Understanding Back EMF and Inductors: A Closer Look at KVL and Induced Voltage

[nDever]

Hi guys,

I'm pretty sure this has been answered somewhere around here.

So, consider a pure inductance (zero resistance) connected in series to a battery. We know that the magnitude of the induced voltage across the inductor is given by

$E=L\normalsize$$\frac{di}{dt}$.

According to KVL, the sum of the voltage drops must equal the sum of the source emfs, so according to the equation, the current in this circuit must increase at a rate such that the induced electromotive force constantly equals the battery voltage. Now with respect to polarity, if the applied voltage across the inductor is positive then the polarity of the induced emf is negative. All of this, I think I understand...correct me here if anything is wrong.

What I'm not understanding is this. If the induced voltage across the inductor is always equal and opposite the source emf, how can there possibly be a current at all? If there is no difference in electric potential, how can there be any work done?

 

[sophiecentaur]

Haha
You specified an inductor with NO resistance, which is impossible. So you can't complain when the result of your calculation doesn't make sense. As soon as you introduce a finite resistance then a current can flow and everything's alright.

 

[NascentOxygen]

Your equation is correct for both magnitude and sign (polarity). The battery maintains E constant, so di/dt is held constant. This means that current increases linearly with time, forever increasing, and at a constant rate.

The exciting part is when you finally decide to disconnect the inductor from the battery! A quick disconnect implies that current is quickly disrupted, meaning that di/dt will have a very large magnitude ...

 

[sophiecentaur]

Your equation is correct for both magnitude and sign (polarity). The battery maintains E constant, so di/dt is held constant. This means that current increases linearly with time, forever increasing, and at a constant rate.

The exciting part is when you finally decide to disconnect the inductor from the battery! A quick disconnect implies that current is quickly disrupted, meaning that di/dt will have a very large magnitude ...

Be very careful of that. It can be literally lethal.

 

[Crazymechanic]

and even if your inductor would have zero resistance in the conductor, still current would flow with applied dc after the back emf would drop.
Because as it is said , an inductor opposes the CHANGE in current not current itself, because for the opposing electric field to be there it needs a cahanging magnetic field which can only be there if the input current is changing , once it reaches a certain level say the emf of the battery then it just stays there and the current starts to flow through the inductor.

 

[sophiecentaur]

and even if your inductor would have zero resistance in the conductor, still current would flow with applied dc after the back emf would drop.
Because as it is said , an inductor opposes the CHANGE in current not current itself, because for the opposing electric field to be there it needs a cahanging magnetic field which can only be there if the input current is changing , once it reaches a certain level say the emf of the battery then it just stays there and the current starts to flow through the inductor.

In the 'gorblimey' case, the connection has to be to a true voltage source. In which case no current can even start to flow. My reasoning here is that you would have to give a timescale for this to happen. It would have to be zero time or infinite time (the only two value that are not arbitrary). It cannot be zero so it has to be infinite time.

To put it 'properly, the RL time constant is L/R, which is infinite if R is zero.

 

[Crazymechanic]

I see your approaching this from a pure calculations viewpoint, fine with me as the OP was interested in that too.:)

 

[Dale]

Now with respect to polarity, if the applied voltage across the inductor is positive then the polarity of the induced emf is negative.

I don't know what you are talking about here. The voltage across the inductor is the applied voltage. There is no separate induced EMF to worry about in circuit theory.

If your battery is a perfect 1.5 V source with no resistance and if the negative terminal is grounded then the positive terminal is at 1.5 V. The E in the equation above is 1.5 V, and that is the voltage across the inductor with the higher voltage being on the side attached to the positive terminal.

Btw, for a two element circuit there is no difference between parallel and series.

 

[sophiecentaur]

@Crazymechanic
Well, in my first post, I pointed out that the ideal case is actually a nonsense scenario.

Totally counter intuitive that adding resistance can make something happen quicker!

 

[NascentOxygen]

In the 'gorblimey' case, the connection has to be to a true voltage source. In which case no current can even start to flow.

I see no problems with the use of an ideal voltage source powering an ideal inductor. It will work just fine without the need for any resistance.

 

[sophiecentaur]

I see no problems with the use of an ideal voltage source powering an ideal inductor. It will work just fine without the need for any resistance.

I was trying to find a suitable mechanical analogue to this one. Pushing against something with no resistance and it just stays there!

I have it, of course. It's trying to push a slippery object with a tangential force.