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A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration isA. zeroB. `(3)/(2) CV^(2)`C. `(35)/(6) CV^(2)`D. `(9)/(2) CV^(2)` |
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Answer» Correct Answer - B Net charge `Q =Q_(2)-Q_(1)` potential is `V_(l)` `:. V_(1)=((C_(0))/(C+C_(0)))V_(0)` Similarly after nth operation , `E = 1//2C^(1) V^(2)`. |
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