In figure, the switch is in the position 1 for a long time. Then the switch is shifted to position 2 at $t=0$. At this instant, the value of $i_1$ and $i_2$ is

enter image description here

Well I'm confused over what will happen here. So the current through the inductor in middle wire will be $\frac{E}{R} $ just after shifting the switch. But what about the inductor in the right most wire?

There are three possibilities for the rightmost inductor according to me-

  1. The current in the closed loop now is same throughout. So the current through the inductor also equals $\frac {E}{R} $.

  2. The inductor opposes the current in the circuit. So the current through it is zero at $t=0$

  3. The total flux of both the inductors remains constant.

So what exactly happens here and why?

  • $\begingroup$ Are you asking what is the current in the circuit at the instant $t=0^+$ ? $\endgroup$ Jul 1, 2018 at 21:40

1 Answer 1


Option 1 would not work, since it would require more energy than is available: $2$x$\frac {LI^2} 2 > \frac {LI^2} 2$

Option 2 is fine as the initial condition, but it cannot last, since the current in the left inductor has to go somewhere. So, there will be a voltage spike across the right inductor, which will force the current through it to rise very quickly, possibly producing a spark across the inductor in the process.

Option 3 should help determine the current in the right loop right after the spike. Over time the current will decay due to losses in the resistor.

  • 1
    $\begingroup$ How is option 2 fine? For the circuit to have 0 current, the rate of change in current of the inductor would have to be infinite. $\endgroup$ Jul 1, 2018 at 21:42
  • $\begingroup$ @Goldname "...fine as the initial condition", i.e., before it starts rising. Also, if the circuit was ideal (no parasitics, no sparks), the rate of the current change through the inductor would be infinite, because it would have to be zero and I at the same time. $\endgroup$
    – V.F.
    Jul 1, 2018 at 22:06

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