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 =Ui−Uf=600μ J Hece, option (b) is the correct answer.

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