How can there be a current flow when back emf equal and opposite of suply voltage? Isn't there a cancellation of voltage to zero volts? I have attached a video regarding this. Please answer this intuitively, I have already wasted enough time on this but couldn't find satisfactory answer.


  • \$\begingroup\$ When you apply a voltage to a resistor, the resistor produces an equal and "opposite" to the source voltage (in the same way the back-emf of the inductor is opposite). And yet current still flows through the resistor. Indeed KVL says that no matter what we attach to a voltage source, it must have an "opposite" voltage in this way. \$\endgroup\$ – The Photon Jul 30 '16 at 15:43
  • \$\begingroup\$ @ThePhoton then what makes current to flow with 90 degree phase shift only in inductor but not in resistor.It is back emf isn't which is the voltage different from what you mentioned \$\endgroup\$ – Abhishek Jul 30 '16 at 15:56
  • \$\begingroup\$ Why the current lags the voltage is a separate question. Please search the archives and see if it has been asked before, and if it hasn't, post a new question. Actually a couple of answers here seem to have answered this question and missed what I thought was the main one about the direction of the back emf. \$\endgroup\$ – The Photon Jul 30 '16 at 15:59

Don't think of the back emf as being caused by the source voltage. Think of the back emf as being caused by the changing current.

The source produces a current. That (changing) current causes the inductor to produce a back emf. The back emf limits (but doesn't eliminate) the current produced by the source. If it eliminated the current, then there would be no back emf. The two things don't oppose each other so much as they balance each other.

By KVL, the back-emf of the inductor must be be equal and opposite to the voltage produced by the source, or else energy conservation would be broken.

Isn't there a cancellation of voltage to zero volts?

There is, in the sense that if you follow the path of the circuit and add up all the voltages you pass (the source and the inductor, in a simple case), then the sum of all those voltages will be zero once you've followed the complete circuit.

But of course that's just KVL, and it applies to any circuit elements (resistors, capacitors, or whatever) and not just to inductors.

I think what's confusing (at least to me) is the use of the word "opposite". In a normal drawing of this circuit, we draw the source with its positive terminal at the top of the page, and then the inductor will also produce a back emf that's positive at the top of the page. But if you consider the voltage in the direction of traversing the circuit (for example, always going in a clockwise direction around the circuit), then the two potential differences (source emf and inductor back-emf) will be opposite.

edit RE the video

I think the video is confusing because he never defines the sign convention he is using to define his voltages across the different elements.

In order for the description to be correct, he must have defined opposite conventions for the two types of elements, like this:


simulate this circuit – Schematic created using CircuitLab

Keep in mind Kirchoff's voltage law. With these conventions, \$V_r\$ must be equal to the V1 source voltage (regardless of what type of resistor is used or even if you replaced the resistor with a different element like a capacitor or inductor). And \$V_l\$ must be the opposite of the V2 source voltage (again, it would still be opposite even if you replaced the inductor with a different element like a resistor or capacitor.

Again, this is simply because of KVL and the choice of sign convention for the voltage across the two elements. But in the video he never shows the chosen sign convention, so this makes it impossible to understand what he's trying to teach.

Either that or the guy in the video has no understanding of circuit theory and believes that a circuit with an inductor in it can violate KVL, which is simply not correct.

  • \$\begingroup\$ But alternating current is caused by alternating voltage so back emf can be compared with source voltage right? \$\endgroup\$ – Abhishek Jul 30 '16 at 16:04
  • \$\begingroup\$ I'm not sure what you mean by "can be compared with". The sum of potential differences around the circuit is 0 at every instant (in a lumped circuit). So if you just have a source and an inductor in the circuit, the potentials across them are equal at every instant. \$\endgroup\$ – The Photon Jul 30 '16 at 16:05
  • \$\begingroup\$ You are right that voltage across inductor equal to source voltage.I understood law of conservation when applying to resistive circuit but I am not able to know how law of conservation takes place for inductive circuit all because of back emf \$\endgroup\$ – Abhishek Jul 30 '16 at 16:11
  • \$\begingroup\$ Conservation of energy says that the power provided by the source (V*I) must be equal to the change in energy stored in the magnetic field (\$\frac{\mathrm{d}}{\mathrm{d}t}\left(\Phi^2/L\right)\$). \$\endgroup\$ – The Photon Jul 30 '16 at 16:21
  • \$\begingroup\$ But then in the video which I have mentioned back emf is equal at every instant to supply voltage you can inverted waveform doesn't it cancel out?please do check out the video explanation. \$\endgroup\$ – Abhishek Jul 30 '16 at 16:26

Voltage across an inductor rules rate of change of current, not its value.

i.e. voltage across is zero? then current do not change, what was it befor? 1A? Okay it stays at 1A constant.

  • \$\begingroup\$ youtube.com/watch?v=ae0fy435zJA \$\endgroup\$ – Abhishek Jul 30 '16 at 8:31
  • \$\begingroup\$ Sir i understood what you said.but in this video you can see back emf is head to head equal to supply voltage.that means net voltage is zero.But still there exist current which is 90 degree out of phase wrt supply \$\endgroup\$ – Abhishek Jul 30 '16 at 8:34
  • \$\begingroup\$ how does the current stay at 1A considering this as simple circuit when the net voltage is zero across inductor? \$\endgroup\$ – Abhishek Jul 30 '16 at 8:52
  • \$\begingroup\$ Net voltage is zero? The applied voltage to the coil is equal to the back EMF of the coil, it is the same voltage, there is no net zero voltage here. It is exactly the same as in a resistance, you apply a voltage V, which is equal to IxR. IxR does not cancel the applied voltage V, it is equal to it. \$\endgroup\$ – Claudio Avi Chami Jul 30 '16 at 9:07
  • \$\begingroup\$ The video is about AC, where the voltage and current are continuously changing sinusoidally - hence the induced back emf is also sinusoidal (differentiate a sinusoid and you get another sinusoid) \$\endgroup\$ – Chu Jul 30 '16 at 9:10

The back EMF actually is equal and opposite of supply only at the moment the supply is connected to an inductor that was without current before.

After that it more or less slowly dereases and current starts to flow.

  • \$\begingroup\$ Suppose supply is 5A Ac and current starts to build up at the same time back emf starts to build up till it reaches the supply voltage but as soon as the back emf equals supply there is again change in supply voltage which the back emf tries to catch.I hope i am right up to this point.So there will always be some difference voltage for which current is established.I hope i am right upto this.But ideally i.e according to the video i have attached it shows net voltage is zero.How on earth is this possible? \$\endgroup\$ – Abhishek Jul 30 '16 at 8:41
  • \$\begingroup\$ I didn't watch the video. As I described I assumed the case that an inductor was switched to a DC voltage source. In the steady state AC case the back EMF is not equal to the equal to the external voltage (except for single moments). \$\endgroup\$ – Curd Jul 30 '16 at 18:59

The voltage applied to a coil cannot be cancelled with the coil voltage. The coil voltage is EQUAL to the applied voltage.

  • \$\begingroup\$ No claudio you are wrong. \$\endgroup\$ – Abhishek Jul 30 '16 at 9:13
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    \$\begingroup\$ Well if you think that you know what happens then don't ask for explanations. Is the first time in all my career that I hear somebody say that a voltage applied in parallel can be cancelled by the load. \$\endgroup\$ – Claudio Avi Chami Jul 30 '16 at 9:14
  • \$\begingroup\$ Please don't take it in negative way.see the YouTube link which I have attached.you can see back emf is EQUAL as well as OPPOSITE.but my question is not that. \$\endgroup\$ – Abhishek Jul 30 '16 at 9:20
  • \$\begingroup\$ You have an ideal AC voltage source (i.e. it has no internal resistance) connected to a coil. The voltage on the coil terminals must be exactly the applied voltage. There can be no cancellation of voltage in such a simple circuit. \$\endgroup\$ – Claudio Avi Chami Jul 30 '16 at 9:27
  • \$\begingroup\$ If voltage across coil is equal to applied voltage then what is back emf? \$\endgroup\$ – Abhishek Jul 30 '16 at 11:28

Think of current in an inductor as having inertia.

You need to apply a large force (voltage) to accelerate an inertial mass like a car (or increase current flow), and the only way to stop it once it's moving and has gained momentum, is to apply a large opposing force (voltage) like the brakes.

Attempting to stop it suddenly with a brick wall (or opening a switch) results in large and possibly destructive forces (or a high voltage spike).

  • \$\begingroup\$ I think that way too but mathematical statement say that back emf is equal and opposite to applied voltage.How is this possible sir? \$\endgroup\$ – Abhishek Jul 30 '16 at 11:32
  • \$\begingroup\$ Push the stationary car. Instead of instantly moving, it pushes back against you. That opposing force is similar to back EMF. \$\endgroup\$ – user_1818839 Jul 30 '16 at 11:39
  • \$\begingroup\$ And if I keep pushing more than the frictional force the car moves without it pushing me back \$\endgroup\$ – Abhishek Jul 30 '16 at 11:42
  • \$\begingroup\$ Exactly. And if you try to stop the car, can you do it all at once? Only if you put a large and strong barrier in front of it. Same thing with an inductor. Current, once flowing, will continue to flow, and if the voltage across the inductor has to change to make that happen, then it will change. \$\endgroup\$ – mkeith Jul 31 '16 at 5:27

Very simple:

  1. Linear solenoid DC fed,switched, Back-EMF(volts)=constant x di(amperes)/dt,negaive sign, so if dt tends to 0,Back-EMF tends to infinity.
  2. I saw this curve volts(EMF) vs dt in cents of a sec,seems to be of the form 1-e exp -kt,charge of a Capacitor form.I got a numerical real example 500 volts easy.
  3. Have we a switch time off controlable.Is so important as resonance.

I just upped the answer that said, "think of the back emf as being caused by the changing current." That's the correct view, I think.

Note: If you truly want to understand the details, you really need to read a book on physics. There are two really good sources I think you might consider, but which one is better will depend on your interests and background. One is Feynman's Lectures on Physics, available here (or just buy the set of three lecture books.) The second is a wonderful and more modern book on physics by Chabay and Sherwood, called "Matter & Interactions," 3rd edition (or later.) It's superb and provides a useful set of mental models to think with (more directly and better than any other introductory physics book I know about.)

As the current increases for example, then the non-Coulomb electric field curls around the region of changing flux in the inductor, opposing the increase in the current and polarizing the inductor. The new surface charges produce a Coulomb electric field that follows the wire and points opposite to the non-Coulomb field. (Note that the Coulomb voltage due to surface charges approximates the non-Coulomb voltage due to time-varying magnetic fields when the wire resistance is insignificant. In other cases where the wire resistance is significant, then those two values will no longer be so nearly close to each other.)

  • \$\begingroup\$ Thanks jonk.which books give info on inductor and other electrical component by understanding with the help of law of energy conservation as Base? \$\endgroup\$ – Abhishek Jul 31 '16 at 4:20
  • \$\begingroup\$ Matter & Interactions, 3rd ed, is especially good for gaining insight into magnetic and electric fields. It covers much more than just the main three kinds of electric and magnetic effects on electrons in a wire: the Coulomb electric fields due to surface charges, the non-Coulomb electric fields due to time-varying magnetic fields, and the magnetic forces when the wire itself is moving. (All can be present simultaneously.) But I'm most impressed by how well it provides intuition and then tests those insights with quantitative estimations and comparisons. Nice book. \$\endgroup\$ – jonk Jul 31 '16 at 4:34

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