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I am confused whether voltage across inductor is $$ \ V = - L* di/dt\, $$ or $$ \ V = L* di/dt\, $$ Does it have to do with inductor being in AC or DC?

EDIT: okay based on answers I found, confused me that if there is a negative sign in the formula, it is back emf that is been generated by the current flowing through a coil and according to Lenz's law some force have to oppose that change! Now if $$ \ V = - V_{emf}\,$$ than how do I calculate actual back emf generated by an inductor?

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marked as duplicate by Chupacabras, winny, Sparky256, Adam Haun, laptop2d Jan 22 '18 at 6:31

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  • \$\begingroup\$ This is a very basic thing that is explained all over the internet. What is your source of confusion? \$\endgroup\$ – Matt Young Jan 21 '18 at 4:19
  • \$\begingroup\$ oh, i just see what the problem is! As back emf generated by self-inductance have negative sign according to Faraday's law and the voltage across an inductor is same as shown in second equation (without negative sign). Am I right @MattYoung? \$\endgroup\$ – Rajan Jan 21 '18 at 4:33
  • \$\begingroup\$ As long as you remember that the back emf opposes any change in current, you don’t need to worry about the sign. Just put the emf arrow on the inductor in the opposite direction to the assumed current flow and the value of the emf in that direction is \$L\frac{di}{dt}\$ \$\endgroup\$ – Chu Jan 21 '18 at 23:46
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When you apply a DC voltage source across a perfect inductor current ramps up at a rate equal to V/L. This ties in with the equation V = Ldi/dt.

However, there is a controlling mechanism inside an inductor that prevents the current rise being infinite straight after voltage is applied. This is usually referred to as a back emf. That back emf opposes the applied voltage and is therefore negative. This gives rise to the equation V = -Ldi/dt.

You can visualize this back emf if instead of a single coil inductor you had two perfectly coupled windings sharing a common ground. When you apply (say) +10 volts to the driven winding, +10 volts appears on the terminal of the "secondary" and it is +10 volts with respect to ground so, if you accept that the same +10 volts is also induced in the driven winding then it has to oppose the driving voltage.

That voltage is produced by the changing flux of the forward current in the driven coil rising positively. There is no reason to suppose that the driven coil doesn't also have this opposing voltage because it is subject to the same changing flux. That bit confuses engineers a lot but it is true for electrical engineering.

Now I'm potentially in deeper water because, you could then argue that if it is exactly +10 volts then surely no current can flow in the first place. The bottom line is (without going in too deep) is that the back emf and forward voltage are not quite the same. This then becomes a physics question!

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  • \$\begingroup\$ That's where I am confusing! Now you said that there is some bottom line on the back emf and applied voltage. if correctly I have to calculate back emf how do I do that? I mean I get that there is back emf generated when the voltage applied but is $$\ V = - L* di/dt\,$$ the right equation to calculate the back emf? \$\endgroup\$ – Rajan Jan 21 '18 at 22:03
  • \$\begingroup\$ Yes, it is the right equation but it’s a case of physics where the difference might be seen as important. For EE it isn’t of measurable importance. \$\endgroup\$ – Andy aka Jan 21 '18 at 22:18
  • \$\begingroup\$ Some are providing an answer's link and what I understood from that is while the inductor is consuming the power equation is without the negative sign but when is releasing the power it has negative power or the vice-versa is true? \$\endgroup\$ – Rajan Jan 21 '18 at 22:58
  • \$\begingroup\$ @Rajan if you mean an inductor can store energy and release it then yes. \$\endgroup\$ – Andy aka Jan 21 '18 at 23:16
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In DC, it behaves as a linear devices (for ideal cases, nonlinear when it comes to magnetostatics), so the voltage across an inductor is \$V=L\frac{di}{dt}\$. Things get nonlinear when there is resistance in the same circuit (e.g. an RL Circuit).

In AC, the impedance relies on the frequency of the voltage going through it, thus in steady state, the impedance can be characterized as \$Z_L = j\omega L\$. Therefore, the voltage across the inductor would be \$V=IX_L\$, where \$X_L\$ would just be \$Im\{Z_L\}\$ or \$\omega L\$. If you were to talk about the instantaneous voltage across an inductor, you would get: \$V(t) = V_{RMS}\sin (\omega t)\$. The voltage also leads the current by \$90 ^{\circ}\$.

You asked about emf. For emf, the equation is \$V=-L\frac{di}{dt}\$. I'm not going to delve into Faraday's Law because there's no reason to talk about that math and things get more complicated in AC. But if you're curious, go to this question if you want to learn about why there's confusion. Sergei's answer is actually a very good explanation.

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  • \$\begingroup\$ just one small thing if voltage induced in an inductor is normal equation without negative sign than V = -Vemf right? I mean the value of the voltage across the inductor and the back emf are same! \$\endgroup\$ – Rajan Jan 21 '18 at 6:39
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    \$\begingroup\$ Oh my friend, what a mess you wrote down. Inductor is linear by all means (at least in the ideal modelling considered here) and also telling apart AC and DC case is nonsense. Inductors only care their own voltage and current, it's just we humans that find easier different analysis approach upon applied stimuli. \$\endgroup\$ – carloc Jan 21 '18 at 7:58
  • \$\begingroup\$ I am afraid you got a rather peculiar idea of linear . I'd suggest you have a look back to your favorite books ;) \$\endgroup\$ – carloc Jan 23 '18 at 9:03

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