Consider a circuit as shown in this figure below-
Assuming switch S1 is closed such that current flows in loop ABCDE for a long time and steady state is reached. Now, my question is simple, suppose (after when the steady state is reached) the switch S2 is flipped, thereby disconnecting from point D and reconnecting to point F then, what will be the current in the loop BFGH?
Here's my problem, After the steady state is reached a nearly constant current of magnitude V/R1 will flow in loop ABCDE.
Suppose there is a current (say i) in the loop BFGH just after the switch is flipped(say at t=5sec), The inductor L2 will oppose this change in current. But if there was NO current in the loop BFGH at t =5- sec (minus sign denotes 'just before' 5seconds) and a finite current (i) at t = 5+ sec this would mean that derivative of current at t = 5 is infinity in wire GH! As a result the magnitude of 'potential drop' across inductor will become infinite! This can't happen so I concluded that the current in the loop BFGH must be zero.
But wait! If there's zero current in loop BFGH at t = 5+ and a finite current of magnitude V/R1 in wire BC implies that current derivative is again infinite in wire BC! which would mean that the magnitude of 'potential drop' across inductor L1 must be infinity! So the current in loop BFGH must not be zero.
So you guys can see that I'm in a fix. Can someone tell what's the problem here? What will be the current in wire GH and where am I going wrong?