# The idea of voltage in the ideal inductor equation

In a book that I have been reading (Practical Electronics for Inventors 4th ed by Paul Scherz and Simon Monk), I came across the equation on pg 129 for for inductor $$V_L = L \frac{di}{dt}$$

where $$V_L$$ is described as voltage across an inductor = induced EMF.

This is where I get really confused as to the usefulness/practicality of the equation. Let us take a circuit that consists of a series of voltage source, resistor, inductor and a switch. During the period of closing the switch, does this equation describe the applied voltage across an inductor? Or the induced EMF in the inductor as a result of changing magnetic flux? Because it seems to me the two are distinct and they are not equal as shown by https://physics.stackexchange.com/a/308992/289350

So what exactly is this V that I am subbing into the equation? Is it the voltage I read across the inductor using the voltmeter at the instant I switch on the circuit? Or is it the net sum of the applied voltage and negative induced EMF?

I have also referred to this The sign in the formula for the voltage across the inductor to understand that the derivations are not from Faraday's law but from energy equation of an ideal inductor + lumped model analysis. From the derivations, it seems to be implied that this V is the applied voltage. So is the author wrong? I am so confused.

• VL is just the voltage across the inductor, wether it's applied by an external source or induced. When an external voltage is applied it causes the current to change. When the current is changed by an external source, a voltage is induced. Context is what's important here.
– Bart
Sep 18 at 10:06
• Does that mean there is no accurate model of the voltage across the inductor, seeing as the rate of change in current is hard to compute at tiny time steps since t=0? It can be a result of the initial applied current and the induced current as a result of that initial applied current. Meaning to say would I be able to plot exactly how the voltage is across the inductor at tiny time steps without measuring? Sep 18 at 10:21
• In EE we are using the passive sign convention, thus, voltage across an inductor is defined to be positive at the terminal where the symbolic current enters. And this is why we are using $$V_L = L \frac{di}{dt}$$ without the minus sign. youtu.be/DutVZLUvPZc?si=zPy7BTnjOuu2hz_6&t=119
– G36
Sep 18 at 11:16

does this equation describe the applied voltage across an inductor? Or the induced EMF in the inductor as a result of changing magnetic flux?

It's both. These two quantities must be equal, because both are equal to the potential difference between the terminals of the device.

That means if you apply an ideal voltage source across the inductor, the changing magnetic flux will always be enough to produce an EMF equal to the voltage you applied. Meaning the change in the current will be enough to produce a changing magnetic flux sufficient to produce a back EMF equal to the applied voltage.

Put another way, the more voltage you apply, the faster the current will change, just like the equation says.

• Are there cases where the back EMF is not equal to the applied voltage? I think the confusion could be understanding that current can still flow even if net PD is 0. And that the back EMF is induced by a change in the applied current which means the back EMF can never "block" the applied voltage because it is produced by the CHANGE in applied voltage Sep 19 at 0:36
• In an ideal inductor, no, they must be equal just as I wrote. As for the current flowing with 0 voltage, that's absolutely possible. The voltage depends only on the derivative of the current, not on the current itself. Sep 19 at 2:00

During the period of closing the switch, does this equation describe the applied voltage across an inductor?

Yes it does.

Or the induced EMF in the inductor as a result of changing magnetic flux?

If it was for the induced emf then there would be a minus sign: -

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

So what exactly is this V that i am subbing into the equation? Is it the voltage i read across the inductor using the voltmeter at the instant i switch on the circuit? Or is it the net sum of the applied voltage and negative induced emf?

If it's an ideal inductor then, you are measuring both the applied voltage and the negative of the induced voltage. For a non-ideal inductor it can only truly represent the applied voltage.

If you want to measure the induced voltage you would need a 2nd winding that is identical to the driven winding AND, that 2nd winding would need to be 100% coupled to the 1st winding.

So is the author wrong? I am so confused.

The answer given is sloppy and shows a lack of understanding. In particular, he misses the point when this is stated: -

The back emf can never exactly equal the applied voltage as then the current would be zero and not changing which would mean that there cannot be an back emf.

It's just plain wrong. For instance, consider a resistor with a voltage applied to it and consider the way we show the resistor voltage with an arrow; that arrow opposes the applied voltage but, do we say current can never flow in a resistor?

• What do you mean by the part about the non ideal inductor with two windings? Could you elaborate on that? Sep 18 at 13:56
• @Iberico the 2nd winding (100% magnetically coupled and identical) cannot, by its very nature contain the applied voltage hence, any voltage seen on this 2nd winding is truly the induced voltage. Sep 18 at 15:34
• The comment you say is wrong I take to be describing the parasitic R. Applied voltage is back EMF + resistive drop with resistive drop being the difference between the two. Sep 18 at 23:41
• I see. To sum it, the V I am measuring is the voltage across the inductor. In an ideal inductor, the voltage across will be 0 because the induced EMF increases as the applied current increases without limit (short circuit and fry your compnents/wire). For a non ideal inductor, you will measure the applied voltage as back EMF increases. It's kind of like net 0 volt and then it reaches the steady voltage based on the internal impedance value of inductor and the applied voltage value. Hence you will never be able to measure the true induced EMF, unless you couple a second winding 100% Sep 19 at 0:23
• @asdf30 - Hi, (a) At the moment, it's unclear who your comment is directed to above. (b) It seems you are also referring to another comment, but it's unclear which one. || Please remember these features to (a) address a comment, as I did here, with @ followed by a username (which will offer autocomplete options on compatible devices, when starting to type a username), and (b) please consider quoting the relevant part of a comment you refer to (I use "*...*"). Linking to a comment is also possible (the date/time is a link). Sep 19 at 8:10