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A capacitor having initial charge `q_0 = CE//2` is connected to a cell of emf E through a resistor R as shown. Find the total heat generated in the circuit after the switch S is closed. A. `(1//12) CE^2`B. `(1//8)CE^2`C. `(1//4)CE^2`D. none of these |
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Answer» Correct Answer - B b. Final charge on capacitor `q_(f) = CE` Charge flown `= CE-CE//2 = CE//2` Work done by battery: `W_(b) = E(CE//2) = 1/2 CE^2` Change in energy of capacitor is `DeltaU = 1/(2C)[q_(f)^(2) - q_(0)^(2)]` ` = 1/(2C) [(CE)^2 - (CE//2)^2] = (3CE^2)/8` Heat `= W_(b) - Delta U = 1/8 CE^2` . |
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