`4muF` capacitor is charged to 150 V and another capacitor of `6muF` is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Options

A. 0.005J
B. 0.500J
C. 0.003J
D. 1.003J

Answer

C

Explanation

`4mu f` charged to 150V would have `q_(1)= C_(1)V_(1)= 600mu C`
`6mu f` charged to 200V would have `q_(2)= C_(2)V_(2)= 1200mu C`
After connecting them across each other, they will have a common potential difference V.
Charges will readjust as `q_(1). and q_(2).`
`V= (q_(1).)/(C_(1)) = (q_(2).)/(C_(2)) = (q_(1). + q_(2).)/(C_(1) + C_(2)) = (q_(1) + q_(2))/(C_(1) + C_(2)) = (1800mu C)/((4 + 6) mu f)` [Conservation of charge]
V= 180 volt
Initial energy
`U_(i) = (1)/(2) C_(1) V_(1)^(2) + (1)/(2) C_(2)V_(2)^(2) = (1)/(2) (4mu f) (150V)^(2) + (1)/(2) (6mu f). (200V)^(2)= 0.165J`
Final energy
`U_(f) =(1)/(2) [C_(1) + C_(2)] V^(2)`
`=(1)/(2) (4mu f + 6mu f) .(180)^(2) = 0.162J`
Heat produced `= |U_(f)- U_(i)|= 0.003J`