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Complex impedance problem

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  • #1
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parallell.png


For the problem on the left, the 2 capacitors are parallel to each other so the 1/z = 1/-2j + 1/-2j = 2/-2j

so 1/z = 1/-j = -1/j

so the total impedance of the 2 capacitors, z = -j

Now if you add this to 5j + 4 (the total impedance of the resistor and inductor) you get 4j+4. However this doesn't give the right answer..
 

Answers and Replies

  • #2
berkeman
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parallell.png


For the problem on the left, the 2 capacitors are parallel to each other so the 1/z = 1/-2j + 1/-2j = 2/-2j

so 1/z = 1/-j = -1/j

so the total impedance of the 2 capacitors, z = -j

Now if you add this to 5j + 4 (the total impedance of the resistor and inductor) you get 4j+4. However this doesn't give the right answer..
The capacitors are not in parallel. They have an inductor between them. You need to write the KCL equations and solve them simultaneously...
 
  • #3
gneill
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parallell.png


For the problem on the left, the 2 capacitors are parallel to each other
No they are not. One pair of their leads (the bottom ones) are connected together, but there's an inductor in the way of the other connection. So, not parallel.


EDIT: Oops. Berkeman beat me to the punch!
 
  • #4
berkeman
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The capacitors are not in parallel. They have an inductor between them. You need to write the KCL equations and solve them simultaneously...
Actually, it's easier than using KCL equations. Do series-parallel combinations of the complex impedances to get the overall input impedance...
 
  • #5
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Actually, it's easier than using KCL equations. Do series-parallel combinations of the complex impedances to get the overall input impedance...
So if they are not in parallel they must be in series? In which case we just add them? -2j +-2j =-4j?
 
  • #6
berkeman
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So if they are not in parallel they must be in series? In which case we just add them? -2j +-2j =-4j?
For the left circuit, you first make a series combination, then a parallel combination, then a final series combination. You fold the circuit up from right to left in this case...
 
  • #7
166
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For the left circuit, you first make a series combination, then a parallel combination, then a final series combination. You fold the circuit up from right to left in this case...
Ah I see..

For the one on the right, I done the following but I don't get the answer correct?

the 4 ohm resistor and the 1j ohm inductor are in series so you just add them up to get 4+1j, then this combined total is in parallel with the 2j ohm inductor so

1/(4+1j) + 1/(2j) = 1/Z

so Z = 16/25 + 38j/25 (once simplified and rationalised)

then you add -1j ohm to the answer of the above (as the -1j capacitor is in series with our above answer) so

Z = 16/25 + 13j/25 (2)

Then the remaining 1 ohm resistor (the one to the left) is calculated in parallel with (2)

so 1/Z = 1/1 + 1/(16+13j/25)

But simplifying the above and finding it in the form r<theta, gives me a different answer to the given answer. Where have I gone wrong?

Thanks
 
  • #8
berkeman
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Ah I see..

For the one on the right, I done the following but I don't get the answer correct?

the 4 ohm resistor and the 1j ohm inductor are in series so you just add them up to get 4+1j, then this combined total is in parallel with the 2j ohm inductor so

1/(4+1j) + 1/(2j) = 1/Z

so Z = 16/25 + 38j/25 (once simplified and rationalised)

then you add -1j ohm to the answer of the above (as the -1j capacitor is in series with our above answer) so

Z = 16/25 + 13j/25 (2)

Then the remaining 1 ohm resistor (the one to the left) is calculated in parallel with (2)

so 1/Z = 1/1 + 1/(16+13j/25)

But simplifying the above and finding it in the form r<theta, gives me a different answer to the given answer. Where have I gone wrong?

Thanks
It looks like you are using the correct steps. Can you post your work in detail so we can check it?
 
  • #9
166
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It looks like you are using the correct steps. Can you post your work in detail so we can check it?
I just made a small algebraic mistake!

Thanks anyway.. :)
 
  • #10
berkeman
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Sweet! :smile:
 

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