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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 parallel combination of two capacitors. |
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Answer» Solution :On DISCONNECTING the capacitors from their respective batteries and JOINING them in parallel , the capacitors acquire a COMMON potential V , which is given as : `V = ("TOTAL charge")/("Total CAPACITANCE") = (Q_(1) + Q_(2))/(C_(1) + C_(2)) = (C_(1) V_(1) + C_(2) V_(2))/((C_(1) + C_(2)))` `therefore` Total electric energy stored in parallel combination of capacitors : `U_(f) = (1)/(2) (C_(1)+ C_(2)) V^(2) = ((C_(1) V_(1) + C_(2) V_(2))^(2))/(2 (C_(1) + C_(2)))` |
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