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Find an expression for loss of electrical energy when two capacitors (or conductors ) maintained at different potential are allowed to share their charges . |
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Answer» Solution :Consider two capacitors having capacitances `C_1` and `C_2` and maintained at the potentials `V_1` and `V_2` respectively . Their TOTAL initial electrostatic energy is `U_(1) = (1)/(2) C_(1) V_(1)^(2) + (1)/(2) C_(2) V_(2)^(2)""... (i) ` On BRINGING them in electric constant potential acquired by them will be `V = (C_1 V_1 + C_2 V_2)/(C_1 + C_2)` `therefore` Net final electrostatic energy of the two capacitors `U_(2) = (1)/(2) (C_1 + C_2) V^(2) = (1)/(2) ((C_(1) V_1 + C_2 V_2)^(2))/((C_1 + C_2))` `therefore` LOSS of electrostatic energy on account of sharing of charge `Delta U = U_(1) - U_2 = 1/2 C_1 V_(1)^(2) + (1)/(2) C_(2) V_(2)^(2) - (1)/(2) ((C_(1) V_(1) + C_(2) V_(2))^(2))/((C_(1) + C_(2)))` `= (C_(1)^(2) V_(1)^(2) + C_(1) C_(2) V_(1)^(2) + C_(1) C_(2) V_(2)^(2) + C_(2)^(2)- (C_(1)^(2) + V_(1)^(2) + C_(2)^(2) V_(2)^(2) + 2 C_(1) C_(2) V_(1) V_(2)))/(2 (C_(1) + C_(2)))` `= (C_(1) C_(2) (V_(1)^(2) + V_(2)^(2) - 2V_(1) V_(2)))/(2 (C_(1) + C_(2))) = (C_(1) C_(2))/(2 (C_(1) + C_(2))) (V_(1) - V_(2))^(2)` This is term is always positive irrespective of the fact whether `V_(1) gt V_(2)` or `V_(1) lt V_(2)` |
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