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

So charge flow in path 1 `=120-60=60muC`
So charge flow in path 2= `60(60-30)=90muC`
So charge flow in path 3 `=(60-30)=30muC=1200muJ`
W work done by battery of 10V `=-30 xx 10muJ=-300muJ`
Initial energy of capacitors `=[1/2((60)^(2))/(6)+1/2 ((60)^(2))/(3)]muJ=900muJ`
Final energy of capacitor `=[1/2 6 xx (20)^(2)+1/2xx3 xx (10)^(2)] muJ=1350muJ`
So heat loss `=(1200-300)-(1350-900)muJ=450muJ`