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A capacitor of capacitance C1 is charged...

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.

Text Solution

Verified by Experts

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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