MH's ideal coil and voltage question

[verpies]

However it is possible to externally change the magnetic flux penetrating a shorted ideal inductor. Doing so will instantaneously cause a current to circulate through it ^*, in order to maintain the previous flux level penetrating its windings.  This is a voltageless current! - it cannot be measured by a voltmeter and it was not caused by a voltage source.

Better yet, the current induced in the above scenario is independent of the rate of change of flux (dΦ/dt) penetrating that inductor.

On the other hand, the emf induced across an open inductor is not independent of  the rate of change of flux (dΦ/dt) penetrating that inductor.

[poynt99]

Brad,

To be clear, I am in agreement with MH. It makes sense to me now, and apparently when I answered the question years back on OUR, I also got the answer correct.

Regarding my simulation, when using such small resistance values without changing some settings in SPICE (LT Spice must already be set to handle this), the simulation engine runs out of computational precision, which is why it "flatlines" with very low values. When the value is too low, the sim runs out of gas and starts making gross approximations, which is evident below with R=1f Ohm. But you do see that it is honing in on the 2.4A value? Any smaller in value and the trace just flatlines.

For an ideal inductor, yes Tau is infinite, but this has does not preclude current flow through the inductor. If however the inductance was an unrealistically large value like 1 million Henries, then yes the current would essentially be zero for a relatively long duration of time. That was my confusion. Tau simply determines the rise time, and since it is infinitely long, the trace becomes a nice straight line rather than the curve we normally see.

I was wrong in my analysis, but it is clear to me now.

[tinman]

Better yet, the current induced in the above scenario is independent of the rate of change of flux (dΦ/dt) penetrating that inductor.

On the other hand, the emf induced across an open inductor is not independent of  the rate of change of flux (dΦ/dt) penetrating that inductor.

That is because one is closed,and the other open.

If-as in the question i posted in my last post,the inductor is a closed loop,in that it has an ideal voltage source that completes the loop. Being an !ideal! voltage source,it too must be void of resistance and resistive losses,and so becomes part of the ideal inductor.

So now that the inductor and voltage source are a complete ideal loop,void of resistance,across no two points in that loop can a voltage exist.
And hence,once again,you cannot place a voltage across an ideal inductor,when current is flowing through that closed inductor loop.


Brad

So the question asked is referring to a closed loop scenario.

[tinman]

Brad,

To be clear, I am in agreement with MH. It makes sense to me now, and apparently when I answered the question years back on OUR, I also got the answer correct.

Regarding my simulation, when using such small resistance values without changing some settings in SPICE (LT Spice must already be set to handle this), the simulation engine runs out of computational precision, which is why it "flatlines" with very low values. When the value is too low, the sim runs out of gas and starts making gross approximations, which is evident below with R=1f Ohm. But you do see that it is honing in on the 2.4A value? Any smaller in value and the trace just flatlines.

For an ideal inductor, yes Tau is infinite, but this has does not preclude current flow through the inductor. If however the inductance was an unrealistically large value like 1 million Henries, then yes the current would essentially be zero for a relatively long duration of time. That was my confusion. Tau simply determines the rise time, and since it is infinitely long, the trace becomes a nice straight line rather than the curve we normally see.

I was wrong in my analysis, but it is clear to me now.

Rise time is an actual value,not a trace on a scope.
Is verpies also wrong? Quote: Since an ideal inductor must have a zero resistance, this means that it must be shorted (if it ain't shorted, it ain't ideal) and it becomes physically impossible to connect any real voltage sources in series with it.

Otherwise, I agree with the above statement.  Not only an ideal inductor is devoid of an asymptotic V/R current limit but also the current through an inductor of infinite inductance, that is somehow connected to an ideal voltage source, could never change because of the implied zero di/dt at any voltage.
How long will it take for the current to reach it's peak in an ideal inductor(regardless of inductance value),when that inductor has no resistance,and is supplied with an ideal voltage?

Once you have this value,you can divide it by 5 to obtain your time constant for the rise of current in that ideal inductor. Mh is using math that applies to an inductor on the understanding that that inductor will reach a maximum current value in a finite time. This cannot be applied to an ideal inductor,where the current will never reach a maximum value when an ideal voltage is placed across it.


Brad

[allcanadian]

 . . . "Mad Hatter: "Why is a raven like a writing-desk?"
"Have you guessed the riddle yet?" the Hatter said, turning to Alice again.
"No, I give up," Alice replied: "What's the answer?"
"I haven't the slightest idea," said the Hatter"