Two capacitors `C_(1) and C_(2)` are connected with two batteries of emf `epsilon_(1)` and `epsilon_(2)`. The circuit components are connected with a switch S as shown in. Initially the switch is open and the capacitors are charged. Find the heat produced when switch S is closed.
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Answer

`40 muJ`

Explanation

When switch id opened, let us consider consider an isolated system as shown. In this isolated system, net charge should be zero.
`3(x-10)+2(x+5)2=0` or `x=4 `
Hence charge on `C_(1)` is `q_(1)=2(4+5)=18 muC`
And charge on `C_(2)` is `q_(2)=3(4-10)= -18 muC`, or left plate is negative.
Hence, charge on `C_(2)` is `q_(2)=18 muC` (right plate is positive )
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When switch is closed, the final charge is should in the figure. Note that `Deltaq_(1)=8muC` is charging battery, hence work done by `5V` battery will be negative.
`W_(5V)=-(8muC)(5V) = - 40 mu J`

`Deltaq_(2)` is moving from positive terminal of `10 V` battery Net work done by batteries is work done by `10 V` battery is `(120 muC) (10 V) = 120 muJ`.
Net work done batteries is
`W_("battery") = - 40 + 120 =80 muJ`
Change in potential energy of the capacitors
`DeltaU = U_(f)-U_(i)`
`=((1)/(2)3[10]^(2)+1/(2)2[5]^(2))-((1)/(2)3[6]^(2)+1/2[9]^(2))=40 muJ`.