t
field
ance,
probe
n the
curious
THIS IS NO GAME YOUR HEADS UP MAY HELP THE GROUP THOUGH
I HAVE ALREADY GIVEN MY INPUT TO THIS TOPIC
I AM PROTEUS
t
field
ance,
probe
n the
curious
THIS IS NO GAME YOUR HEADS UP MAY HELP THE GROUP THOUGH
I HAVE ALREADY GIVEN MY INPUT TO THIS TOPIC
I AM PROTEUS
-- Finally owning up to your ignorance, huh?
-- As usual, more of your evasion since, clearly, the reference was to your waste of bandwidth caused by the unhelpful abuse you heap on JT while condemning PROTEUS for the waste of bandwidth caused by his unhelpful abuse of Fester Bestertester. True hypocrisy.
--- OK.
A passive clamp-on ammeter is essentially the secondary of a transformer wound on a core that can be opened or closed in order to get it around a conductor so the current in that conductor can be measured without cutting it and using a conventional ammeter.
A transformer is used to transfer _power_ from a source into a load, thus the power, P2, required by the load, will be that power, P1, supplied by the source.
In an ideal transformer there will be no losses, and then P1 and P2 will be equal.
Next, the voltages on the primary and the secondary will be directly proportional to the ratio of the number of turns on the primary to the number of turns on the secondary, and the currents in them will inversely proportional to the turns ratio.
With that in mind, let's say we have a transformer with a one turn primary and a 1000 turn secondary, across which is connected a 1000 ohm resistor which is dropping one volt.
The current in the load will then be:
E 1.0V I = --- = ------ = 1e-3 ampere = 1 milliampere R 1e3R
and the power dissipated by the load:
P = IE = 1e-3A * 1V = 1e-3 watt = 1 milliwatt
Now, since the turns ratio is 1000:1, the current in the primary is inversely proportional to the current in the secondary, and since the current in the secondary is 1 milliamp, the current in the primary must be:
Is * nS 1e-3A * 1000t IP = --------- = -------------- = 1 ampere nP 1t If we now double the current in the primary, the current in the secondary will be doubled as well, causing the 1000 ohm resistor to drop
2 volts.If we triple the primary current, the secondary current will be tripled as well, the resistor will drop 3 volts, and so on...
So, what we have is a device which will have an output voltage which is directly proportional to the input current and which we can use to determine the input current by measuring the output voltage.
JF
.
Jim Thompson is notoriously enthusiastic about plonking people he disagrees with. Someone who took him seriously might be upset by it. This strikes me as information that Proteus IIV might find useful.
If I felt like heaping abuse on Jim Thompson I'd be more explicit about it.
The image of the incorrigibly offensive John Fields ever being put on the defensive is amusing. Not as amusing as the idea that you might ever have consulted a text-book, but distinctly comical all the same.
But we can rely on you for seriously side-splitting - if unintentional
- comedy. The clown who still thinks that the 555 is the answer to every circuit problem imagines that he can call somebody else a fatuous ass.
-- Bill Sloman, Nijmegen
Since the original claim was
" >Wrapping some turns around the power company's lines will get you many, many
This isn't the reason - lines is plural and the nett current through the lines as a bunch balances out to zero.
Wrapping a clamp-on meter around one line means that there is current circulating around the clamp - the current that goes through the selected line in one direction is matched by equal and opposite current flowi g through the other lines in the other direction. The coupling coefficient is unlikely to be good, but it is finite.
-- Bill Sloman, Nijmegen
al
If one can rely on that "passive". Someone who knew a little more than you might extract some of the power circulating through the wire and produce an active solution without an external power source.
-- Bill Sloman, Nijmegen
FINALLY an answer on-topic. Thank you.
After watching the 3 Stooges act that is aee / sed...
Sheesh!
I AM FBt
Admitted that the S/N ratio on Usenet can be frustrating. But did you stop to consider the possibility that a) you failed to grasp other attempts to explain it to you; b) your question was so elementary for *this* group that few people bothered; c) your last post might be taken as a slap in the face by those who tried to help.
-- Since that\'s obvious to the most casual observer, the context of his statement must have been about wrapping some turns around [one] of the power company\'s lines, which I addressed by referring to it as "the power line".
-- Bullshit, liar. What you were doing was telling P3 to clean up his act or you\'d plonk him, and then, out of the blue, you just _had_ to take another little snipe at Jim. I guess you just can\'t get over his reporting you to the FBI, huh?
-- Can\'t come up with anything, huh? What a surprise!!!
Right you are.
A big thank you to those responders who gave answers to my question. Much appreciated.
My comment was addressed to the "noise". :-)
FBt
te:
at
aelf.
ur
I don't plonk anybody, and I certainly wasn't threatening that I'd plonk him
If he'd known how much pleasure I was going to get out of pointing how how seriously far out of touch with reality that proved him to be, he probably wouldn't have done it. I know he is conscious of his duty to protect his country, but it isn't as if he is fanatical about it.
Even you will have to admit it has to be the funniest prat-fall we've ever had around here.
kNice to see you sending yourself up from time to time. There have been times when I've imagined that you lack a sense of humour, but that really is funny. Nice one.
eOr any more. Your judgement isn't all that great outside of electronics either.
-- Bill Sloman, Nijmegen
...
rNice try.
-- Bill Sloman, Nijmegen
field
nce,
robe
the
r aYou asked about the Fluke i200s current clamp probe. I've never used one, and I'm not sure that I've even seen one, which discouraged me from trying to improvise an explanation.
John Field's response - now that he has finally got around to making the kind of useful post that he claims to represent the bulk of his output - does seem to be plausible.
From time to time we get responses from the people who designed the gear under discussion, but you don't seem to have been that lucky.
-- Bill Sloman, Nijmegen
Yes. Classic AC clamp-on ammeters are simply transformers. One "turn" through the clamp, many turns in the fixed coil for output. The output feeds into a voltmeter.
Those are AC-only devices. There are also Hall-effect clamp-on ammeters, and those work for both AC and DC. These have been available for a decade or so, and pricing is now down as low as $60. I used to have one that could read down to about 500mA DC, and it only cost $129. Very useful in robotics and controls work.
John Nagle
Which make & model would that $129 model be? It's always useful to know someone else's favorite tools...
Dave
? "Fester Bestertester" ?????? ??? ?????? news: snipped-for-privacy@news.eternal-september.org...
There's nothing fancy about that, the electricity meters of a medium-voltage consumer (real and reactive energy) are powered from the two potential and the two current transformers, without any other power supply (medium voltage=15 kV in Crete).
-- Tzortzakakis Dimitrios major in electrical engineering mechanized infantry reservist hordad AT otenet DOT gr
-- Of course, you fraud, since by snipping the rest you sidestep the issue, which is your ignorance in believing that a solenoid wound around an alternating current carrying conductor can be used to extract power from the varying magnetic field surrounding that conductor. Nothing could be further from the truth, as demonstrated here: news:26iag5hjpub42ookl0nk74vc3ffgs7316q@4ax.com Since, conveniently, you don\'t have access to abse, I\'ll take the liberty of emailing you the photos as soon as I post this. Enjoy. :-) JF
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