C
T
=
C
1
C
2
C
1
+C
2
C
T
=
2×3
2+3
μF
C
T
=1.2 μF
Step II: To find the amount of charge, Q
Given, C= 1.2μF=1.2× 10
-6
F
Charge, Q is given by
Q= C×V
= 1.2× 10
-6
F×10V
= 1.2× 10
-5
C
Since the capacitors are in series, charges on
the plates are equal and are 1.2× 10
-5
C
Example 03
A certain electronic circuit has three
capacitors connected in series, if the charge,
Q is stored in each plate of the capacitors.
Find the value of the capacitance of one
capacitor that can be used instead of using
the three capacitors.
Solution
Step 1: To find the total potential difference,
V.
V= V
1
+ V
2
+V
3
Step 2: To find the p.d on each plate by
using the same (equal amount of) charge.
V=
Q
C
T
,V
1
=
Q
C
1
, V
2
=
Q
C
2
and V
3
=
Q
C
3
Step 3: To substitute the value above in step
one
Q
C
T
=
Q
C
1
+
Q
C
2
+
Q
C
3
Step 4: Fact out Q on either side
Q
1
C
T
= Q
1
C
1
+
1
C
2
+
1
C
3
Step 5: If we divide by “Q” both sides, we
get.
1
C
T
=
1
C
1
+
1
C
2
+
1
C
3
The total capacitance is given by:
C
T
=
C
1
×C
2
+
C
2
×C
3
+
C
1×
C
3
C
1
×C
2
×C
3
Example 04
What value of capacitor could be used to
replace a set of 5μF, 10μF and
15μFcapacitors connected in series?
Solution
1
C
T
=
1
C
1
+
1
C
2
+
1
C
3
1
C
T
=
1
5
+
1
10
+
1
15
1
C
T
=
6+3+2
30
1
C
T
=
11
30
C
T
=
30
11
μF
C
T
= 2.73μF
The value of a single capacitor that could
replace them is 2.73μF capacitor.
Capacitors in parallel
One capacitor of a certain value can be
replaced by a number of capacitors joined
together in parallel to give the same
effective capacitance as shown below.
Points to note
When capacitors are in parallel, the potential
difference, V across the plates is the same
but each capacitor receives different
charges.
If more than one capacitor are arranged in
parallel, the effective capacitance is very
large (greater than the value of the largest
capacitance)
The effective capacitance in series is given
by:
C
T
= C
1
+ C
2
+C
3
How to derive the formula
Step 1: To find the total charges in the
system
Q
T
= Q
1
+ Q
2
+Q
3
Step 2: To find the charge on each plate by
using the same potential difference, V.
Q
T
=C
T
V, Q
1
=C
1
V, Q
2
=C
2
V, Q
3
=C
3
V
Step 3: Substitute the values into equation 1
C
T
V= C
1
V+C
2
V+ C
3
V
Step 4: Fact out “V” both sides
V
C
T
=V
C
1
+C
2
+ C
3
Step 5: If we divide by “V” both sides we
get:
C
T
= C
1
+C
2
+ C
3
Example 01
The figure below shows two capacitors in a
circuit. Work out for their effective
capacitance.
Solution
The two capacitors are arranged in parallel,
hence
C
T
= C
1
+C
2
C
T
= 4μF+3μF
C
T
= 7μF
The effective capacitance is 7μF
Example 02
An electric circuit has three capacitors
arranged in parallel to a cell. If the potential
difference, V across each plate is the same,
calculate the total capacitance.
Solution
Step 1: To find the total charges in the
system
Q
T
= Q
1
+ Q
2
+Q
3
Step 2: To find the charge on each plate by
using the same potential difference, V.
Q
T
=C
T
V, Q
1
=C
1
V, Q
2
=C
2
V, Q
3
=C
3
V
Step 3: Substitute the values into equation 1
C
T
V= C
1
V+C
2
V+ C
3
V
Step 4: Fact out “V” both sides
V
C
T
=V
C
1
+C
2
+ C
3
Step 5: If we divide by “V” both sides we
get:
C
T
= C
1
+C
2
+ C
3
Example 03
What value of capacitor could be used to
replace a set of 5μF, 10μF and
15μFcapacitors connected in parallel?
Solution
The three capacitors are arranged in parallel,
hence
C
T
= C
1
+C
2
+ C
3
C
T
= 5μF+10μF+ 15μF
C
T
= 30μF
The single capacitor to replace the three is a
capacitor of value 30μF
Example 04
Three capacitors of capacitance values
5μF, 10μF and 15μF are available in the
laboratory, with the aid of diagram, explain
how they should be connected in order to
obtain:
(a) The maximum value of capacitance
Solution
To obtain the maximum value of the
capacitance, the three capacitors should
be connected in parallel as shown
below.
The effective resistance can be obtained
as follows.
C
T
= C
1
+C
2
+ C
3
C
T
= 5μF+10μF+ 15μF
C
T
= 30μF
(b) The minimum value of the capacitance
Solution
To obtain the minimum value of the
capacitance, the two capacitors should
be connected in series as shown below.
The effective value of the capacitance
can be obtained as follows.
1
C
T
=
1
C
1
+
1
C
2
+
1
C
3
1
C
T
=
1
5
+
1
10
+
1
15
1
C
T
=
6+3+2
30
1
C
T
=
11
30
C
T
=
30
11
μF
C
T
= 2.73μF
Example 05
Three capacitors of values 2μF, 3μF and
6μF are connected in series and then in
parallel. What is the equivalent capacitance
in each case?
Solution
CaseI: Series connection
1
C
T
=
1
C
1
+
1
C
2
+
1
C
3
1
C
T
=
1
2
+
1
3
+
1
6
1
C
T
=
3+2+1
6
1
C
T
=
6
6
C
T
= 1μF
Case II: Parallel connection
C
T
= C
1
+C
2
+ C
3
C
T
= 2μF+3μF+ 6μF
C
T
= 11μF
Combining parallel and series connection
If both series and parallel connections are
combined, we first deal with the
combination which has many capacitors and
join them together in order to simplify the
circuit.
Example 01
Work out for the effective capacitance in the
circuit below
Solution
Step 1: Consider the parallel circuit of 1μF
and 2μF capacitors only.
C
p
= C
1
+C
2
C
p
= 1μF+2μF
C
p
= 3μF
Step 2: Join the 3μF and the 6μF together,
and the new circuit will be in series. Then:
