dum question: v = Ldi/dt 4

[lakevillethor]

Fellas,

I am having trouble understanding a basic concept. What I would like to know is how, via what mechanism, the voltage spikes in an inductive load. Lets say that you have an inductor hooked up to 240 VAC. The voltage will raise according to the rate of change of current times the inductors value. This obvious. But how does the voltage increase if the voltage source is a transformer. The transformer does not care what is load side of it, right (assuming you're not drawing a boat load of current)? If that is the case, how, again via what mechanism, does the voltage increase?
-AT

 

[peebee]

There is inrush, or at least there can be (depends on the exact phase angle of the voltage when you flip the switch on).

See

http://www.physics.brandeis.edu/phys29a_2003/AC.pdf

, page 252 (262 in the PDF file), Section 9.8.5., for a nice description of inrush currents. Wherever it says "transformer", just think "inductor."

The di/dt is happening BECAUSE of the voltage you applied. YOU applied the voltage, and as you connected it to the inductor, you DEFINED the voltage. The voltage is the cause, the di/dt id the EFFECT. The inrush current won't cause a voltage "ping", it is the RESULT of the voltage you applied.

 

[SidiropoulosM]

On energization, the current builds up slowly. Remember, current through an inductor cannot change instantaneously. Mathematically, the current is the time integral of the applied voltage. Another way of saying this is that the inductor accumulates volt-seconds and the current increases in proportion to the volt-seconds received. An inductor behaves very much like a capacitor, but instead of storing ampere-seconds (coulombs), it stores volt-seconds.

 

[peebee]

I guess part of the problem is how you define terms like "sudden" and "slowly" (and "is&quot.

As a power engineer, I tend to think of inrush as a rather instantaneous event that happens, like, BANG, when you turn something on, and then quickly decays out. Viewed in terms of steady-state operation, it's "sudden".

In terms of the waveform, though, yes, it rises smoothly and essentially exponentially from zero. Viewing the first few cycles, it is "slow".

 

[rbulsara]

lakevillthor:

Yes, if you quickly make and break the circuit of an inductor, it will cause a high voltage on open circuited inductor, the very phenomena used to advnatage in striking the starting voltage to 'start' a fluorescent lamp by switching a inductor called magnetic ballast. (its another matter that now a days electronic ballsts have taken over but performing the same task). The initial strike voltage is in the range of 1500 to 2000V, while the supply voltage is rated for 120 or 220V.

 

[peebee]

Clarification to rbulsara's post:

You can get a voltage spike on opening the circuit. But you won't get one on energization.

Rapid sequential make-break-make-break can add an overvoltage each time, up to something like 5x system voltage.

 

[jbartos]

Suggestion to lakevillethor (Electrical) Oct 10, 2003 marked ///\\Thank you all for posting - this discussion is really helping me understand it thoroughly. I always knew the mathematical formulas - sadly this was enough to get my by school without having to understand it thoroughly.
///Some formulae in modern engineering and science are so much detachec from reality that are dificult to explain them in something real, e.g. various transformations.\\ Let me set up an example and see if this is exactly analagous to the voltage spike phenomenon. Lets say you grab a hold of your friend and you shake him, slowly pushing him, and then slowly pulling him. After you have pushed him, he begins to move away from you (bear in mind that you are keeping your hands on him the whole time). As your arms extend and he begins to exert a force, pulling you with him. Exactly at that time, you start to pull him towards you. It takes more force to pull him towards you at this point because the momentum is carrying him the other direction. This larger force you exert is exactly the same as the voltage spike. lastly, is V = L Di/Dt like saying you cannot change velocity instantaneously in nature? You guys are really helping me put everything together here.
///It is highly commendable to seek thorough understanding of the mathematical relationships. However, the advent of software and software calculations changed the way the engineering and design are performed. Often, the most important part is to correctly interpret the input data for the software, correctly input data into software, which then guarantee the software correct output, if the software was properly verified and validated. This is a current trend. It means that to produce the required work, the software used formulae are not necessary to know, may be very difficult to understand and not really needed. However, it is convenient to know how to spot check the software output. That is where the understanding of relationships, analogies, proportions, etc. become good to know.\\
-Andrew Thoresen

 

[jghrist]

When you switch in an inductor, you instantaneously changing the voltage across it, not the current through it. The current will depend on the integral of the voltage. If the voltage is V·cos(wt), then the current is V·sin(wt)/wL. wL is the inductive reactance X[sub]L[/sub]. Note that sin(wt) lags cos(wt) by 90°.

 

[lakevillethor]

Gentlemen,

Thanks for all your responses. After going over this time and time again, one thing is clear...the voltage spike will occur when the circuit is opened.

One more question...In steady state conditions will an inductor act as a short circuit? I know that an inductor acts a short circuit in DC sitautions, but I am talking about steady state AC here. I think it does.
-AT

 

[SidiropoulosM]

No, the inductor will not work like a short circuit in steady state AC. A short circuit has zero impedance. The inductor will have an impedance which will be a phasor quantity, i.e. will have a magnitude and a phase displacement. The phase displacement will be 90 deg. The magnitude will be (frequency x L), where L is the inductance.

P.S. You can see that for DC the frequency is zero, so the impedance is also zero, therefore the inductor acts as a short circuit.

 

[lakevillethor]

I didn't really articulate what I meant very well. What you said is what I meant.
-AT

 

[jbartos]

Suggestion to lakevillethor (Electrical) Oct 13, 2003 marked ///\\Gentlemen,
One more question...In steady state conditions will an inductor act as a short circuit?
///Not for AC, only for DC or the DC part of composite of AC and DC, e.g.
i=(Vmax/|Z|)x[sin(sin(wt+alpha-theta)-exp(-Rt/L)xsin(alpha-theta)]
The component with exp is decaying because of the inductance causing the short circuit to the DC and resistance, in the circuit.\\
I know that an inductor acts a short circuit in DC sitautions, but I am talking about steady state AC here. I think it does.
///No, it does not since XL = 2 x pi x f x L. For DC at steady state, the XL = 0 Ohms.\\\

 

[peebee]

The higher the frequency, the more a capacitor looks like a short and the more an inductor looks like an open circuit.

The lower the frequency, the more an inductor looks like a short and the more a capacitor looks like an open circuit.