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I recently encountered this problem dealing with inductors, similar to the diagram below.enter image description here

The switch is closed for a long period of time. I know that eventually the inductor and act like a short, this is no problem. The problem I have is calculating the direction and magnitude of the current after the switch is opened. When the switch is closed, current flows down through the inductor. I know that, after the switch is opened the flow of current abides by this equation:

$50i-.1\frac{di}{dt}=0$

by KVL. I have no problem solving this differential equation for the general case; I'm just having difficulty determining the initial conditions of the current once the switch is flipped open. I'd have to assume that the current through the inductor does not change in magnitude or direction after the switch is flipped open, or else the induced EMF by the inductor would be momentarily infinite. So the current would flow clockwise throughout the right loop. Is this line of reasoning correct? Thanks for the help.

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Just after the switch is opened, the magnitude and direction of current through the inductor are to be taken as same just before the switch is opened (which is I think $\frac {12}{30} A$ in clockwise direction). Then use the DE you formulated above.

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  • $\begingroup$ Alright, that's what I guessed would happen. Thanks. $\endgroup$ – Lance Lampert May 18 '19 at 19:02

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