Ok, I gave him the wiki reference. Harry
----- Original Message ---- > From: Michel Jullian <[email protected]> > To: [email protected] > Sent: Thu, March 18, 2010 7:34:49 PM > Subject: Re: [Vo]:circuit diagram > > Nothing mysterious about this circuit, it's a silly boost converter without a > load. See: > target=_blank >http://en.wikipedia.org/wiki/Boost_converter 2010/3/18 > Harry Veeder < > href="mailto:[email protected]";>[email protected]>: > > > > > > ----- Original Message ---- >> From: Jed Rothwell < > ymailto="mailto:[email protected]"; > href="mailto:[email protected]";>[email protected]> >> > To: > href="mailto:[email protected]";>[email protected]; > ymailto="mailto:[email protected]"; > href="mailto:[email protected]";>[email protected] >> Sent: Thu, > March 18, 2010 5:22:20 PM >> Subject: Re: [Vo]:circuit > diagram >> >> Stephen A. Lawrence wrote: > >>By > the way, I should say "Thanks!" for >> taking the time to post all > these >>here. It's interesting, even if I >> don't believe > for a minute that it's OU. > > Someone should communicate > the >> gist of the comments here to the > author of the video. > Tell him to invest in >> an ammeter, for crying out > loud. > > - Jed > > I am ignorant about electronics but > I don't see what the fuss > is about since it is all DC current. If you > know the resistance and the voltage can't you safely infer that as the > voltage > rises and falls > so does the current? No, V=R*I works only on a > pure resistor. An inductor or a capacitor obey different laws. > I > still think that in certain "simple" circuits voltage measurements can serve > as > a pretty good indicator of current and power. Not > here. Michel __________________________________________________________________ Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com
