Find out current `I_1 , I_2, I_3, ` charge on capacitor and ` (dQ)/(dt) ` of capacitor in the current when it is initially uncharged in the following situations.

(a) Just after the switch is closed
(b) After a long time when switch is closed.

Explanation

(a) Initially the capacitor is uncharged so its behaviour is like a conductor
Let potential at A is zero so at B and C also Zero and at F it is `in . ` Let potential at E is x so at D also x. Apply Kirchoff's ` 1^(st) ` law at point E:
` (x-in )/( R) +( x-0)/(R) +(x-0)/( R) =0`
` (3x)/(R) =(in )/(R) `
` x= ( in )/(3) " " Q_(in ) =0 `
` therefore " "I_1 =(-in //3 +in )/( R) =(2in )/(3R) `
` I_2 =(dQ)/(dt) =(in )/(3R) , " " I_3= ( in )/(3R) `
Alternatively
` i_1 =(in )/( R_(eq) ) = (in )/( R+ (R)/( 2)) =( 2in )/( 3R) `
`i_2 = i_2 =(i_1)/(2) =(in )/( 3R) `
(b) at ` t= infity ` (finally )
Capacitor is completely charged.So their will be no current through it.
` I_2 = 0, I_1,=I_2= (in )/( 2R) `
` V_(E) - V_(E) = V_(D) -V_C=( (in )/(2R)) R= (in )/(2) `
` (##MOT_CON_JEE_PHY_C25_SLV_036_S01.png" width="80%">
` rArr " "Q__c =(in C)/( 2 ), (dQ)/( dt) =I_2 =0 `
` (##MOT_CON_JEE_PHY_C25_SLV_036_S02.png" width="80%">