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Question

A 2μF capacitor is charged to a potential 10 V. Another 4μF 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. Then,

A
Final energy in 2μF capacitor is 100 μJ
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B
The net amount heat of dissipated in the circuit is 600 μJ
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C
Final energy in 4μF capacitor is 200 μJ
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D
The net amount of heat dissipated in the circuit is 200μJ
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Solution

The correct option is C Final energy in 4μF capacitor is 200 μJ
Before connection
Q1=2×10=20μC
Q2=4×20=80μC

Ui=122(10)2+124(20)2=900μJ

Since connected as shown

After connection,
Qnet =20+80=60μC
V=602+4=10Volt
Uf=12×6×(10)2=300μJ
|ΔU|=UfUi=600μJ

Uf for 2μF capacitor =12×2×(10)2=100μJ

Uf for 4μF capacitor =12×4×(10)2=200μJ

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