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!

What's wrong with this calculation simplified transformation circuit?

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Use source transformation on the voltage source and series-connected impedance for the circuit shown here to find the equivalent current source and parallel-connected impedance. Continue the simplification by combining the two parallel current sources into an equivalent current source, and by combining the three parallel impedances into a single equivalent impedance.

    2. Relevant equations

    1/((1/Zr)+(1/Zl))
    I=V/R
    3. The attempt at a solution

    First we simplify the voltage source into a current source
    I=V/R
    30<-90/15
    =2<-90 A
    to find the new current source we have I1+I2
    =2<-90 A+4<90
    =6<0A
    Now to combine the resistor with the inductor
    1/((1/Zr)+(1/Zl))
    Therefore the new Z is
    3+j9
     

    Attached Files:

  2. jcsd
  3. May 18, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    That is not the correct addition. Express the two current sources as complex numbers and add per the rules of complex addition.
    What happened to the 15 ohm resistor?
     
  4. May 18, 2013 #3
    Because we are simplifying we combine the resistor 15
    Using


    15 + (ZR*ZL)/(ZL+ZR)
     
  5. May 18, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    " ... and by combining the three parallel impedances into a single equivalent impedance." That's what the problem said. So it gave you a hint right there.

    The 15 ohm resistor is now in parallel with the 30 ohm and the inductor, so why are you adding its impedance to the parallel impedance of the other two components?
     
  6. May 19, 2013 #5
    so is this correct calculation

    30V<-90 = I * 15 ohms
    I = 2A<-90

    The 15 ohm || 30 ohm = 10 ohms,

    the 2A<-90 || 4A<90 = -j2 +j4 = j2 = 2<90

    the 10 ohm || +j10 ohm = ((10) * (+j10)) / (10 +j10) = (+j100) / (10 +j10) = 5 +j5
     
    Last edited: May 19, 2013
  7. May 19, 2013 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Ah, much better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: What's wrong with this calculation simplified transformation circuit?
Loading...