    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|=Uf−Ui=600μJ Uf for 2μF capacitor =12×2×(10)2=100μJ Uf for 4μF capacitor =12×4×(10)2=200μJ  Suggest Corrections  0      Similar questions  Related Videos   PHYSICS
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