Mesh Analysis

  • Thread starter eehelp150
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  • #1
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Homework Statement


PK88BSE.png


Use Mesh analysis to find voltage of Node A and B

Homework Equations


V=IR
KVL

The Attempt at a Solution


Are my mesh equations correct?

Mesh1:
-j30V + (-j50 + 10)I1 - 10I2 = 0

Mesh2:
(10-j20)I2 - 10I1 - j50V - I3(-j20) = 0

Mesh3:
(-j20+30+j10)I3 - (-j20)I2 = 0
 
Last edited:

Answers and Replies

  • #2
gneill
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Yes, your mesh current equations look fine.
 
  • #3
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Yes, your mesh current equations look fine.
How would I get the voltage of Node A and B?
 
  • #4
gneill
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How would I get the voltage of Node A and B?
Ohm's law. You've solved for the currents so now you can find the potentials across any components you wish. Choose a path from the reference node to the location where you wish to know the potential and sum up the PD's along the way.
 
  • #5
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Ohm's law. You've solved for the currents so now you can find the potentials across any components you wish. Choose a path from the reference node to the location where you wish to know the potential and sum up the PD's along the way.
So if I wanted the voltage at Node A, I would multiple (I1-I2) with 10ohm resistor?
 
  • #6
gneill
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So if I wanted the voltage at Node A, I would multiple (I1-I2) with 10ohm resistor?
That would work, yes. Any path from the reference node to a would work, and that happens to be a pretty convenient choice :smile:
 
  • #7
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That would work, yes. Any path from the reference node to a would work, and that happens to be a pretty convenient choice :smile:
I ended up getting:
14.329 < -71.74 degrees for Va
Is that right?


Would Vb be: (I2-I3)*(-j20) -> 4.48 + 36.39j?
 
Last edited:
  • #8
gneill
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I ended up getting:
14.329 < -71.74 degrees for Va
Is that right?


Would Vb be: (I2-I3)*(-j20) -> 4.48 + 36.39j?
They look good to me.

Note that for Vb you could also have just added j50 to Va; there is a fixed source of that amount tying them together.
 

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