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Question

A 2μ F capacitor is charged to a potential 10 V. Another 4μ F capacitor is charged to a potential 20 V . The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. Find heat developed in the circuit.

A
300μJ
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B
600μJ
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C
900μJ
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D
450μJ
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Solution

The correct option is B 600μJ

Before connection , charges in each capacitor are given by Q=CVQ1=2×10=20μC and Q2=4×20=80μC

And energy stored in capacitors before the connection is given by
Ui=12C1V21+12C2V22
Ui=12×2×(10)2+12×4×(20)2=900μJ

When the capacitors are connected,


Equivalent capacitance of circuit is ,Ceq=2+4=6μ F

And the net charge of circuit is
Qnet=20+80=60μ C

Now the common voltage of capacitors is given by
V=QnetCeq

V=602+4=10 V

Energy stored in circuit is given by , Uf=12CeqV2

Substituting the data given in the question we get,

Uf=12×6×(10)2=300μJ

Heat generated in circuit is
=UiUf=600μ J

Hece, option (b) is the correct answer.

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