Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: AC Circuit Superposition

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    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).

    2. Relevant 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?

    3. 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!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 20, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    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.
     
  4. Nov 20, 2011 #3
    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?
     

    Attached Files:

    • EQ1.jpg
      EQ1.jpg
      File size:
      13.5 KB
      Views:
      157
    Last edited: Nov 20, 2011
  5. Nov 20, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    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.
     
  6. Nov 20, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    I'm not clear on what you're asking here.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook