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QwertyXP
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Join Date: Oct 2013
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Back EMF Across Solenoid

10/26/2013 3:31 AM
Consider an ideal solenoid (no resistance, no leakage reactance etc.) connected across an AC supply. The back EMF induced in it will be exactly equal and in opposite direction to the source voltage (which means that when a certain terminal of the AC supply is positive, the side of solenoid connected with it would also be positive, and vice versa).

My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out? It's like having having a circuit with only two batteries and terminals of similar polarities shorted with each other.

The equation below doesn't appear to be balanced:

V(source)=Back EMF (which is equal to source) + CurrentxReactance

when back EMF is equal to source, the CurrentxReactance part should be zero!?

I've read quite a few explanations on the internet but have yet to fully understand what's happening here.
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Tornado
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#1

Re: Back EMF across Solenoid

10/26/2013 4:10 AM

Keep reading. Maybe you will still never understand it.

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Re: Back EMF across Solenoid

10/26/2013 4:13 AM

Does it by chance have anything to do with c < ∞, and/or Ω > 0?
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Winfield Hill
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#3

Re: Back EMF across Solenoid

10/26/2013 10:02 AM

Your ideal solenoid has to have some real-world property remaining, you can't arbitrarily zero out all of them! Let's say you grant it only winding and core inductance. Then there certainly will be current flowing, but it'll be 90º out of phase with the ac line. Let's say you grant it only winding copper resistance (usually the dominant parameter in a typical say 10-watt solenoid), then you'll have ... well, you know.

A real solenoid will have both inductance and resistance, some of which will be related to the ac core losses. In all three of the above cases you can posit that the solenoid works, but it'll certainly have a considerable ac current flowing to make it work!

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lyn
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#4

Re: Back EMF across Solenoid

10/26/2013 11:49 AM

Maybe this guy knows:

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TrevorM
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#5

Re: Back EMF Across Solenoid

10/27/2013 1:36 AM

You don't understand the terms you use well enough.

In your idealised case the terms "back emf" and "current reactance" are the same thing This means that your equation is wrong.

The correct formula is that (when added as vector amounts) the source volts equals current reactance volts plus resistive volts.

You should know that the current reactance terms and resistive terms are at 90 degrees to each other, and that in inductive circuits the resultant current always lags the driving voltage.
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dvmdsc
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#6

Re: Back EMF Across Solenoid

10/27/2013 2:02 AM

When your solenoid is a coil there is also a "Q"- factor that counts.

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jhhassociates
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#7

Re: Back EMF Across Solenoid

10/27/2013 6:26 AM

Apply the standard Ohm's law formula to your idealised solenoid V = I x R.

But in your case R = 0 therefore I must =∞. That is quite a lot of current.

Unfortunately there are some very strange results in mathematics when dealing with infinity so you are going to have to do this experimentally.

To achieve such a large current you will need a perpetual motion machine running at slightly over 100% (101% is perfectly adequate). Feed the output of the machine back to the input to create a cascade positive feedback effect. Once the current has built up to an infinite level, connect it to your solenoid and observe the results.

WARNING Do not try his experiment without proper supervision, it could be dangerous. For details of where to find qualified supervisors contact your local lunatic asylum and explain your requirements.
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Rixter
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#8

Re: Back EMF Across Solenoid

10/27/2013 8:10 AM

Whatever the load, the voltage across it will equal the applied voltage, obviously. If the load is a resistor, the voltage will be proportional to the current flowing through the resistor. If the load is an inductor (solenoid), the voltage will be proportional to the rate of change of the current flowing through the solenoid.

Since you are applying the voltage, the current is the dependent variable. In other words, if you apply a voltage to a solenoid, the current through the solenoid will change at a rate proportional to the voltage applied.
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WoodwardDL
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#9

Re: Back EMF Across Solenoid

10/27/2013 12:46 PM

The power dissipated by a perfect inductor is zero for any applied AC voltage. The power curve (V x I) is 90 out of phase with applied voltage, and is halve positive and half negative, resulting in zero! See this URL:

http://www.allaboutcircuits.com/vol_2/chpt_3/2.html
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