Recent content by slh3410

I have no idea how to solve what I think is the proper voltage division setup: (120∠0) [(10∠90) / (23.7∠35.75)]

So for multiplication of phasors I use a cross product? What do I do for division?

I thought I needed the current to eventually use V=IZ, but I guess not. I followed you up until you got to current. Why is Vs only over L1+Z1? Because you are calculating voltage drops over that outermost loop? Where did Vn come from as well?

Homework Statement Where Vs = 120 ∠0° V Homework Equations All values are represented in ohms, so they are treated as resistors. The Attempt at a Solution Series inductor + resistor 30+j10 ohms = 31.62∠18.435° Parallel resistor and inductor+resistor (1/50)= 0.02∠0°...

Thanks very much gneill

So assuming all those values in the picture are good, it's correct to say the power delivered by the independent source is 80V*21.6A=1728W? I really don't get why my teacher would go out of their way to essentially say "watch your sign!" when it's all positive and straightforward. Makes me...

Thanks a lot for the reply. I guess my question is, how do I know which way the current is actually flowing? If I reverse all the values then the circuit still seems like it works. Do I just start the current flow out of the positive independent voltage? Do you agree that "The power...

I believe I have solved this circuit correctly. Each loop and node satisfies KVL and KCL, respectively. All I was given was Vg=528V. The question is as follows: "Find the power delivered by the dependent voltage source." Currently my answer is 80V*21.6A=1728W, but I want to be sure because...

Alright, I'll double check the arithmetic because that is certainly not your job :) Thank you to everyone.

Am I on the right track now? I took the 'just use nodal-voltage analysis' advice. For Va) [(Va-Vb)/8]+[Va/40]-[5.4]=0 For Vb) [(Vb-Va)/8]+[Vb/80]+[Vb/120]+[1.5]=0 Solving that, Va=96V and Vb=72V Solving for V1 and V2 in the original problem, I again get V1=96V and V2=72V. This...

I'm given that i1=5.4A and i2=1.5A. I know that I need to apply mesh-current and then nodal-voltage analysis, but I have never dealt with two current sources. It seems like however I set it up, I am missing a piece. My first attempt at a KVL equation using mesh analysis incorrectly...