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Question

Three capacitors 6 μF, 3 μF and 4 μF are connected to a battery of emf 12V as shown in the Figure. If the point b is grounded, then :

44163.jpg

A
the charge on 3 μF capacitor is 24 μC.
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B
the potential at a is 4V.
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C
the potential at c is 8V.
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D
the energy dissipated in the battery is 432 μJ
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Solution

The correct options are
A the potential at a is 4V.
B the energy dissipated in the battery is 432 μJ
C the charge on 3 μF capacitor is 24 μC.
D the potential at c is 8V.
Potential diff across ac is Vac=12V
We know when capacitors are in series charge present on them must be equal.
charge on 6μF=chargeon3μF
q6μF+q3μF=12
q(36μF)=12
q=24μFC
Vdiff across 6μF capacitor VaVb=24×1066×106
Va=4.
Vdiff across 3μF capacitor V0Vc=24×1063×106=8.Vc=8
total charge flown from battery =72μc.
work done by battery
=72×106×12
=864μJ
energy stored =12CV2
=12(12)2×(4+2)
=3×44=432.
energy dissipated =864432=432μJ.

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