Find the charge flown through the path 1, 2, 3 as shown in figure after closing switch S and heat generated in the circuit.

Explanation

` (##MOT_CON_JEE_PHY_C25_SLV_034_S01.png" width="80%">
So charge flow in path 1= 120- 60`= 60 mu C `
So charge flow in path ` 2= 60 + ( 60-30) = 90 mu C `
So charge flow in path ` 3= (60 -30 ) = 30 mu C `
So wark done by battery of ` 20 V =60 xx 20 mu J = 1200 mu J`
W work done by battery of ` 10 V =- 30 xx 10 mu J =- 300 mu J `
Initial energy of capacitors = `[(1)/(2) ((60)^(2))/( 6) + ( 1)/( 2) ((60)^(2))/( 3) ] mu J= 900 mu J`
Final energy of capacitor
`= [(1)/(2) 6xx (20 )^(2) +(1)/(2) xx 3xx (10 ) ^(2) ] mu J = 1350 mu J `
So heat loss = `( 1200 -300) -(1350 -900) mu J = 450 muJ`