AC nodal analysis help needed

[javier]

Hello fellow engineering students and engineers,

This week I had a fast 2 class introduction to AC and today my teacher sent me 12 exercises due for next Friday, good to mention that I’m so freaked out because I am not familiar with AC yet so I don’t know to work and accomplish the problems, I see angular functions and other variables for the voltage sources and I don’t know how to deal with them, I would be really appreciate any clues, recommendations, heads up you guys can give me, all I need is to figure out how to solve one and the I will go by myself. Thanks in advance.

[spitfire]

Hi there javier! Let's see if I can help you understand the problem!

I am going to assume you know nodal analysis method already so I'll jump into how to work it for AC.

So first thing is to make sure you have impedances for inductors and capacitors in the network, which has been provided to you it looks like.

What you'll need to do is do your work in polar form. It will be good to practice converting to and from polar/complex forms.

First, 10cos(t - 45) = 10sin(t + 45) , sine leads cosine by 90deg

Your sources become this: 10 V < 45 and 5 V < 30

Then set up your 1 nodal equation for Va as you would DC:

(Va-{10<45}/{3<0}) + (Va/{1<-90}) + (Va-{5<30}/{1<90}) = 0

(Va/{3<0}) - ({10<45}/{3<0}) + (Va/{1<-90}) + (Va/{1<90}) - ({5<30}/{1<90}) = 0

When dividing in polar, the new angle becomes the difference between the top and bottom angles, hence 30 - 90 = -60 for the last term.

(Va/3) + ({Va<90}) + ({Va<-90}) = ({3.33<45}) + ({5<-60})

Va<90 and Va<-90 is equivalent in rectangular form as [j + (-j)] and thus cancel each other out.

Va/3 = {3.33<45} + {5<-60}

convert to rectangular form.

Va/3 = 2.35 + j2.35 + 2.5 - j4.33

Va/3 = [4.85 - j1.98]

Va = 3*[4.85 - j1.98]

Va = 14.55 - j5.94

Va magnitude = [(14.55^2)+(5.94^2)]^(1/2)

Va angle = tan^-1(-5.94/14.55)

So Va = 15.72 < -22.21

The thing to keep in mind is when it is best to convert to polar form, or complex, and vice versa. Mind your j's and you should move through your AC class without too much issue.

Hopefully you were able to follow my post, if there is any questions or concerns, let me know and I'll try to clear them up.

[Robinet]

I disagree with spitfire, I like doing the calculations in rectangular form and convert the answer to Polar, but that's just a matter of taste I assume. This also depends on what you are trying to find out and so on. This is exactly like DC but just with angles in the calculations, keep that in mind. It starts to differ from DC when calculating phase compensation and 3 phase systems, but other than that its pretty much alike.

[Farit]

Ok this is the result that I obtain.

AC and DC Nodes are resolved in the same way, same rules are applied, the only difference it that you will work with polar and rectangular numbers and also that you will work with the admittance of the component.

Step 1: I found all the admittances of the components.

Step 2: I identified the nodes. You must know, that each time that you find two components connected one to each other in the junction formed between them, you will find a node. So beside Node A, I found other 2 nodes called as VB and VC. As part of step 2, I changed the power supply VC to a "cos function" by substracting 90° to the angle of the equation.

Step 3: I created the equation for Node VA and by replacing VA and VC values into the equation, finally I got the result. 

I used the virtual TI Nspire CX CAS software to resolve the equation.

[javier]

Thank you very much guys, you made me notice about a couple mistakes I was doing, hey Farid where did you get that calculator? I was looking for one earlier to solve a Matrix but I couldn't find one, so I had to go with the determinant method lol. Thanks again guys, you rock!

[Farit]

Javier,

I bought TI CX CAS some time ago for AC Analysis course and the simulator is part of the software that came as an accessory with the calculator, but you can download a trial version from Texas Instrument web page, it gives you 30 days of free usage.

Also you can do a search of the HP50G, that is a simulator of real HP50G and you can install it in your computer and if you have a smart phone or tablet you can download the App also.

[Robinet]

But as a disclaimer the app for android costs a couple of €. The 48g is free tough, I have the 48g on my phone and own the 50g.