# Mesh Analysis

1. Oct 11, 2016

### eehelp150

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

Use Mesh analysis to find voltage of Node A and B

2. Relevant equations
V=IR
KVL

3. 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: Oct 11, 2016
2. Oct 11, 2016

### Staff: Mentor

Yes, your mesh current equations look fine.

3. Oct 12, 2016

### eehelp150

How would I get the voltage of Node A and B?

4. Oct 12, 2016

### Staff: Mentor

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. Oct 12, 2016

### eehelp150

So if I wanted the voltage at Node A, I would multiple (I1-I2) with 10ohm resistor?

6. Oct 12, 2016

### Staff: Mentor

That would work, yes. Any path from the reference node to a would work, and that happens to be a pretty convenient choice

7. Oct 12, 2016

### eehelp150

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: Oct 13, 2016
8. Oct 13, 2016

### Staff: Mentor

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.