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```					                         SKH St. Mary’s Church Mok Hing Yiu College
Physics Experiment
Form 6S

Student Name : _________________ ( )                                           Date : ____________
Group Number : _____
Marks : ___________
E19 Capacitors in Series and in Parallel

Objective : To measure the capacitance of some real capacitors in series and in parallel
combinations using a reed switch.
Apparatus : Capacitors (1 F. 2.2 F)                                     2
Reed switch                                                   1
Signal generator (low impedance)                              1
d.c. voltmeter (0 – 10 V)                                     1
d.c. micrometer(-10 – 100 A) with 10 mA shunt                1
Battery holder with 4  1.5 V dry cells                       1
Variable resistor (0 – 5 k)                                  1
Theory : Capacitors are one of the standard components of electronic circuits. Complicated
combinations of capacitors often occur in practical circuits. It is, therefore, useful to
have a set of rules for finding the equivalent capacitance of some general
arrangement of capacitors. It turns out that we can always find the equivalent
capacitance by repeated application of two simple rules. These rules related to
capacitors connected in series and in parallel.
Consider two capacitors connected in parallel; i.e., with the positively charged
plates connected to a common input wire and the negatively charged plates attached
to a common output wire. What is the equivalent capacitance between the input and
output wires?                            C1
+

+   C2
In this case, the potential difference V across the two capacitors is the same, and
is equal to the potential difference between the input and output wires. The total
charge Q, however, stored in the two capacitors is divided between the capacitors,
since it must distribute itself such that the voltage across the two is the same. Since
the capacitors may have different capacitances, C1 and C2, the charges Q1 and Q2
may also be different. The equivalent capacitance Ceq of the pair of capacitors is
simply the ratio Q/V, where Q = Q1 + Q2 is the total stored charge. It follows that
Q Q  Q2 Q1 Q2
Ceq   1                                          (1)
V        V        V     V
giving         Ceq = C1 + C2                                        (2)
Here, we have made use of the fact that the voltage V is common to all three
capacitors. Thus, the rule is:
The equivalent capacitance of two capacitors connected in parallel is the sum of the
individual capacitances.
For N capacitors connected in parallel, Eq. (2) generalizes fairly obviously to
N
C eq   C i
i 1

E19 - Capacitors in series and in parallel - 1
Consider two capacitors connected in series; i.e., in a line such that the positive
plate of one is attached to the negative plate of the other. In fact, let us suppose that
the positive plate of C1 is connected to the input wire, the negative plate of C1 is
connected to the positive plate of C2, and the negative plate of C2 is connected to the
output wire. What is the equivalent capacitance between the input and output
wires?
C1       C2
+   +

In this case, it is important to realize that the charge Q stored in the two capacitors is
the same. This is most easily seen by considering the internal plates; i.e., the
negative plate of C1 and the positive plate of C2. These plates are physically
disconnected from the rest of the circuit, so the total charge on them must remain
constant. Assuming, as seems reasonable, that these plates carry zero charge when
zero potential difference is applied across the two capacitors, it follows that in the
presence of a non-zero potential difference the charge +Q on the positive plate of C2
must be balanced by an equal and opposite charge -Q on the negative plate of C1.
Since the negative plate of C1 carries a charge -Q, the positive plate must carry a
charge +Q. Likewise, since the positive plate of C2 carries a charge +Q, the negative
plate must carry a charge -Q. The net result is that both capacitors possess the same
stored charge Q. The potential drops, V1 and V2, across the two capacitors are, in
general, different. However, the sum of these drops equals the total potential drop V
applied across the input and output wires; i.e., V = V1 + V2. The equivalent
capacitance of the pair of capacitors is again Ceq = Q/V. Thus,
1    V V  V2 V1 V2
  1                                         (3)
Ceq Q           Q       Q Q
1      1     1
giving                                                      (4)
C eq C1 C 2
Here, we have made use of the fact that the charge Q is common to all three
capacitors. Hence, the rule is:
The reciprocal of the equivalent capacitance of two capacitors connected in series
is the sum of the reciprocals of the individual capacitances.
For N capacitors connected in series, Eq. (4.16) generalizes to
N
1       1

Ceq i 1 Ci
Procedure : (A) Measuring the capacitance of a single capacitor
1. Connect the circuit as shown below.

Signal
generator
V

1F                  mA

E19 - Capacitors in series and in parallel - 2
2. Adjust the variable resistor to its maximum resistance value.
3. Connect the reed switch to the low impedance output of the signal generator.
Set the frequency of the signal generator to 100 Hz. Adjust the output of the
signal generator such that a buzzing sound can just be heard.
4. Reduce the resistance of the variable resistor gradually until the current does
not increase any more.
5. Record the readings of the microammeter, voltmeter and the frequency of
the signal generator.
6. Without changing the potential difference, repeat the experiment by
changing the frequency of the signal generator in steps from 100 Hz to
300 Hz. Tabulate the results.

(B)    Measuring the equivalent capacitance of two capacitors in series
7. Repeat the experiment by connecting two capacitors in series as shown below.

Signal
generator
V

1F                       mA
2.2  F

(C)    Measuring the equivalent capacitance of two capacitors in parallel.
8. Repeat the experiment by connecting two capacitors in parallel as shown
below.

Signal
generator
V

mA
1F           2.2  F

Results : (A) Measuring the capacitance of a single capacitor
Voltmeter reading V = ____________ V

Frequency of signal generator f /Hz
Capacitance C1           I     /F
(      )
fV

Mean value of C1 /F

E19 - Capacitors in series and in parallel - 3
1.   Plot a graph of the I against f. How is the current related to the frequency?

______________________________________________________________
2.   Calculate the slope of the graph. From the graph, determine the capacitance of
the capacitor.
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
3.   Compare the measured value of the capacitance of the capacitor with that
marked on the capacitor. Account for any discrepancy.
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
4.   Repeat the procedures 1 to 6 to find the capacitance of the other capacitor
provided.
Voltmeter reading V = ____________ V

Frequency of signal generator f /Hz
Capacitance C2           I     /F
(      )
fV

Mean value of C2 /F

E19 - Capacitors in series and in parallel - 4
5.   Plot a graph of the current I against f and hence find the capacitance of the
capacitor.

______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________

(B) Measuring the equivalent capacitance of two capacitors in series
Voltmeter reading V = ____________ V

Frequency of signal generator f /Hz
Capacitance Ceq          I       /F
(      )
fV

Mean value of Ceq /F

1. Calculate the theoretical value of the equivalent capacitance of the combination.
[Remark : 1  1  1 , where C1 and C2 are the values obtained from (A)]
C eq     C1   C2

_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________

E19 - Capacitors in series and in parallel - 5
2.   Compare the theoretical value of the equivalent capacitance with the measured
value. Account for any discrepancy.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
(C) Measuring the equivalent capacitance of two capacitors in parallel
Voltmeter reading V = ____________ V

Frequency of signal generator f /Hz
Capacitance Ceq          I     /F
(      )
fV

Mean value of Ceq /F

1. Calculate the theoretical value of the equivalent capacitance of the combination.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
2.   Compare the theoretical value of the equivalent capacitance with the measured
value. Account for any discrepancy.
_______________________________________________________________
_______________________________________________________________
_______________________________________________________________
Discussion : 1. State the main precautions in the above experiments.
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
2. The variable resistor in the discharge circuit is used to protect the
microammeter. The value of this resistor is therefore made as large as
possible, but there is a constraint on the maximum value which can be used.
What is this constraint?
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________

E19 - Capacitors in series and in parallel - 6

```
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