Two inductors of self inductances `L_(1) and L_(2)` and of resistances `R_(1) and R_(2)` (not shown) respectively are conneted in the circuit as shown. At the instant t = 0, key k is closed. Obtain an expression for which the galvanometer will show zero deflection at all times after the key is closed.

Explanation

Since there is no current through `BD` therefore points `B` and `D` are at same
potential, `V_(B)=V_(D)`
Potential difference, `V_(AB)=V_(AD)`
`L_(1)(di_(1))/(dt)+i_(1)R_(1)=L_(2)(di_(2))/(dt)+i_(2)R_(2)`………`(1)`
Similarly, `V_(BC)=V_(DC)`
`i_(1)R_(3)=i_(2)R_(4)` ........`(2)`
From equation `(1)` and `(2)`
`L_(1)(R_(4))/(R_(3))(di_(2))/(dt)+i_(2)(R_(1))/(R_(3))=L_(2)(di_(2))/(dt)+i_(2)R_(2)`
`(L_(1)(R_(4))/(R_(3))-L_(2))(di_(2))/(dt)=i_(2)[R_(2)-(R_(1))/(R_(3))]`........`(3)`
At `t=0`, `i_(2)=0`
`:. L_(1)(R_(4))/(R_(3))-L_(2)=0` as `(di_(2))/(dt) ne 0`
`:. (L_(1))/(L_(2))=(R_(3))/(R_(4))`..........`(4)`
At `t=oo`,
`(di_(2))/(dt)=0`, `rArr i_(2)=((epsilon)/(R_(2)+R_(4)))=`constant
`rArr R_(2)-(R_(4)R_(1))/(R_(3))=0`
`:. (R_(1))/(R_(2))=(R_(3))/(R_(4))`.......`(5)`
From `(4)` and `(5)`
`(L_(1))/(L_(2))=(R_(1))/(R_(2))=(R_(3))/(R_(4))`