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AdrianH 21st Jan 2021 8:09 pm

Question on a bit of theory
 
1 Attachment(s)
This is something I will admit is a week subject for me Coils/inductances and induced emf with changing magnetic fields.

I am trying to read a series of articles from Practical Television circa 1957 and here is a partial copy of the text.: -
Attachment 225101

Now I must be missing a point as I can not understand where the formula comes from and how they achieve the figure of 1.5KV they get.
If reading it correctly they say L = 10mH inductance, I = 1 Amp and tr = 7 uS or 0.000007 secs so I actually get 2K8 Volts.

So please where am I going wrong.

Adrian

kalee20 21st Jan 2021 8:37 pm

Re: Question on a bit of theory
 
If the peak-to peak current is 1A, and the retrace time 7μsec, then current is changing at the rate of dI/dt = 1A/7μsec = 143,000 A/sec.

Now the defining equation for inductance is given by V = L x dI/dt.

Putting in the numbers, with L = 10mH, we get voltage = 1,430V or not far off the 1.5kV they claim.

Hope that helps!

ms660 21st Jan 2021 8:41 pm

Re: Question on a bit of theory
 
In the waveform I is peak (say 0.5 Amp) not peak to peak (say 1 Amp) therefore I is doubled (2I) in the formula shown.

Lawrence.

AdrianH 21st Jan 2021 8:52 pm

Re: Question on a bit of theory
 
Hello kalee20 ( sorry do not know your name )

I am not going to disagree with you, it was just the formula given is V = L.2I/tr as if the 1 amp was in fact +1 amp to -1 amp?

Not sure if you follow my thoughts, I will have to check the following months or two's mag for any corrections, hence why I got the 2.8KV.

Adrian

AdrianH 21st Jan 2021 8:58 pm

Re: Question on a bit of theory
 
1 Attachment(s)
Further down in the same article there is this statement-
Attachment 225105

Now again they talk of 1 amp peak to peak and it only works when you use I = 1, so not sure if one can understand my confusion.

Adrian

It is September 1957 PT mag pages 77, 78 in case anyone wants the article.

Also thanks to other responses they are coming in faster then I know.

AdrianH 21st Jan 2021 9:02 pm

Re: Question on a bit of theory
 
And posts disappearing as well!

Radio Wrangler 21st Jan 2021 9:04 pm

Re: Question on a bit of theory
 
It's actually very simple. The induced voltage is an instantaneous thing.

V = -L (di/dt)

di/dt is an expression stolen from calculus, it means the slope of the current versus time waveform in amps per second.

Draw the current/time waveform. Decide on what instant of time you're interested in the voltage. Place a ruler on that point, angled to be asymptotic to the curve. Calculate its slope, amps/second, and multiply by L. You now have the voltage at that instant. the minus sign just tells you which way round the voltage is.

If you're worried about risetimes, then notice that a rise (or fall) is usually an S-shaped curve with slower sections at each end. It goes fastest in the middle, so you get the highest peak voltage there.

If you want an average, place your ruler across the two places which define the beginning and end of the period you want to average, and use that slope.

Does the change of current make the voltage? or does the voltage make the change of current?

Both! they are inseparable, like voltage and current in ohm's law. You get both at once, inseparably and for free!

David

Edit: Crossposted..... not aware of any disappearing posts. I'll check... No, nothing's been done. Sometimes there are delays. Sometimes I forget to click 'post quick reply' and wonder where the occasional one has gone.

ms660 21st Jan 2021 9:09 pm

Re: Question on a bit of theory
 
Quote:

Originally Posted by Radio Wrangler (Post 1333242)
It's actually very simple. The induced voltage is an instantaneous thing.

V = -L (di/dt)

di/dt is an expression stolen from calculus, it means the slope of the current versus time waveform in amps per second.

Draw the current/time waveform. Decide on what instant of time you're interested in the voltage. Place a ruler on that point, angled to be asymptotic to the curve. Calculate its slope, amps/second, and multiply by L. You now have the voltage at that instant. the minus sign just tells you which way round the voltage is.

If you're worried about risetimes, then notice that a rise (or fall) is usually an S-shaped curve with slower sections at each end. It goes fastest in the middle, so you get the highest peak voltage there.

If you want an average, place your ruler across the two places which define the beginning and end of the period you want to average, and use that slope.

Does the change of current make the voltage? or does the voltage make the change of current?

Both! they are inseparable, like voltage and current in ohm's law. You get both at once, inseparably and for free!

David

Edit: Crossposted..... not aware of any disappearing posts. I'll check... No, nothing's been done. Sometimes there are delays. Sometimes I forget to click 'post quick reply' and wonder where the occasional one has gone.

So where did he go wrong then?

Lawrence.

kalee20 21st Jan 2021 9:11 pm

Re: Question on a bit of theory
 
Quote:

Originally Posted by AdrianH (Post 1333237)
Hello kalee20, I am not going to disagree with you, it was just the formula given is V = L.2I/tr as if the 1 amp was in fact +1 amp to -1 amp?

OK - the general formula V = L x dI/dt is correct.

If the current changes from +I to -I in tr, then the change in current is 2I, giving the formula stated. But they do clarify is as 1 amp peak-peak (which would have to be +0.5A to -0.5A).

The formula in your later clipping, energy = 1/2 x L x I^2, is also correct, this is the energy in an inductor when it is passing a current I. Nothing here about peak-peak, or times! It's analogous to a mass m moving at velocity v, having kinetic energy 1/2 x m x v^2, the factor 1/2 creeps in again.

Cobaltblue 21st Jan 2021 9:12 pm

Re: Question on a bit of theory
 
Only one post deleted and that was done by the user probably made a typo or some-such.
Cheers

Mike T

kalee20 21st Jan 2021 9:17 pm

Re: Question on a bit of theory
 
Quote:

Originally Posted by Cobaltblue (Post 1333247)
Only one post deleted and that was done by the user probably made a typo or some-such.

Mine - I posted it, looked at it on-screen, realised it was muddying the waters, and deleted it. Only live for 30 seconds!

AdrianH 21st Jan 2021 9:18 pm

Re: Question on a bit of theory
 
Not a disappearing post of mine but I was sure I saw a brief post from another member, perhaps the same member deleted their post after a rethink, perhaps my mind playing tricks, not an issue.

The part that confused me is in one paragraph 1 amp peak to peak must mean 0.5 to -0.5 Amps so they use 2 x 0.5 and in another paragraph 1 amp (p-p) means 1. I am reading the article as V = L x 2 x 1 over time when it is L x (2 x i ) over time.

AdrianH 21st Jan 2021 9:20 pm

Re: Question on a bit of theory
 
I just can not keep up, the emails notifications are just too slow and my typing and thinking, likewise.

I could do with a chat room at times.

ms660 21st Jan 2021 9:30 pm

Re: Question on a bit of theory
 
Quote:

Originally Posted by AdrianH (Post 1333251)
The part that confused me is in one paragraph 1 amp peak to peak must mean 0.5 to -0.5 Amps so they use 2 x 0.5 and in another paragraph 1 amp (p-p) means 1. I am reading the article as V = L x 2 x 1 over time when it is L x (2 x i ) over time.

Yes, I read it as 2I (as in post#3) It's easy to mix up 1 and I that's printed in the formula in that article.

Lawrence.

DerekG0VDV 21st Jan 2021 9:33 pm

Re: Question on a bit of theory
 
Good evening
The authors text and diagram are inconsistant. The diagram shows a 2 amp pk to pk waveform not a one amp pk to pk as set out in the text. Your calculation is correct for the diagram. The authors is correct for the text.
Regards Derek G)VDV

AdrianH 21st Jan 2021 9:41 pm

Re: Question on a bit of theory
 
Think I can justify it to my own brain, the use of I (capital I as p-p) and i (small i) to signify +i to -i.

Thanks all for the assistance, perhaps I can get myself to move on with the article.

I keep learning as I go, much of this for me would have been taught some 40 plus years ago never to be used again during my working career, just goes to show.

So as such, I thank you all for helping me. Just await the next lot of daft questions I will come up with.

I have one on bounced RADAR signals but will leave that for now.

Adrian


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