Basic Cicuit Analysis - Mesh

  • Thread starter FrogPad
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  • #1
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This is a question from a review for our midterm coming up soon. I don't understand what I am doing wrong and need some guidance. It would be rather tedious for you (the reader) to view an ASCII circuit, so here is a http://static.flickr.com/46/118451767_63645eb724_b.jpg"of my work:

http://static.flickr.com/46/118451767_63645eb724_b.jpg

Sorry for the link to another page. I couldn't resize it to upload to physicsforums because I don't have software on this computer to do it. I PROMISE it is just my work.

Thanks in advance.

EDIT:I forgot to draw [itex]V_x [/itex] on the circuit diagram!!! It is supposed to be over [tex] R_{4k\Omega} [/tex] with the labeled [tex] + [/tex] and [tex] - [/tex] right next to the loop 2 label.
 
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Answers and Replies

  • #2
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By Ohm's Law, [tex]V_x = (I_2 - I_3) \times 4000[/tex], thus [tex]I_1 = I_2-I_3[/tex], and not [tex]I_3-I_2[/tex] as you have written.
 
  • #3
809
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Yeah, I get ohms law, and that the voltage for Vx is a combination of I1 and I2 multiplied by the resistance. I don't understand WHY it is:

[tex] I1 = I2-I3 [/tex]

and not: [tex] I1= I3-I2 [/tex]

oh *shoot*... nevermind I got it now.

The current is going to run from the high potential to the low potential, from positive to negative. If the current is running from positive to negative then the current that runs along side it is a POSTIVE I2 and a NEGATIVE I3.

For some reason I was thinking that the current should run from negative to positive.


cool. thanks.
 

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