1. Feb 2, 2009

### ttiger2k7

1. The problem statement, all variables and given/known data

Use mesh analysis to find $$i_{x}$$.

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

2. Relevant equations

$$V = iR$$
$$\Sigma V=0$$

3. 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.

2. Feb 2, 2009

### LeeroyJenkins

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: Feb 3, 2009
3. Feb 2, 2009

### ttiger2k7

Thanks, but the problem is asking for a mesh analysis.

4. Feb 3, 2009

### LeeroyJenkins

ttiger2k7, I re-did the problem using mesh, hope it helps.