MH's ideal coil and voltage question

[Magluvin]

I just googled a few math sites and they all said that you can not divide any number by 0.  It is impossible and incorrect to do so.

So, now my question is...why is that?  One site said that you would get infinity by doing this which is why you can't do this.  Gee, and I thought I understood simple math, ha ha.

The intuitive answer, at least for me, would have been that division means breaking something into a certain number of equal parts and if there are no divisions into equal parts, then you are left with the original amount undivided.  But, this fails the multiplication test when you multiply the answer to double check it....

10 divided by 0 = 10.  Then 10 times 0 should equal 10, and of course it does not.  So, 10 divided by 0 either equals infinity, 0 or, it is undefined.  Some say it can't be done.  All of this depends upon which math sites you want to believe.

Bill

Maybe we wouldnt have to worry with the 0 value for resistance, where it may be replaced by a time of electron travel over a distance. Can an electron move faster than the speed of light? Is the speed of light infinity or instantaineous?  So there would still be a time thing to work with as I dont think T=0 can even exist. It means no time, nothing happened because that time didnt exist. 

Mags

[tinman]

Thanks MH for correcting me.

Although the question involves t=0 to t=13s, knowing what happens right at t=0 (the moment Vsource connects the inductor) is important.

It's also important to know what results when a number is divided by 0.

Let's see if examining this causes Brad to come to a different answer.

When you divide one number !x! by another number !y!,it is asking you to work out how many !y's! will fit into !x!,so how many 0's in 10?.
So to work out the L/R time constant,when the value of R is 0,then the answer is infinity,and so inductance in this case with the inductor being ideal, dose not play a part in current rise time. That being the case,you are placing an ideal voltage across a perfect dead short.

All L/R time constants are based around the ideal inductor where the resistance value is added in series with the inductor ,by way of a resistor that is external to the inductor. As the current rises,the voltage across the inductor drop's until it reaches a value of 0 volts. The problem here is that we have an ideal voltage that dose not drop,and there for the current will continue to rise to an infinite amount. If we remove the series resistor from our ideal inductor,we have a dead short due to the L/R time constant being infinite,and playing no part in current rise time.

Like MH said--there is !no! time constant,as it is infinite.
And so we now have an ideal voltage-->a voltage that will not change,being placed across a short that will not allow a voltage potential to exist across it.

So i stand by my answer-->you cannot place an ideal voltage across an ideal inductor.
If you did(theoretically),the current would rise instantly to an infinite value.


Brad

[poynt99]

When you divide one number !x! by another number !y!,it is asking you to work out how many !y's! will fit into !x!,so how many 0's in 10?.
So to work out the L/R time constant,when the value of R is 0,then the answer is infinity,and so inductance in this case with the inductor being ideal, dose not play a part in current rise time. That being the case,you are placing an ideal voltage across a perfect dead short.

All L/R time constants are based around the ideal inductor where the resistance value is added in series with the inductor ,by way of a resistor that is external to the inductor. As the current rises,the voltage across the inductor drop's until it reaches a value of 0 volts. The problem here is that we have an ideal voltage that dose not drop,and there for the current will continue to rise to an infinite amount. If we remove the series resistor from our ideal inductor,we have a dead short due to the L/R time constant being infinite,and playing no part in current rise time.

Like MH said--there is !no! time constant,as it is infinite.
And so we now have an ideal voltage-->a voltage that will not change,being placed across a short that will not allow a voltage potential to exist across it.

So i stand by my answer-->you cannot place an ideal voltage across an ideal inductor.
If you did(theoretically),the current would rise instantly to an infinite value.


Brad

You may wish to exercise a few examples to see if your conclusion is correct. Your conclusion being that if the time constant (Tau) is infinite, the load immediately presents itself as a perfect short, meaning the current will be infinite and instantaneous.

Here are a few: (in all cases, L=6H)

1) R=1, Tau=6s
2) R=0.1, Tau=60s (1min)
3) R=1m Ohm, Tau=60ks (16.6 hours)
4) R=1u Ohm, Tau=6M(million)s (1667 hours)
5) etc.

What is happening to Tau as R decreases?
If R could be 0, Tau must be infinite.
What happens to the inductor current after t=0 when Tau=infinity?

[Magneticitist]

then we have the extra confusing idea that whether a real coil or ideal coil, at T=0 the current is at absolute 0 because there is absolutely no resistance to yield a voltage drop, it's at a maximum across the coil. there is absolutely no current flow to yield a magnetic field, because the magnetic field that will not be created without current is opposing the current flow. hmm.

a superconductive coil will yield no external field because it perfectly contains it within the conductor resisting current change, perfectly. with absolute 0 resistance we are approaching an infinite magnetic field force and thus an infinite reactance. since infinity cannot overpower infinity, they simply cancel each other out in practice.

resistance is the factor that prevents a conductor from perfectly resisting an exterior magnetic field because its the factor that resists current.. with 0 resistance there is perfect opposition like a super cooled conductor opposing a magnet. in essence, can't an absolute 0 resistance in a conductor/inductor be considered both a dead short AND open circuit at the same time?

[Magluvin]

then we have the extra confusing idea that whether a real coil or ideal coil, at T=0 the current is at absolute 0 because there is absolutely no resistance to yield a voltage drop, it's at a maximum across the coil. there is absolutely no current flow to yield a magnetic field, because the magnetic field that will not be created without current is opposing the current flow. hmm.

a superconductive coil will yield no external field because it perfectly contains it within the conductor resisting current change, perfectly. with absolute 0 resistance we are approaching an infinite magnetic field force and thus an infinite reactance. since infinity cannot overpower infinity, they simply cancel each other out in practice.

resistance is the factor that prevents a conductor from perfectly resisting an exterior magnetic field because its the factor that resists current.. with 0 resistance there is perfect opposition like a super cooled conductor opposing a magnet. in essence, can't an absolute 0 resistance in a conductor/inductor be considered both a dead short AND open circuit at the same time?

There ya go. Resistance is a necessity for current to flow. Lol 

Mags