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In the circuit shown in Figure, initially K
1
is closed and K
2
is open. What are the charges on each capacitor. Then K
1
was opened and K
2
was closed (order is important), What will be the charge on each capacitor now? [C=1μF]
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Solution
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When K
2
is open and K
1
is closed the capacitors C
1
and C
2
will charge and potential develops across them i.e., V
1
and V
2
respectively which will be equal to the potential of battery 9V
∴V
1
+V
2
=9……..I
∵V=
C
q
=orVα
C
1
Or
V
2
V
1
=
C
1
C
2
V
2
V
1
=
6C
3C
3V
2
=6V
1
V
2
=2V
1
…II
From Eqns. I and II
V
1
+2V
1
=9
3V
1
=9
V
1
=3
Volt
V
2
=2×3Volt=6Volt
∴q
1
=C
1
V
1
=6C×3=18C[(from II C=1μF)
=18×1μF=18μc
q
2
=C
2
V
2
=3C×6
=3×1μF×6=18μC
So, charges on each capacitor i.e., q
1
=q
2
=18μc
When k
1
is open and k
2
is closed then charge q
2
will be distributed among
C
2
and C
3
. Let it be q
2
and q
3
∴q
2
=q
2
+q
3
As C
2
and C
3
are now in parallel combination so their potentials
remain same (V)
∴q
2
=C
2
V+C
3
V
18μC=3×1μFxV+3×1μF×V
18=6V
V=3Volt
So potential on C
2
and C
3
capacitors are 3 Volt each
q
2
′
=C
2
V−3×1μF×3Volt=9μC
q
3
=C
3
V=3×1μF×3Volt=9μC