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AC Circuit Superposition

  • Engineering
  • Thread starter Rurik1
  • Start date
  • #1
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Homework Statement



I need help with determining the current through the 1Ω resistor.

http://img403.imageshack.us/img403/5422/acex22.jpg [Broken]

The trouble I have is that the voltage sources are phased as given in picture, and I got stuck on how to deal with these phases when calculating the current with impendance.

I know I need to count the impendance, but I get the wrong answer when I count the current from every source (superposition theorem).

Homework Equations



If I look at it through the superposition theorem and current divider rule, I should get the correct answer with regards to current through the 1 Ω resistor.

I am absolutely certain that I don't do the voltage right, i.e. 5 Δ 30 and 5 Δ -30 . Is this correct thinking?

The Attempt at a Solution



Back to superposition theorem, I can count the total impendance for every voltage source. Then I divide the voltage, example

(5 Δ 30) / Z Δ ω = I

From there I go to the current divider rule and count the current through the resistor per source. This is repeated with the upper source as:

(5 Δ -30) / Z Δ ω = I

------------------


I've handled these problems well without the voltage phases, but I don't think I handle the voltage phases rather well here. I would really appreciate any pointers on how to proceed!
 
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Answers and Replies

  • #2
gneill
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Hi Rurik1, welcome to PF.

You can put all the phasor quantities in complex form. The complex math takes care of the phases.

As a suggestion, why don't you convert the voltage sources and their networks up to the 1 Ohm resistor into their Norton equivalents? Those can then be combined into a single Norton source in parallel with that resistor.
 
  • #3
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gneill

Thanks for your quick reply! Norton is another solution, however I want to see where I am blinding myself here.

OK, superposition, starting from below:

Z1(3Ohms) = 3Δ0o

Z2(5Ohms) = 5Δ0o

Z3(1Ohms) = 1Δ0o

Z4(L) = 2∏*50*0.05Δ90o

From the source below, the current through 1 Ohm should be as per this picture:

And so on... It amazes me that I can't see it... ?!

EDIT: this means, that the next count from the upper source will be 5Δ-30o, so the impendance angle will be subtracted - is this correct?
 

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  • #4
gneill
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Your picture looks like an exercise in continued fractions :smile: My approach would be to work symbolically until the last stage, defining useful intermediate values as required. So, taking your definitions for Z1 through Z4, I'd define Z5 = Z3||Z4 for example. Then to find the current through Z3 due to V2 (the bottom source), after suppressing V1 I'd note the voltage divider consisting of Z2 and Z5. Thus:
[tex] I_2 = V_2 \left(\frac{Z_5}{Z_2 + Z_5}\right) \frac{1}{Z_3} [/tex]
After that the numbers can be plugged in and the result obtained.
 
  • #5
gneill
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EDIT: this means, that the next count from the upper source will be 5Δ-30o, so the impendance angle will be subtracted - is this correct?
I'm not clear on what you're asking here.
 

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