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Luke1294
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Homework Statement
Use the mesh-current method to find the power dissipated in the 20 ohm resistor
Homework Equations
[tex]p=vi^2[/tex]
The Attempt at a Solution
Looking at the picture above, you can see how I set my current flows, in hopes of making the most currents flow together. If the picture isn't entirely clear, voltage supplied at the far left of the circuit is dependent upon the current flowing from left to right on the 3 ohm resistor. I'm assuming I have some major sign-error issues and that's what's keeping me from the correct answer, provided to be 259.2 W.
My equations-
Current 1-
[tex] 135 + 2I_1 +20(I_1+I_2) + 3(I_1+I_2) = 0[/tex]
[tex] 25I_1 + 20I_2 + 3I_3 = -135 [/tex]
Current 2-
[tex] 10i_0 + I_2 + 20(I_2+I_1) - 4(I_2-I_3) = 0[/tex]
[tex] I_1+I_3 = i_0 [/tex]
[tex] 10(I_1 + I_2) + 20(I_2+I_1) - 4(I_2-I_3) = 0 [/tex]
[tex] 30I_1 + 17I_2 +14I_3 = 0 [/tex]
Current 3-
[tex]5I_3 + 4(I_3 - I_2) + 3(I_1+I_3) = 0 [/tex]
[tex]5I_1 - 4I_2 + 12I_3 = 0[/tex]
Okay, so I throw everything into a matrix, solve, for I1, I2, and I3. I got some very funky answers, so I know SOMETHING is wrong based on that... I got
[tex]I_1 = \frac{35100}{373}
I_2 = \frac{-42930}{373}
I_3 = \frac{-23085}{373} [/tex]
So to find the actual current going through the 20 ohm resistor, I would sum I1 and I2. I wind up with -7830/373...When I plug that value into my power equation, I get 8813.245, a far cry from the expected value.
Any pointers?[tex]
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