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Voltage divider; missing capacitor

  • Thread starter mathman44
  • Start date
  • #1
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Homework Statement



fc4m8p.jpg


I can't seem to get a start on this. Could anyone provide a hint or something to get me started? Thanks...
 

Answers and Replies

  • #2
407
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I don't entirely get this questions either...does the fact that the output is an exact reduced copy mean that if we restrict ourselves to sinusoidal input, for any input frequency, there is no phase difference between the input and output waveforms?
 
  • #3
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Anybody? I tried setting the "right" impedance equal to 1/9th of the "left" impedance to reform the voltage divider but this is a huge mess.
 
  • #4
1,929
228
Anybody? I tried setting the "right" impedance equal to 1/9th of the "left" impedance to reform the voltage divider but this is a huge mess.
You can do it that way. It's easier to work with conductivity: 1/impedance

The "right" conductivity is [itex] 10^{-6} + j \omega 10^{-10} [/itex]

The conductivity between A and B should be 9 times that.
 
  • #5
4,239
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You want the imaginary parts of the complex impedences to be in the same ratio as the real parts; 9 to 1.
 

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