Questions
A capacitor of capacitance 160 F is charged to a potential difference of 200 V and then connected across a discharge tube which conducts until the potential difference across it has fallen to 100 V. The energy dissipated in the tube is
detailed solution
Correct option is D
E=12CV2−V′2=12×160×10−6(200)2−(100)2=12×160×10−6×3×104=2.4JTalk to our academic expert!
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Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
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