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 options are A Final energy in 2μF capacitor is 100 μJ B The net amount heat of dissipated in the circuit is 600 μJ 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