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Question

A 2 μF capacitor is charged to a potential difference of 10 V. Another 4 μF capacitor is charged to a potential difference of 20 V. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of another. What is the amount of heat dissipated in the circuit ?

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

The correct option is B 600 μJ
Before connection:


Initial charges on capacitors are,

Q1=C1V1=20 μC

Q2=C2V2=80 μC

When opposite polarity plates are connected, each plate will have a charge
Q=8020=60 μC (Charge conservartaion)


Let the potential difference be V across plates after connection completes.

Thus, the equivalent capacitor has 60 μC charge on it.

60 μC=Ceq×V

Since, capacitors make parallel combination

60 μC=(4+2) μF×V

V=10 V

Initial potential energy of system,

Ui=12C1V21+12C2V22

Ui=(12×2×100)+(12×4×400)

Ui=900 μJ

Final potential energy stored in the system is,

Uf=12Ceq(V)2

Uf=12×6×100=300 μJ

Heat dissipated in circuit:

H=UiUf

H=900300=600 μJ

Hence, (b) is correct.

Why this question?Tip:In such problems the system after connectioncan be visualized as an equivalent capacitorwith +ve plate indicating where net charge is coming to be positive.

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