Standing Waves and Impedance

Discussion in 'General Electronics Chat' started by pyro, Oct 22, 2006.

  1. pyro

    Thread Starter New Member

    Sep 5, 2006
    3
    0
    Hello!

    I have a little question about standing waves in AC chapter 13!

    There's an example with two pictures:
    [​IMG]
    [​IMG]

    I think I understand them. Because of the wavelength and the frequency, the source sees a different resistance.
    Now my question:
    The line is not terminatet with a 100 Ohm resistor, but with 50 Ohm. This means the wave is reflected inverted. So the voltage at the end is 0,4V (0,5-0,1 reflected). This means 0,4^2/50 Watt will be dissipated as heat. Now the reflected wave comes back. Because of the length of the cable and the wavelength, the voltage at the source now is 0,6V (0,5+0,1, because of Lambda/4 wave length). That means 0,4V (1V-0,6V) is dropped across the source. This means 0,4^2/75 is dissipated into heat there. But now I have a problem with this: Because of the reflected voltage, there's less current drawn from the source (0,4V/75 Ohm). But this would mean that there's more heat dissipated in the smaller resistor at the end of the line than at the source. But according to Maximum Power Transfer Theorem this would be impossible!
    Now what's correct? 0,4V/75 Ohm, that's surely not enough current for creating 0,4^2/50 Watt heat at the end of the line!
    I'd be glad if you could help me
    Thanks
    pyro
     
  2. pyro

    Thread Starter New Member

    Sep 5, 2006
    3
    0
    Does nobody have an idea?
    I thought standing waves aren't that difficult?
     
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