MH's ideal coil and voltage question

[Magneticitist]

Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?

no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.

if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebral-measuring than genuine intentions of shedding light on the topic (that was) at hand.

[MileHigh]

Brad:

I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.

For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.

It seems crazy doesn't it?  All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.

Okay, so after one second the current is 0.8 amperes.

I attached the formula from Hyperphysics for a voltage source energizing an RL circuit.  You can see the red trace for the current and the formula.  You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong.  The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.

So now I am going to prove to you that the first part of the answer given to you is correct.  We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms.   Sound fair enough?  If I am correct the two answers for the current value at one second should be very close to each other.

From the red formula:

i = 4/0.001 * (1 - 2.7182818^(-0.001/5))

i = 4000 * (1 - 2.7182818^-0.0002)

i = 4000 * (1 - 0.99980001999866673333)

i = 4000 * 0.000199980013332666693332

i = 0.799920005333 amperes.

So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor.   You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.

So there is your proof in the pudding.  It's time for you to move on and try to answer the question and more importantly understand the how and the why.

MileHigh

[tinman]

The Tau business is fairly simple to explain.

When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line.  Note that it is a straight line with a constant slope of V/L.

So Tau at infinity just means the current trace is a perfectly straight line.  Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line.  i.e.; "There is no time constant."

So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.

Since we are discussing limits, the only possible way for the current to flat-line at zero "forever" would be for the inductance to be infinity.  Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.

So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.

Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal.

I will once again post the definition of an ideal voltage.

Quote: An ideal voltage source is a two-terminal device that maintains a fixed voltage drop across its terminals. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits.

This is the definition that is used to define behaviours in circuits such as yours.
The voltage of 4 volts from your ideal voltage supply,will not alter during the 3 seconds after T=0.

Regardless of whether it is L/0 or R/0,Tau is always infinite,meaning that the current will not rise in the case of an ideal inductor.
You are trying to use real world calculations to define outcomes in an ideal situation that dose not exist,and then you are surprised by the outcome.

Poynts simulation shows the results expected,and if he ran that simulation for !lets say! 100 years,we would see the current rise expected for such a low value of resistance--there is nothing wrong with his sim.


Brad

[MileHigh]

Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?

no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.

if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebral-measuring than genuine intentions of shedding light on the topic (that was) at hand.

I will repeat to you, the circuit is only "imaginary" if you have a bad attitude and refuse to open up your mind and learn something new and you refuse to apply your knowledge.  I will repeat to you, that there is almost no difference between an inductor with zero ohms resistance and an inductor with 0.001 ohms resistance and I just proved it to you and Brad in my previous posting where I did the "real" calculation.

If you can't understand how closely an ideal inductor matches a real inductor and how much can be learned from trying to understand how an ideal inductor works then read over the thread again.  I am not here to argue that anymore.

<<<  if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged  >>>

The quoted sentence above does not really make any sense and that is the whole point of this exercise.  Understand how an inductor works so you can talk about using one sensibly and also understand how a Joule Thief works and then build better Joule Thieves.

Yes there are different ways of using inductors in circuits.  A good start is right here.  This example applies directly to the energizing phase of a Joule Thief.  If you understand the basics then moving to regular inductors is easier.  And like I told you in many circuits real inductors for all practical intents and purposes act exactly like ideal inductors.

"Spiteful cerebral-measuring" is you just being defensive because you are in the realm of real electronics here, not the usual throwing around of meaningless catch phrases.

The best thing you could do for yourself is have the same spark as Brad and follow it through and learn something real and substantial about electronics.  I will remind you, you read the question and you did not have the slightest idea what to do.  I am hoping that you and Brad arrive at a successful conclusion to this debate.

I am not going to touch any Joule Thief thread, sorry.

[tinman]

Brad:

I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.

For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.

It seems crazy doesn't it?  All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.

Okay, so after one second the current is 0.8 amperes.

I attached the formula from Hyperphysics for a voltage source energizing an RL circuit.  You can see the red trace for the current and the formula.  You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong.  The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.

So now I am going to prove to you that the first part of the answer given to you is correct.  We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms.   Sound fair enough?  If I am correct the two answers for the current value at one second should be very close to each other.

From the red formula:

i = 4/0.001 * (1 - 2.7182818^(-0.001/5))

i = 4000 * (1 - 2.7182818^-0.0002)

i = 4000 * (1 - 0.99980001999866673333)

i = 4000 * 0.000199980013332666693332

i = 0.799920005333 amperes.

So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor.   You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.

So there is your proof in the pudding.  It's time for you to move on and try to answer the question and more importantly understand the how and the why.

MileHigh

I have filled in the blanks in your diagram,and also noted that there will be no voltage curve when the voltage is ideal,and set at 4 volts for 3 seconds-as per your original question.

I am pretty sure Poynt has not yet finished looking into this,and i think you will find that i am correct when i say that your ideal voltage of 4 volts across the ideal inductor for the first 3 seconds from T=0 ,remains 4 volts,as an ideal voltage dose not change in value over time-regardless of the load,and so there is no such voltage curve as you have depicted in your diagram .

Brad