fig. 4. C
2
and C
3
in Series

Names:__________________________________________________________________________
Experiment 3. Capacitors Lab. Laboratory Report Form.
Two capacitors in series
1. V
A
= ___________________V
2. V
2
(measured) = ___________________ V
3. V
2
(expected) = ___________________ V
4. % difference = |measured - expected|/measured x 100 % = __________
C eq
as a parallel combination of C
2
and C
3
C
2
= C
3
5. C eq
= _________

F.
6. V eq
(measured) = ___________ V
7. V eq
(expected) = _________V
8. % difference = __________
C
2
= 10 C
3
9. C eq
= _________

F.
10. V eq
(measured) = ___________ V
11. V eq
(expected) = _________V
12. % difference = __________
C eq
as a series combination of C
2
and C
3
C
2
= C
3
13. C eq
= _________

F.
14. V eq
(measured) = ___________ V
15. V eq
(expected) = _________V
16. % difference = __________
C
2
= 10 C
3
17. C eq
= _________

F.
18. V eq
(measured) = ___________ V
19. V eq
(expected) = _________V
20. % difference = __________
Question 1: Are the measured voltages consistent with the given formulae for calculating equivalent capacitance?
Question 2: When two capacitors are connected in parallel and one of the capacitors is much larger than the other, the equivalent capacitance is close to the value of the (larger or smaller) capacitor.
Question 3: When two capacitors are connected in series and one of the capacitors is much larger than the other, the equivalent capacitance is close to the value of the (larger or smaller) capacitor.