find equivalent capacitance between point a and b in following cases PLEASE SOLUTION WITH EXPLANATION
Answers
Answer:
Part (a)
Here, we need to calculate the equivalent capacitance between the points 'a' and 'b'.
The capacitors with capacitance
C
and
3.00
μ
F
are in series. This combination is further in parallel with
6.00
μ
F
capacitor. The entire combination is in series with
20
μ
F
capacitor.
Equivalent capacitance of the capacitors with capacitance
C
and
3.00
μ
F
is given by:
1
C
s
=
1
C
+
1
3
or,
C
s
=
3
C
3
+
C
μ
F
Equivalent capacitance of the combination
C
s
and the capacitor with capacitance
6.00
μ
F
is given by:
C
p
=
C
s
+
6
μ
F
=
3
C
3
+
C
μ
F
+
6
μ
F
=
9
C
+
18
C
+
3
μ
F
Equivalent capacitance between points 'a' and 'b' is given by:
1
C
e
q
=
1
C
p
+
1
20
or,
C
e
q
=
20
C
p
20
+
C
p
μ
F
=
180
C
+
360
29
C
+
78
μ
F
Hence, the equivalent capacitance between points 'a' and 'b' is
180
C
+
360
29
C
+
78
μ
F
.
Part (b)
We have the formula:
Q
=
C
V
Here,
Q
= Net charge in the circuit
C
=
C
e
q
=
180
C
+
360
29
C
+
78
μ
F
Potential difference across the ends a and b
(
V
)
=
16.0
V
Therefore,
Q
=
180
C
+
360
29
C
+
78
μ
F
×
16.0
V
=
2880
(
C
+
2
)
29
C
+
78
μ
C
Hence the net charge in the circuit is
2880
(
C
+
2
)
29
C
+
78
μ
C
.
In case of charges, charge across the capacitors remain the same when they are in series combination. However, the charges are divided when the capacitors are in parallel combination.
We need to find the charge on the capacitors having capacitance
C
and
3
μ
F
.
The charge on combination
C
p
and
20
μ
F
capacitor will be the same as they are in series.
Therefore, charge on
C
p
, i.e.
Q
p
=
2880
(
C
+
2
)
29
C
+
78
μ
C
Note that when capacitors are in parallel combination, the charge gets distributed such that the charge in one branch will be a fraction of the net charge. This fraction will be equal to the net capacitance of one branch as a part of the sum of the capacitance of all the branches.
So, net charge on the combination
C
s
is given by:
Q
s
=
Q
p
C
s
C
s
+
6
=
2880
(
C
+
2
)
29
C
+
78
×
3
C
3
+
C
3
C
3
+
C
+
6
μ
C
=
960
C
29
C
+
78
μ
C
This charge stays same across the capacitors have capacitance
C
and
3.00
μ
F
.
Therefore,
Charge on capacitor
C
= Charge on
3.00
μ
F
capacitor =
960
C
29
C
+
78
μ
C