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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 μJBefore 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=80−20=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=Ui−Uf ⇒H=900−300=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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