In the circuit shown in figure the switch is closed at t=0.the current I^3,as soon as switch is closed is ???Plzz give full solution steps

At time=0 when capacitor has no charge on it,it behaves as an ordinary conductor. so the circuit reduces to a mere bunch of resistors and a battery. now net current in the circuit at t=0 is 3E/4R so this current equally divides into three halves [ E1,E2,E3 ] and so net E3= E/4R

here we use some terminology...

t=0^- means the time just before the switch is closed/opend depending on the previous position.

t=0⁺ means the time just after the switch is closed/opend depending on the previous position.

now for the above circuit at t=0^- there is no charge on the capacitor. so V^-  = 0 volts

that is voltage just before closing the switch.

we know from the continuity of charge , Q⁺  = Q ^-  that is

the charge present on the capacitor plates just before closing the switch should always be equal to charge that is present on the capcitor plates just after teh switch is closed.

so V ⁺  = V ^- 

so V ⁺  =0 volts.    here V – voltage across capacitor.

At t=0 ⁺    capacitors are replaced by short circuits( V=0) then it becomes resistive network. we can easily solve that resistive network for the current.