Inductance Current 1

[Chezma]

The relation between voltage and current for an inductor is given by:
v = L . di/dt
I think this relation is valid only in steady state periodic cyclic operation and provided that the circuit contains a resistive element.
My question now :Why a resistive element should be contained in the circuit to make this relation valid ?

 

[Skogsgurra]

That relation (usually with a minus sign) is always valid. No exception.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

There have been many failed attempts to wind an induction coil without resistance.
There has been some success with super conductive magnet coils but although these coils are inductive and without resistance the nature of them is such that there is no variation in the current under normal operating conditions. Actually the only way to change the current is to introduce some resistance into the circuit by heating a small portion of the superconductive coil to drive it out of the superconducting region.

Bill
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"Why not the best?"
Jimmy Carter

 

[Chezma]

Even the simulator doesn't work if pure inductance is connected across AC supply unless a resistor in series included.

 

[Skogsgurra]

That is a common problem with simulators. They need a finite resistance so you do not get a divide-by-zero error in the steady state solution.

It is about reality. Not that the relation doesn't hold.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

It doesn't work in the real world either.
But perhaps the simulator is working. Zero resistance would be expected to return a value of zero in many of your equations. You may get a few divide by zero errors also.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Skogsgurra]

Total agreement there, Bill!

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Chezma]

So,why for a steady state cyclic operation we need a resistive load with the pure inductance to get a vlid relation between current and voltage according to the equation:
v = - L.di/dt ???
What is the secret in that resistor ???

 

[Skogsgurra]

As you can see, the di/dt is a derivative and the relation that is responsible for the transient solution.

For the steady state solution, Ohm's law is valid and if you make R in the I=E/R equal to zero, you will get divide by zero.

Nothing to it. Just plain elementary physics and math.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Chezma]

That is ok,but at transient L has infinit impedance and at steady state has a finite impedance and not zero.

 

[Skogsgurra]

If you can switch your simulator to 'transient only', that would work. But all simulators I have worked with always deliver a complete solution. That means transient AND steady state.

You may need to study how simulators work.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Chezma]

Let us explain this phenomena in different way:
Suppose a voltage of train of square wave of pluses (with only positive polarity)applied to a pure inductance,and according to the equation:
V = N.do/dt , whare o is the flux
The flux wave form will start from zero and increases linearly to its positive peak and the decreases to zero again and and then back to its positive peak ,amd never goes to negative polarity.
But because of a resistor in the circuit the flux will swing between positive and negative peaks.
My question again ,why a resistor in the circuit makes that difference ?

 

[electricpete]

When you suddenly apply a periodic excitation, the solution has two components:
1 - a steady state solution
2 - a transient solution

I assume your square wave averages to zero (otherwise if you apply it to inductor alone, the solution would grow to infinity). This problem then acts very similar to sinusoidal excitation

The transient solution has a dc component.
When a resistor is present, the steady state solution does not have a dc component. When the transient component decays to zero with time constant L/R, all that is left is the steady state solution with no dc current.

The simulator is telling you the correct solution. Try to solve the problem yourself and compare. If the square wave problem is too hard, try it with sinusoidal source.

=====================================
(2B)+(2B)' ?

 

[electricpete]

And going to your original question, I agree with the others. v = L*di/dt describes the behavior of an ideal inductor under both steady state and transient conditions.

By ideal inductor, I mean resistance neglected, saturation neglected, hystesis neglected etc. When you make those assumptions what is left is the ideal inductor which follows v = L*di/dt under all circumstances by definition

=====================================
(2B)+(2B)' ?

 

[Chezma]

Yes,correct the square wave averages to zero - it is identical around the horizontal axis.

 

[Chezma]

electricpete,what is the transient response of a pure inductance to a square wave ? With no resistor included.

 

[electricpete]

Let's look at a simpler excitation. Just apply a single pulse with your voltage source and then turn your voltage source back to zero for the rest of eternity.

The initial pulse will be enough to create some current in your R-L circuit. Can that current continu flowing forever?

I hope you answer is No. The resistor is dissipating energy and there is no input from the source, so the system loses energy and must eventually end in zero state. Once the voltage is switched off, the current decays according to I(t) = I(t0)*exp(t-t0)

You can think of your square wave as superposition of a series of pulses. The dc component is a characteristic of the initial startup transient and it decays to zero just like our single pulse. What is left after a long period of time is the steady state component which has no dc in an R-L circuit.



=====================================
(2B)+(2B)' ?

 

[electricpete]

Correction in bold:
Once the voltage is switched off, the current decays according to I(t) = I(t0)*exp(t-t0)
should've been:
Once the voltage is switched off, the current decays according to I(t) = I(t0)*exp((t-t0)*R/L)

=====================================
(2B)+(2B)' ?

 

[Chezma]

I am agree with you electricpete in the case of a resistor exists in circuit ,but what would be the behavior if no resistor is there, just only pure inductance.

 

[Skogsgurra]

Chezma,

Your 23 Jan 11 11:36 does not say anything new. It is the same question as before - simply because there is a linear and direct dependence between current and flux in an ideal coil.

So even if you try to make us understand, which you really do not need to, it would be much better if you tried to understand that what you have observed and what Pete, Bill and I are telling you is the truth.

Endless discussions like this one will only diffuse your field of view and are also rather tedious.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...