Magnetic Flux conservation 


My teacher said that after switch is shifted (after very long time), $\phi_i = \phi_f$ $\implies i_oL  = i3L \implies i = \dfrac{i_o}{3} $ where $i_o$ is $\dfrac{\varepsilon}{R}$
So the initial current in the circuit after switch is shifted is $\dfrac{i_o}{3}$
But, I really didn't understand why the flux should be conserved in this case i.e. why  $\phi_i = \phi_f$. I would like to know about this concept and the reasons involved. 
$\phi_i$ and $\phi_f$ are the total flux in both inductors immediately before and immediately after the switch is shifted?
 A: A better arrangement would be as follows the reason being that at no time will there be an open circuit as the switch position is changing?  

With the switch in position 1 a current flows though the switch but no current flow through inductor $2L$ as it is short circuited by the switch.  
The right loop consisting of the two inductors, the cell and the resistor has a magnetic flux passing through it $\Phi_{\rm i} = L \,i_{\rm i}$
When the switch is moved to position 2 the magnetic flux linked with that loop cannot change instantaneously and so the right loop adjusts itself by having the having the same magnetic flux linked with it but now that magnetic flux is contributed by both inductors with $\Phi_{\rm f} = L \,i_{\rm f}+ 2L \,i_{\rm f}$.  
You may wonder as to the origin of the magnetic flux linked with the circuit.
Some time ago I was trying to explain Faraday's law and why two turns produced double the flux linkage than one turn.
Being unable to draw a satisfactory sketch of the area through which the magnetic field was passing I hit on the idea of using a soap film to represent the area.  
 
There is an equivalent circuit for capacitors which perhaps is intially easier to analyse?  

In this case it would be charge which is conserved and on moving the switch to position 2 the initial voltage across the two capacitors would drop to $\frac v 3$.
Note that in both case energy is not conserved 
