From: email@example.com (Gene De Guzman)
Subject: Re: HELP WITH INDUCTION SETUP
Date: 17 Nov 2002 13:14:23 -0800
References: <3DD40746.4C3B0E33@xympatico.ca> <3DD5A5A7.717E1C22@sympatico.ca>
NNTP-Posting-Date: 17 Nov 2002 21:14:24 GMT
Thanks for the warning and thank you very much for the clear and
simple explanation ... it was very enlightening.
"Walter Harley" wrote in message news:...
> "Joseph Legris" wrote in message
> > You had better find someone who can supervise you directly if you really
> > want to build a high-power pulse transformer - it is much too
> > complicated for a beginner.
> Joseph is quite right about finding someone to supervise you directly. But,
> to answer your question about what he meant about the resistances:
> The simple answer is "550 amps is one heck of a lot of current."
> Ordinary-sized wires just aren't going to work. To carry that kind of
> current you need big thick bars of metal, not little wires.
> The longer answer is: Ohm's Law, one of the basic laws of electricity, says
> that when a certain amount of current flows through a certain resistance, it
> creates a certain amount of voltage drop. It's like saying that when you
> push water through a pipe at a certain speed, the friction of the pipe makes
> the pressure at the output be less than the pressure at the input.
> (Resistance is like friction, water current is like electrical current,
> voltage is like pressure.) Mathematically, it's written E = IR, where E is
> voltage (in volts), I is current (in amps), and R is resistance (in ohms).
> By the way, this "hydraulic" analogy is not quite perfect. But it's good
> enough for this simple case.
> Wire, unless it's superconducting, has finite resistance. Small, but
> finite. Suppose you have a length of wire that has one ohm of resistance
> (easy to do). Then, to get 550 amps to flow through it, you need to have
> 550 volts to push with. If you only have 1.2 volts, the most current you
> can get is 1.2 amps. If you have 1.2 volts and you want to get 550 amps,
> you need to have lower resistance: specifically, 1.2 / 550 = 0.0022 ohms.
> That's where Joseph's number came from.
> Your windings are probably a fraction of an ohm or even an ohm. So Ohm's
> Law says there's just no way you can get the current you want without having
> a lot higher voltage. (Which, by the way, is risky.)
> There's also a law that says that power (in watts) equals current (amps)
> times voltage (volts): P = EI. So, if you had 1.2 volts pushing 1.2 amps
> through a 1-ohm length of wire, it would dissipate 1.2 * 1.2 = 1.44 watts of
> power, as heat. That would not get it very hot. But if you had 550 volts
> pushing 550 amps through that same wire, it would be 550 * 550 = 302.5
> kilowatts, which is a decent fraction (say, 0.1%) of the output of a power
> station, and the wire would instantly vaporize.