MH's ideal coil and voltage question

[MileHigh]

The waveform is correct.

[tinman]

I might remind you MH,that your question asks 'what happens from T=0!
I have answered that question in way of the current trace graph,as that is about all that happens-rising and falling current flows. You have already stated that i need not worry about the magnetic fields'although there value would just rise and fall with the current. As we are dealing with ideal components,then no heat or power is dissipated from the closed loop.

You question dose not ask as to how i calculated the answer,only what happens through the entire cycle at each time period.

So you have my answer,and im asking you if it is correct.
Yes or no is all thats needed.


Brad

[Johan_1955]

We must thank the likes of MH. poynt etc. for their patience in inching us along.

Yep, you''''re right, present is prepared out of thank, a fan-site is now running, 682 members, also including:

[tinman]

The waveform is correct.

I am getting too much of a Jiffy Pop popcorn vibe due to the seemingly instant magical appearance of that waveform.  Please explain the whole waveform.

There wasnt any Jiffy Pop (what ever that is) popcorn moment MH. As i stated to you some time back,a generic calculated answer was quite simple--refer to post 569,where i posted the formula used from T=5 seconds to T-7 second's,as also posted below.We just subtract or add the calculated value to that of the previous value--depending on what polarity you stated in your question. Then it's just a matter of playing-join the dots. But my argument was and still is,that is not the correct answer/current trace when using ideal's.

The 2.4 amps flowing through the loop at T=5 seconds,will be an impedance against the current that is induced at T=5 seconds,that is of the opposite polarity. The calculated current peak value at T=7 seconds,is calculated based around the inductance value,time and a starting current of 0-->I=Io. As the polarity is now reversed,that current value is actually I=I-2.4amps.
So i am not sure that simply subtracting(due to it being the negative voltage part of the cycle) the calculated peak current value at T=7 seconds from the value at T=5(2.4 amps) seconds is correct,as the impedance value of the coil at T=5 seconds is higher than would normally be seen by the current flow,should the current start with a value of 0.


Brad

[MileHigh]

Well, you are seemingly making progress, and I am going to ignore the various errors in what you posted because what you are saying is more important in the "error glitches" in your posting.

But I have a question for you:  Way back very early in the thread I posted that formula and told you that it was basically the short answer to the question.  When I did that you went into a small frenzy and you insulted me like a drunk sailor.  So what gives, how do you explain using the formula now when before when I showed you the very same formula you insulted me repeatedly like a drunk sailor?

Don't even bother saying that you don't drink, it's just an expression.