1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
      99
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: AC Circuit Superposition
  1. Superposition circuits (Replies: 1)

  2. Ac circuit (Replies: 7)

  3. AC circuit (Replies: 3)

  4. Ac circuits (Replies: 18)

Loading...