Find Node A and B Voltage with Mesh Analysis

[eehelp150]

Homework Statement

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

 

[gneill]

Yes, your mesh current equations look fine.

 

[eehelp150]

Yes, your mesh current equations look fine.

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

 

[gneill]

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.

 

[eehelp150]

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?

 

[gneill]

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

 

[eehelp150]

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

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

 

[gneill]

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.