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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 

What load for maximum power transfer?
Poll ended at Wed Jan 06, 2010 11:21 am
50+j0: 13%  13%  [ 1 ]
0-j100: 0%  0%  [ 0 ]
50+j100: 13%  13%  [ 1 ]
50-j100: 50%  50%  [ 4 ]
100-j50: 25%  25%  [ 2 ]
Total votes : 8

 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
m0jha wrote:

hi steve, the correct answer os #4.. 50-j100 for maximun power transfer. so isnt my 0-j100 the same thing?? needing to do nothing with the R as theres already R50 there but only adding 100 ohm of capacitance..

i understand the final value reading would be 50-j100 but my choice was based on waht would be needed to get to this figure from original reading of 50j+100

don't get me wrong im not trying to argue a point i just thought this may be the same thing adjustment wise..

billy

The way I see it the only 50 mentioned is in the source, which is 50+j100.
The question asks what the load needs to be i.e. 50-j100.
If the load was 0-j100 there would be cancellation of the reactive component and no resistance to generate a voltage across. Hence no power transfer.
All the power would be dissipated in the generator.
Effectively what causes your PA to blow up under mis-match conditions.

Please point out any errors here Steve. It's a long time since I opened this part of the brain. :lol:

Cheers

Tony

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Thu Jan 07, 2010 11:13 am
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Quote:
hi steve, the correct answer os #4.. 50-j100 for maximun power transfer. so isnt my 0-j100 the same thing?? needing to do nothing with the R as theres already R50 there but only adding 100 ohm of capacitance..

i understand the final value reading would be 50-j100 but my choice was based on waht would be needed to get to this figure from original reading of 50j+100

don't get me wrong im not trying to argue a point i just thought this may be the same thing adjustment wise..

billy

Hi Billy,

As Tony said, you have to have a 50 ohm resistive component in the load, as well as the 50 resistive component in the source; so the load impedance needs to be a total of 50-j100. In effect you then have two 50 ohm resistive components as you work your way around the series "loop" from one side of the voltage generator and back to the other side, sort of: 50+j100-j100+50. If you don't have the 50 ohm resistive component in the load there's no way for the load to dissipate power.

I hope that's clear - sorry if I've misunderstood what you were saying!

73,
Steve G3TXQ

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Thu Jan 07, 2010 11:51 am
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
cheers chaps.. it was me not looking at the diagram and reading the question properly and jumping in head first.. all clear now..

billy

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Thu Jan 07, 2010 7:32 pm
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