Phasor Mesh Current-what am I doing wrong?

  • Thread starter Thread starter teh_dude
  • Start date Start date
  • Tags Tags
    Mesh Phasor
AI Thread Summary
The discussion revolves around a homework problem involving phasor mesh current analysis. The user attempts to calculate mesh currents I1 and I2 using KVL and complex numbers but receives incorrect feedback from the homework site. Key points of confusion include the incorporation of a 600V source at -30 degrees and the correct handling of mutual inductance terms. Participants suggest ensuring proper transformation of the voltage source into the correct format and verifying the signs of the mutual inductance terms based on loop current directions. The user expresses frustration over the persistent errors despite following class demonstrations.
teh_dude
Messages
11
Reaction score
0
Phasor Mesh Current--what am I doing wrong??

Homework Statement


Circuit diagram:
meshphasor.jpg

Calculate the phasor mesh currents, I1 and I2.

Homework Equations


KVL/Mesh Current analysis with phasors/complex numbers.

The Attempt at a Solution



Mesh 1: (-550@0deg) + j75(I1) + 30(I1 - I2) - j40(I1) - j24(I2) = 0

Mesh 2: (600@-30deg) + 30(I2 - I1) + j50(I2) + 80(I2) - j24(I1) = 0

(30 + j35)I1 - (30 + j24)I2 = 550
-(30 + j24)I1 + (110 + j50)I2 = (-600@-30deg)

I got:

I1 = (9.45503221553 @ -38.8333374187 degrees) A

I2 = (2.79176352576 @ 93.1719060416 degrees) A

This problem was even demonstrated in class. Yet the homework site keeps telling me I'm wrong. I don't see where I've made a mistake...?
 
Physics news on Phys.org


How did you incorporate the 600V at -30 degrees in your equations? You didn't show us that, so we can't find your mistake for you, if there is one.
 


Maybe check that the signs of the mutual inductance terms respect the dot notation and choice of loop current directions?
 


rude man said:
How did you incorporate the 600V at -30 degrees in your equations? You didn't show us that, so we can't find your mistake for you, if there is one.

Mesh 2: (600@-30deg) + 30(I2 - I1) + j50(I2) + 80(I2) - j24(I1) = 0

How else can I demonstrate? It's there.

If you mean how did I mix polar notation with rectangular notation, I punched it all into the TI-89, in which the modes were set to degree, polar, and approximate.

Otherwise, I don't know what you're getting at.
 
Last edited:


gneill said:
Maybe check that the signs of the mutual inductance terms respect the dot notation and choice of loop current directions?

I don't entirely understand how to deal with the polarity marks, but I've tried switching plus and minus for the j24Ω term. It's my understanding that they should both be -j24Ω, multiplied by I1 or I2 respectively, but I've tried both minus, both plus, one minus, one plus, the other way around, and none of the answers are correct.
 


teh_dude said:
I don't entirely understand how to deal with the polarity marks, but I've tried switching plus and minus for the j24Ω term. It's my understanding that they should both be -j24Ω, multiplied by I1 or I2 respectively, but I've tried both minus, both plus, one minus, one plus, the other way around, and none of the answers are correct.

I've taken a closer look at your circuit and I believe that you have them correctly oriented. Sorry for the bother.

The idea is that a positively increasing current flowing into one coil at its dot produces a positive polarity of induced voltage at the dot of the second coil. So you have to be attentive as to whether a positive increase in the mesh current will create a positive increase in the current flowing into the coil dot, or whether it will in fact cause a decrease.

attachment.php?attachmentid=41401&stc=1&d=1322725464.jpg
 

Attachments

  • Fig1.jpg
    Fig1.jpg
    7.5 KB · Views: 611


teh_dude said:
Mesh 2: (600@-30deg) + 30(I2 - I1) + j50(I2) + 80(I2) - j24(I1) = 0

How else can I demonstrate? It's there.

If you mean how did I mix polar notation with rectangular notation, I punched it all into the TI-89, in which the modes were set to degree, polar, and approximate.

Otherwise, I don't know what you're getting at.

That is not a viable equation. You must transform the "600V@-30deg" the same way you transform components. Example: you didn't write "wL@+90deg" for the inductive impedance, did you? You wrorte jwL. jwL is the transform of wL@+90deg. Do the same for the 600V@-30 deg. I don't know what your TI-59 is doing & I can't guess.

I checked & your mutual inductance polarities are correct and in fact your equations are correct, except the second equation is not an equation. :frown:
 
Last edited:

Similar threads

Back
Top