Simple Circuit Analysis: Finding i_{x} with Mesh Analysis

AI Thread Summary
The discussion focuses on using mesh analysis to find the current i_{x} in a given circuit with specified resistances and a voltage source. Initial calculations for currents i_{1} and i_{2} yield values of 0.15 A and 0.09 A, respectively. The mesh analysis equation leads to a calculation of i_{x} resulting in -0.0125 A, raising concerns about the negative value's validity. Another participant suggests re-evaluating the mesh equations and provides an alternative calculation yielding i_{x} as approximately 0.014371 A. The conversation emphasizes the importance of correctly applying mesh analysis to arrive at accurate results.
ttiger2k7
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Homework Statement



Use mesh analysis to find i_{x}.

http://img16.imageshack.us/img16/3384/circuitwd7.jpg

Where R_{1} = 100 ohms , R_{2} = 50, R_{3} =100, R_4 = 220, and R_{5} = 470. V_{s} = 15 V.

Homework Equations



V = iR
\Sigma V=0

The Attempt at a Solution



First of all, the current orientation of the left (i_{1}) and middle (i_{x}) loops are clockwise, and the right (i_{2}) loop current is counter clockwise.

Step 1) Finding i_{1} and i_{2}

i_{1} = 15/100 = .15 A

i_{2} = 9/100 = .09 A

Step 2) Mesh Analysis

i_{x}*50+(i_{x}+i_{2})*470+(i_{x}-i_{1})*220=0

i_{x}*50+(i_{x})*470+(.09)(470)+(i_{x})*220+(.15)(220)=0

740i_{x}+9.3=0

i_{x} = -.0125 A

*******

The only thing that is bothering me is that I am getting an negative answer. Is this the correct answer using mesh analysis? Thank you.
 
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I1 is first mesh, I2 second ... (all 3 clockwise)

sum of voltages around a loop = 0

-15 + 100I1 + 220I1 - 220I2 = 0
220I2 - 220I1 + 50I2 +470I2 - 470I3 = 0
470I3 - 470I2 + 100I3 + 9 = 0

simplfy and solve using augmented matrix

I2 = Ix = 0.014371 amps (if numbers are correct)
 
Last edited:
Thanks, but the problem is asking for a mesh analysis.
 
ttiger2k7, I re-did the problem using mesh, hope it helps.
 
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