We have an ideal inductor with the dc source in figure A.

When switch is closed we get equal and opposite voltage across the inductor against the source voltage and a constant rate of rise of current given by di/dt = V/L. This constant rate of rise/change in current will also produce the rising magnetic field (shown in green colour).

Suppose some how we reverse the battery instantly, shown in figure B. That reversed battery will cause the reversed rise of current (shown in figure B in brown colour). But at the same time the magnetic field (which was caused by the current when battery was not reversed) is collapsing and is inducing the voltage and current opposite to the reversed current.

My question is will there be no reversed current flowing as long as the magnetic field is collapsing as that collapsing magnetic field is causing the current in the opposite direction of the reversed current.

Plus there will also be electric and magnetic field produced by the reversed voltage.

How will these two collapsing and rising magnetic field are going to interact with each other ?

enter image description here


Simple mechanical analogy

You have an object that can travel only forwards or backwards.

  • If the object's velocity was +1 m/s (i.e. forwards) then, after 10 seconds the distance traveled will be +10 metres.
  • If the forward speed slowed, the object's distance from the starting point would still increase even though the object's velocity slowed down.
  • If the speed fell to 0 m/s then the object would still be at a positive distance from the starting point and, that distance will remain fixed until the object started moving again.
  • If the object then moved (backwards) at -1 m/s, then the positive distance would start to fall towards zero.
  • If the object continued to move backwards, the distance from the starting point would eventually become negative.

In this analogy: -

  • Velocity behaves like inductor voltage
  • Distance behaves like inductor current

Back to the question

Suppose some how we reverse the battery instantly, shown in figure B. That reversed battery will cause the reversed rise of current

No, current will start to fall from its previous positive value just like the analogy above where distance traveled falls from a positive value when velocity reverses to a negative value.

Here is your homework: -

enter image description here

Study it, then study it again.......

  • 2
    \$\begingroup\$ Current isn't instantly reversed - it falls but doesn't reverse. Just as distance from the origin doesn't become negative; it becomes smaller but is still a positive number and eventually reaches zero AND, if the velocity is still negative then distance reverses but it isn't instant. \$\endgroup\$ – Andy aka Apr 30 '20 at 10:32
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    \$\begingroup\$ @Alex Read my comment again - I added to it and you might have missed that addition. \$\endgroup\$ – Andy aka Apr 30 '20 at 10:35
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    \$\begingroup\$ @Alex, then I strongly advise you to get used to the time-function / differential equation formula of Energy stored in an inductor, and voltage across an inductor. It's really just one line of calculation from there, and I even did that for you in my answer. If you don't learn how to deal with these equations and answer these questions yourself, you will have a very hard time when it comes to exams and the rest of your EE studies. This is ultra-basics! Wikipedia is your friend, but it's better if you learn it from your study material, as that will have a more coherent notation. \$\endgroup\$ – Marcus Müller Apr 30 '20 at 10:53
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    \$\begingroup\$ @Alex - the magnetic field falling is because the current is falling - I wouldn't use the word "collapsing" in this context because that word (collapsing) is generally reserved for an uncontrolled fall in field due to an open circuit. Not all will agree with what I have just said of course. So, the falling field is caused by the falling current and both will reach zero at the same time AND if the voltage remains negative, both will pass through zero and become negative. That rate of change (on both) is V/L. \$\endgroup\$ – Andy aka Apr 30 '20 at 10:53
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    \$\begingroup\$ when battery is reversed the reverse current will rise at the rate of 10 Amps - repeat ten times this: When the voltage reverses across an inductor, the current does not instantly reverse - it's very, very important to get the wording correct (you don't need to repeat that bit of course) @Alex \$\endgroup\$ – Andy aka Apr 30 '20 at 11:13

The situation will look like this:

enter image description here

From t = 0s to t = 2s we have a "charging phase" and the flux will "follow the current". Thus the magnetic field will rise.

But from t = 2s to t = 4s the current will start to decrease and we have a "discharge phase" and magnetic field collapse too (flux follow the current).

But at time t= 4s all the energy stored in the inductor was released (I = 0A and Flux = 0Wb). But due to the input voltage is still present across the inductor. The inductor will start a charging phase and it will start to store the energy in the form of a magnetic field.

  • \$\begingroup\$ from t = 2 to t = 4. voltage is reversed. current falls. magnetic field collapses. but at the same time the current will not flow in reverse direction due to reversed voltage ? @G36 \$\endgroup\$ – Alex Apr 30 '20 at 10:11
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    \$\begingroup\$ The current through an inductor cannot change instantaneously. The rate of change of current in an inductor is proportional to the applied voltage. In order for the current to change instantaneously, you would need to apply infinitely great voltage for a short time--an impulse of voltage, in other words. Thus, first you need to do is "discharge" the inductor. \$\endgroup\$ – G36 Apr 30 '20 at 12:49
  • \$\begingroup\$ PLEASE GIVE ME YOUR FB LINK \$\endgroup\$ – Alex Apr 30 '20 at 12:53
  • \$\begingroup\$ Guess what, I do not have a FB account. Social media are not for me. \$\endgroup\$ – G36 Apr 30 '20 at 12:57
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    \$\begingroup\$ I'm from Poland, are you wanna come? \$\endgroup\$ – G36 Apr 30 '20 at 13:00

First of all, idealized inductors, and instantly switched around ideal batteries would lead to an infinitely high voltage across the inductor. Don't do that, you'll damage your universe.

Then: yes, the differential equation description of an inductor is quite clear on this.

$$v(t) = L\frac{d\, i(t)}{dt}$$

So, let's assume you're not "reversing a battery infinitely fast", but simply attache a fast alternating current (AC) source: Then your inversion would indeed lead to a large time derivative of voltage.

Taking an integral from \$t=0\$ to some current point in time \$T\$ to the equation above leads to:

\begin{align} \int_0^T v(\tau)\, d\tau &= L \left(i(T) - i(0) \right)\\ i(T) &= \frac1L\left(\int_0^T v(\tau)\, d\tau + i(0)\right) \end{align}


hat reversed battery will cause the reversed rise of current (shown in figure B in brown colour).

no, it can't. At least not instantly: the integral's value doesn't "jump", the current through a inductor can only change slowly.

It's exactly the energy stored in the magnetic field that counteracts that sudden change. Speaking in terms of energy, the power source must first completely deplete the one magnetic field there is before building up one with an inverse sign.

  • \$\begingroup\$ so are you saying that first the magnetic field caused by the current will deplete and then the the magnetic field caused by the reverse current should rise ? so as long and the magnetic field is depleting, there would be no reverse current. right ? \$\endgroup\$ – Alex Apr 30 '20 at 10:16
  • \$\begingroup\$ no, that's not what I said. I used equations to describe something that can only be described by equations. \$\endgroup\$ – Marcus Müller Apr 30 '20 at 10:17
  • \$\begingroup\$ the power source must first completely deplete the one magnetic field there is before building up one with an inverse sign. please explain this. \$\endgroup\$ – Alex Apr 30 '20 at 10:32
  • \$\begingroup\$ that sentence is about the magnetic field. I don't know what to explain about this. \$\endgroup\$ – Marcus Müller Apr 30 '20 at 10:36
  • \$\begingroup\$ Also, not electronics.stackexchange.com/questions/495862/… already dealt with all the math \$\endgroup\$ – Marcus Müller Apr 30 '20 at 10:44

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