Two coils of self inductance `L_(1)` and `L_(2)` are connected in parallel and then connected to a cell of emf `epsilon` and internal resistance - R. Find the steady state current in the coils.

Answer

1

Explanation

`E=L_(e)(di)/(dt)RightarrowE=L_(1)(di_(1))/(dt)`
`i=i_(1)+i_(2)`
`(di)/(dt)=(di_(1))/(dt)+(di_(1))/(dt)Rightarro(E)/(L_(e)=(E)/(L_(1))+(E)/(L-(2)`
`(1)/(L_(e)(1)/(L_(e))=(1)/(L_(1))+(1)/(L-(2))Rightarrow(L_(e)=(L_(1)L_(2))/(L_(1)+L_(2)`