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A capacitor of capacitance C_1 is charged to a potential V_1 , while another capacitor of capacitance C_2 is charged to a potential difference V_2 . The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other Find the total energy stored in the two capacitors before they are connected . |
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Answer» Solution :INITIALLY capacitor of capacitance `C_1` is charged to a POTENTIAL `V_1` , hence energy stored in it `U_(1) = (1)/(2) C_(1) V_(1)^(2)` . Similarly for SECOND capacitor of capacitance `C_2` charged to a potential `V_2` the stored energy `U_2 = (1)/(2) C_(2) V_(2)^(2)` `therefore` Total electric energy stored in the COMBINATION before they are connected `U_(2) = U_(1) + U_(2) = (1)/(2) [C_(1) V_(1)^(2) + C_(2) V_(2)^(2)]` |
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