In almost all physics books I've seen it says that the back-emf generated across inductor causes the delay in current (current is shifted 90 degrees with respect to the voltage across inductor). I just think this is wrong. There's no difference between so-called back-efm in inductor and a simple voltage drop across resistor, meaning that if we have two circuits:

1) AC voltage source + resistor,

2) AC voltage source + inductor,

then the voltage across the resistor and inductor, respectively, will be the same with respect to the source voltage! The existence of back-emf in the first case doesn't explain why the current is lagging behind the voltage across inductor.

You can agree that considering this in terms of cause and effect is pretty pointless. The back-emf is proportional to the rate of change of final current (the current that has been already modified by that back-emf). It's like an 'egg or chicken' problem. However, the formula is really simple.

One more problem here - why I think 'back-emf' is a bad name? Because it's not a electromotive force. In a circuit like 2), the Kirchoff law says the voltage across the inductor (back emf) is exactly opposite to the source electromotive force (both added give zero). Suppose back-emf was a real emf. Then there would be no current as the net emf would be zero!

Do I understand this correctly?

  • $\begingroup$ The back-emf of an inductor is generated and adjusts as the current goes through the wire. This is not the same as a voltage drop across a resistor. Like the physics texts tell you, it absolutely is the cause of the current phase difference. See this link $\endgroup$
    – Jim
    Aug 10, 2015 at 18:49
  • $\begingroup$ My point is that in circuit 2), the voltage drop across the resistor is equal the back-emf in circuit 1), assuming identical source voltage wave. If back-emf and voltage drop is the same thing, then it doesn't explain the delay in current. Yes I know the formula, V=dI/dt*L, but that's another story. Are you telling me that by "back-emf" we also mean this formula? $\endgroup$ Aug 10, 2015 at 18:57
  • $\begingroup$ the back-emf doesn't immediately manifest itself. It's a response to an imposed current, which means that it will be proportional to the way the current is changing. The voltage dropped across the inductor must be equal to the voltage supplied by the source (if alone in the circuit), which means the current must be offset in phase from that by 90 degrees, because of that derivative that represents the way the current changes and, thus, produces the back-emf $\endgroup$
    – Jim
    Aug 10, 2015 at 19:01
  • $\begingroup$ There's another issue here - the back-emf across the inductor is proportional to derivative of current AFTER the current has been changed by the inductor. That just feels weird to me. The inductor delays the current and then it 'looks at' this modified current, taking derivative of it and setting back-emf which has to be equal to source voltage (Kirchoff voltage law). $\endgroup$ Aug 10, 2015 at 19:06
  • $\begingroup$ not after, while the current is changed by the inductor. It's a mutual thing. Happens simultaneously $\endgroup$
    – Jim
    Aug 10, 2015 at 19:07

2 Answers 2


First of all, don't mix up voltage with current. In your examples 1 and 2 it is certainly true that the voltages across the resistor and inductor are the same w.r.t. the source voltage. This is just Kirchhoff's voltage law. However, this still results in a current lag in the inductor compared to the resistor. Say the source voltage is

$\Delta V_S = V_0 \sin(\omega t)$

So the resistor voltage is

$\Delta V_R = - V_0 \sin(\omega t)$

so that $\sum \Delta V = 0$ as Kirchhoff's law requires. The same goes for the voltage across the inductor, $\Delta V_L$.

But for the resistor we have Ohm's Law

$\Delta V_R = IR$ so the current through the resistor is just

$I_R = -\frac{V_0}{R} \sin(\omega t)$

But for the inductor we have

$\Delta V_L = - L \frac{dI}{dt}$.

So to get the current, $I_L$, you need to integrate $\Delta V_L$ w.r.t. time so in this example you will get a cosine instead of a sine. Thus, we see a phase shift in the current (but not the voltage). It is worth going through the integral yourself, but it is also in most elementary circuits textbooks.

  • $\begingroup$ What I meant was that the notion of 'back-emf' is just voltage drop, nothing else. It doesn't contain the formula for voltage in an inductor. However, $\Delta V_L = - L \frac{dI}{dt}$ is the property of the inductor. It's important to remember that voltage causes current. $\endgroup$ Aug 10, 2015 at 19:01
  • $\begingroup$ @user4205580 voltage causes current through a closed circuit, but through an inductor it is the change in current that causes a voltage $\endgroup$
    – Jim
    Aug 10, 2015 at 19:03
  • $\begingroup$ @user4205580, I agree that the term "back-emf" is not very useful. I don't use that term myself because as far as I'm concerned $\Delta V_L$ is a voltage just like any other. $\endgroup$ Aug 10, 2015 at 23:09

This question and your explanation are total nonsense...

First of all, an inductor generates reactance not resistance, which is why in the frequency domain, we have an "i" in the equation for an inductor's impedance. This is where the 90 degree phase shift comes from. This is a basic manipulation involving complex numbers. This comes from taking a Fourier transform of Faraday's law with self-inductance.

Second, the back-EMF generated across the inductor depends on the frequency of the AC signal, which is why frequency appears in the equation for the reactance of an inductor. So you can't compare what happens in a resistor (which is frequency-independent) to what happens in an inductor (which is frequency-dependent).


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