Two inductors `L_(1)` and `L_(2)` are connected in parallel and a time varying current flows as shown.
the ratio of current `i_(1)//i_(2)`

Options

A. `(L_(2))/(L_(1))`
B. `(L_(1))/(L_(2))`
C. `(L_(2)^(2))/((L_(1) + L_(2))^(2))`
D. `(L_(1)^(2))/((L_(1) + L_(2))^(2))`

Answer

A

Explanation

As the inductors are in parallel, induced emf across the two inductors is the same, i.e,
`e_(1)=e_(2)`
`L_(1) ((di_(1))/(dt))= L_(2) ((di_(2))/(dt))`
On integrating both sides, we get
`L_(1) int (di_(1))/(dt) = L_(2) int (di_(2))/(dt)`
`L_(1)i_(1)= L_(2)i_(2)`
`rArr (i_(1))/(i_(2))= (L_(2))/(L_(1))`