TransWikia.com

Inductor's back-emf vs voltage drop across resistor

Physics Asked by user4205580 on September 5, 2021

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?

2 Answers

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.

Answered by gleedadswell on September 5, 2021

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).

Answered by sparaps on September 5, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP