Calculate the charge on C3=5uf in the given circuit
Answers
Answer:
If we look at the given figure, we can see C
1
and C
2
are in series, so the resultant capacitance for this combination is given by C
4
⇒
C
4
1
=
C
2
1
+
C
1
1
⇒C
4
=2μF
now the circuit will be look like as new drawn figure. now we can see C
4
is in Parallel with C
3
, So the equivalent capacitance between A and B is given by C
AB
⇒C
AB
=C
3
+C
4
⇒C
AB
=10μF+2μF=12μF
so the total charge flowing through the battery is Q
total
⇒Q
total=
C
AB
V
⇒Q
total=
12×50=600μC
Assume that charge flowing in chain 1 is Q
1
and in chain 2 is Q
2
and we can see both the chains are in parallel with each other then the potential difference will be also same at the ends.
⇒Q
total
=Q
2
+Q
1
=600μC...(1)
⇒
C
4
Q
1
=
C
3
Q
2
⇒
2
Q
1
=
10
Q
2
⇒Q
2
= 5Q
1
from equation (1) and (2)
⇒Q
1
=100μC and Q
2
=500μC
now the charge on C
1
will be q
1
equal to Q
1
because in a series of capacitors same amount of charge will flow so the charge on C
1
is
⇒q
1
=100μC