Engineering Nodal analysis of an AC circuit

[Morgz129]

Homework Statement

Hi I have this question as part of my assignment but cannot get my head around part b, I have worked out part a with an answer of -9.2+17.3i amps which from a previous thread I know is correct, but can't understand why the equations given in the previous thread are different to what I have.

Homework Equations

Kirchhoffs current law

The Attempt at a Solution

V20-v30=v3

((V20-V1)/Z1)+((v20-v30)/z2)+(v20/z4)=0
((v20-v1)/z1)+(v3/z2)+(v20/z4)=0

((v30-v20)/z2)+((v30-v2)/z2)+(v30/z5)=0

 

[osilmag]

For the bottom equation, I think that middle term should be $\frac {V2 - V30}{Z3}$. And that would be equal to the other two terms.

 

[Babadag]

https://www.physicsforums.com/threads/mesh-analysis-ac-solving-simultaneous-equations.958998/

 

[Morgz129]

For the bottom equation, I think that middle term should be $\frac {V2 - V30}{Z3}$. And that would be equal to the other two terms.

My mistake it should be divided by z3, but if I may ask why would it be ((v2-v30)/z3) instead of ((v30-v2)/z3)?

 

[osilmag]

It depends on how you define your directions. If current is flowing into the node from z3 and out of it at z5 and z2 then $v2-v30$ is the difference and $I,z3 = I,z5 + I,z2$