Sinusodial Steady State/Phasor Problem

Discussion in 'Homework Help' started by Kel915, Apr 11, 2014.

1. Kel915 Thread Starter New Member

Apr 11, 2014
2
0
The problem is in the image attached. I tried doing nodal analysis at A and B, and then solving, getting

j*ω*C*Va - I1 + (Va-Vb)/(R+j*ω*L1) - I3 = 0
Vb/(j*ω*L2) + (Vb-Va)/(j*ω*L1+R) -I2 + I3 = 0

plugging in

(Va-Vb)/(.5+j*.5) + 2x*j = 2
Vb/j + (Vb-Va)/(.5j+.5) = 0

0.20 cos(t-81.9)
0.40 cos(t-63.4)

, but the answer is only half correct (10/20 pts). I'm not sure if it's the amplitude or angle that's wrong.

I used this (x= Va, y= Vb, i = j) in the link below

http://www.wolframalpha.com/input/?i=(x-y)/(.5+i*.5)+++2x*i+=+2,+y/i+++(y-x)/(.5i+.5)+=+0

Then for the second part I tried ignored I1 and I2 and did the same thing with nodal analysis

x*(4j) + (Va-Vb)/(.5+j) = 1
(Vb-Va)/(.5+j) + y/(2j) = - 1

but I end up with an answer that's completely wrong.

(x= Va, y= Vb, i = j) in the link below

http://www.wolframalpha.com/input/?i=x*(4i)+++(x-y)/(.5+i)+=+1,+(y-x)/(.5+i)+++y/(2i)+=+-+1

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2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I'm not sure this will provide a useful insight or not.

In the case where I1, I2 and I3 are identical one can ask the question: "What current is actually injected into the passive elements linked to node B from sources I2 & I3?"

A little reflection might lead one to conclude that the answer is none and therefore the two sources are injecting the same value into node A in parallel with the current from source I1. The three sources could then be reduced to a single source injecting current into node A. I believe this would make the analysis rather more easy.

3. WBahn Moderator

Mar 31, 2012
18,085
4,917
You might consider using mesh analysis. You have four meshes, but three of them are trivial.

I guess we have to assume that t is in seconds since the idiot that wrote the problem doesn't comprehend units.

4. WBahn Moderator

Mar 31, 2012
18,085
4,917
I agree with your observation. I don't know how much easier it makes the analysis in the end (for this particular problem). But looking for and exploiting these kinds of simplifications can frequently have major time and effort saving impacts.

5. Kel915 Thread Starter New Member

Apr 11, 2014
2
0
Thanks, I figured out all I was doing wrong was not converting the complex number to polar form to get the amplitudes.