a. At `t=0` Three capacitors my be assumed like batteries of emf `40 V , 20V` and `0V`.ltbr. ` i=(Net emf)/(Total resistance)`
`=(40+20-10-20)/(1+2)`
`=10A` (anti clock)
At `t=oo` When charge redistribution is complete and loop is balanced by Kirchoff's second equation of potential current in the loop becomes zero, as insulator is filled between the capacitors.
b. Redistribution curren was anti clockwise. So, we can assume that `+q` charge rotates anti clockwise in the loop (between time` t=0` and `t=oo`) After this rotation of charges final charges on different capacitors are as shown below.
Applying loop equation in loop `abcda`,
`((40-q))/1+10+((40+q))/2-20+q/4=0`
Solving this equation we get `q=120/7`C
c. Final charges
`q_(1F)=40-q=40-120/7=160/7C`
`q_(2F)=40+q=40+120/7=400/7C`
`q_(4F)=q=120/7C`
d. Total loss of energy during redistribution
`SigmaU_i=1/2xx(1)(40)^2+1/2xx(2)(20)^2=1200J`
`SigmaU_=1/2xx((160/7)^2)/1+1/2xx((400/7)^2)/2+1/2xx((120/7)^2)/4`
`=1114.3J`
`/_\U=Sigma _f-SigmaU_i=-85.7J`
20 V battery will suply energy but `10 V` batter will consume energy. So,
Total energy supplied `=20xx120/7-10xx120/7=171.4J`
Total heat produced =Energy supplied `-/_\U`
`=171.4-(-85.7)=257.1J`
e. Resistors are in sereis. Hence,
`HpropH` or `H_1/H_2=R_1/R_2=(1Omega)/(2Omega)=1/2`
`H_1=(1/(1+2))(257.1_=85.7J`
`H_2=(2/(1+2))(257.1)=171.4J`
