CH26 Capacitors Lab
In this experiment we will determine how voltages are distributed in capacitor circuits and explore
series and parallel combinations of capacitors. 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).
There are two ways to connect capacitors 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:
C2
C1
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, Ceq, from
Q · 1/Ceq = V = V1 + V2 = Q/C1 + Q/C2 = Q [1/C1 + 1/C2]
(3)
So
SERIES:
1/Ceq = 1/C1 + 1/C2
(4)
In the parallel connection, the components are connected together at both ends as shown below:
C1
C2
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:
CeqV = Q = Q1 + Q2 = C1V + C2V = [C1 + C2] V
So
PARALLEL:
CeqC1 + C2
NOTE: Here we will use AC, the voltage is actually V = I/C, where I is the 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.
(5)
(6)
Procedure
1. Turn on the power supply and
set the AC voltage to 10 V.
Measure the accurate voltage with
the multimeter and record it below:
Vac = ___________________V
2. Connect two 0.1 F capacitors
in series. Measure V2 (across C2)
and record it below.
V2 (measured) = ____________ V
3. Compute the expected value of
V2 using Vac, the values of C1 and
C2 with equations 3 and 4.
V2 (expected) = ____________ V
% difference = |measured expected| / measured x 100 % =
__________
4. Connect a third 0.1 F capacitor in parallel with C2.
Compute their equivalent capacitance.
Ceq = _________ F.
Measure and compute the voltage across the equivalent capacitance.
Veq (measured) = ___________ V, Veq (expected) = _________V, % difference = __________
5. Now remove the third capacitor and replace it with a 0.01 F capacitor.
Compute their equivalent capacitance.
Ceq = _________ F.
Measure and compute the voltage across the equivalent capacitance.
Veq (measured) = ___________ V, Veq (expected) = _________V, % difference = __________
6. Now connect the 0.1 F and the 0.01 F capacitor in series. Compute the equivalent capacitance.
Ceq = _________ F.
Measure and compute the voltage across the equivalent capacitance.
Veq (measured) = ___________ V, Veq (expected) = _________V, % difference = __________
7. This method can be used to find an unknown capacitance. Replace C1 with the unknown value
capacitor and determine its capacitance by measuring V2 and using equations 3 and 4.
V1 = ___________ V ,
C1 = _________ F.