A capacitor of capacitance C is charged by connecting it to a battery of emf epsilon. The capacitor is now disconnected and reconnected to the battery with the polarity reversed. Calculate the heat developed in the connecting wires.
Text Solution
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From figure Net charge flow through battery = `=Q_("final")-Q_("initial")=epsilon C-(-epsilon C)=2 epsilon C` `:. ` work done by battery (W) `=QxxV=2 epsilon C xx epsilon =2 epsilon^(2)C` or Heat produced `= 2 epsilon^(2)C`
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