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I am having trouble understanding the behavior of current through an inductor when the direction of applied current is the opposite of what is initially flowing through it. An example is that when an inductor is initially connected to a current source that drives the current clockwise, and then at time t=0, the switch flips such that it is now connected to a current source that drives the current counter-clockwise. I know that the current cannot instantaneously switch directions and I think this is where my confusion stems.

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    \$\begingroup\$ What if a capacitor initially charged at 1V is connected to a -1V voltage source? Try comparing this.... \$\endgroup\$
    – carloc
    Dec 12, 2020 at 14:58

3 Answers 3

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An example is that when an inductor is initially connected to a current source that drives the current CW

Here's your first problem; connecting an inductor (a pure inductor) to a current source requires that to instantly drive the current, an infinite voltage must be present and things fall apart when that happens.

However, let's say you can overcome this initial problem by having the current source ramp up from zero amps...

then at time t=0, the switch flips such that it is now connected to a current source that drives the current CCW.

And here's your next problem; current cannot change instantly in an inductor without an infinite voltage being present. It's basically as I said earlier but it's now unresolvable without the presence of infinite voltage i.e. no amount of fiddling with ramping up or ramping down currents can avoid infinite voltage without the "experiment" becoming flawed in what you are trying to achieve.

However, if you said that you applied a voltage across an inductor and after a short time 1 amp was flowing and then you reversed the voltage then that's OK. At the instant you reversed the voltage, 1 amp would still be flowing in the inductor (despite the voltage being reversed). Over a short time that 1 amp would ramp down to zero amps and, then begin climbing in value (at the same ramp rate) but in the opposite direction: -

$$V = L\cdot\dfrac{di}{dt}$$

It's all embedded in the above equation. Try this little diagram: -

enter image description here

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    \$\begingroup\$ +1. I enjoy your perspectives and your writing on inductors/transformers. \$\endgroup\$
    – jonk
    Dec 13, 2020 at 4:50
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The inductor will produce a voltage impulse at the instant the step change in current occurs.

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Qualitatively, anything that tries to reduce the current in the inductor is also trying to collapse the magnetic field in the inductor. When the magnetic field collapses the energy has to go somewhere and that energy goes towards keeping maintaining current through the inductor as best it can.

For example, suddenly introducing a high impedance in the current path to interrupt/reduce the current to the inductor will cause that stored energy in the collapsing magnetic field to produce a massive voltage spike to continue to drive that same current through that impedance so that the inductor current is maintained. This is like trying to rapidly, but passively, dump energy from the magnetic field.

Trying to instantaneously reverse the current through the inductor is even worse than that so the spike would be even higher. The energy released in the collapsing magnetic field of the inductor will not allow the current to change instantaneously no matter how hard you try. How hard you drive that opposing current only affects whether the the inductor holds out longer or shorter before it exhausts the energy stored in its magnetic field trying to resist and the current direction reverses. This is like trying to actively suck energy out of the magnetic field.

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