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#1
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Jim Kelley wrote:
Cecil Moore wrote: Yep, and for the same power level, a higher impedance usually means a lower current and vice versa. But the impedance *at* such points does not affect the current *at* those points? Cause and effect, Jim. Hint: The impedance equals v/i and is completely virtual, i.e. clearly a result, not a cause. Even the feedpoint impedance of an antenna is a ratio of v/i which is often simply the result of interference between forward and reflected waves. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#2
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![]() Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: Yep, and for the same power level, a higher impedance usually means a lower current and vice versa. But the impedance *at* such points does not affect the current *at* those points? Cause and effect, Jim. Hint: The impedance equals v/i and is completely virtual, i.e. clearly a result, not a cause. Little v over little i, Cecil? Are we talking instantaneous impedances now also? ;-) 73, Jim AC6XG |
#3
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Jim Kelley wrote:
Cecil Moore wrote: Cause and effect, Jim. Hint: The impedance equals v/i and is completely virtual, i.e. clearly a result, not a cause. Little v over little i, Cecil? Are we talking instantaneous impedances now also? ;-) Give me a break, Jim. I'm reading a book titled, "The Matter Myth", so my mind is blown and you are mostly empty space. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#4
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Jim, as I read the arguement, (and these guys are RIGHT), you are confusing
current being constant, in a media, with POWER being constant in a media (minus insertion loss's, of corse). for example, visualize a 1/4 wave (or a 1/2 wave dipole) , at the feedpoint, what is the current? even at a 1:1 swr, in coax, the current is constant, as you predict! But, WHAT, PREY, is the current, at the far END (S) of that 1/4 wave (DIPOLE) ? answer is 0 ! but the voltage has increased to that necessary to equal the amount of power impressed on it (I realize that any number times, 0 = 0, but, obviously, there is a physical limit approaching this, and also, for the power to stay the same, you would have an INFINANT voltage!! This is the stuff that R.F. BURNS are made of! And, when SWR is measured in a cable, current (and voltage) will vary, depending upon WHERE , in the cable that you measure current, (or voltage, even tho, the POWER must stay the same, again considering losses in the transmission line: As information, another Jim, NN7K -- No trees were killed in the sending of this message. However, a large number of electrons were terribly inconvenienced "Jim Kelley" wrote in message ... Cecil Moore wrote: Jim Kelley wrote: Cecil Moore wrote: Yep, and for the same power level, a higher impedance usually means a lower current and vice versa. But the impedance *at* such points does not affect the current *at* those points? Cause and effect, Jim. Hint: The impedance equals v/i and is completely virtual, i.e. clearly a result, not a cause. Little v over little i, Cecil? Are we talking instantaneous impedances now also? ;-) 73, Jim AC6XG |
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