In the connection shown in the figure the switch K is open and the capacitor is uncharged. Then we close the switch and let the capacitor charge up to the maximum and open the switch again. Then
(a) the current through `R_(1)` be `I_(1)` immediately after closing the switch,
(b) the current through `R_(2)` be `I_(3)` immediately after reopening the switch, Find `I_(1)/(I_(2) I_(3))` (in `ampere^(-1)`) (Use the following data: `V_(0)=30VR_(1)=10 K Omega R_(2) = 5 k Omega)`

Answer

`0750`

Explanation

`(Cq_(1))/k=(Q_(0)-q_(1)) Rightarrow q_(1)=Q_(0)/((1+C/K)) Rightarrow (Cq_(2))/K=K(q_(1)-q_(2))`
`q_(2)=q_(1)/((1+C/K)) Rightarrow q_(10)= Q_(0)/((1+C/K)^(10)) Rightarrow Q_(0)/2 = Q_(0)/((1+C/K)^(10))`
`(1+C/K)=10 sqrt(2) Rightarrow Q_(0)/3=Q_(0)/((1+C/K)^(10+n))`
`10sqrt(2)=10+nsqrt(3)`
`(l n2)/10=(l n3)/(l n2)`