Mesh/Nodal analysis with superposition, dependent source and diff frequencies

careless25

Hi,

I am preparing for my circuits final and I am having trouble with this question. Can someone guide me through it?

I am trying to solve this using phasors.

I know I will have to use superposition since the circuit contains sources with different frequencies.

So first I start of with converting the Voltage source to a cosine function and then into phasor notation. So I get V = 5∠-90 where ω = 10000 rad/s.

Then 1mH inductor = 10j Ω, 20μF capacitor = -5j Ω and I open circuit the current source.

If I do nodal analysis now, I get 2 equations but dont know how to solve them.

eq1: V1 - V2 = Vx
eq2 at node V2: V2/-5j + (V2-V1)/5 + (V2 - (2Vx+V1))/10j

and of course V1 = 5∠-90.

Now I tried to solve for Vx and I got 0....Which doesnt make sense since that would create a short circuit. Help?!!?

Thanks

C25

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The Electrician

Your equation 2 is not an equation; where's the equals sign?

But your main problem with equation 2 is that you haven't dealt with the current supplied by the current source. It injects a current into node 2; where is it in equation 2?

careless25

Sorry that equation should be:

V2/-5j + (V2-V1)/5 + (V2 - (2Vx+V1))/10j = 0

As I mentioned in my post, I am doing this using superposition so I open circuited the current source. I am trying to find Vx due to only the Voltage source first.

Eventually I can say Vx(due to voltage soure) + Vx(due to current source) = Vx. (aka superposition)

The Electrician

You only need to solve for V2 since V1 is known.

Introduce symbolic variables: let Vac represent the independent voltage source and Iac represent the current source.

Make the substitutions V1=Vac and Vx=(Vac-V2) in your equation 2 and you should be able to get this solution:



But if you also add the term for the contribution of the current source Iac to your equation you can get a solution for V2 that amounts to a superposition solution. You can set Vac in this expression to zero and see the contribution of Iac alone. Similarly, you can set Iac to zero and see the contribution of Vac alone.

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careless25

I see what you meant by your fisrt comment now.

The textbook I am learning from teaches that we "open circuit" the current sources and ignore them in any of the KCL/KVL equations, solve for the unknows and then do the same with voltage sources.

Now that I have seen your solution it makes a lot more sense to do it the way you do (which is technically the same way.). makes life easier in the end since its just one equation to deal with both the sources.

Thanks for all the help!!

C25

P.S. I finished writing my final today morning, would have been nice to know this. But oh well, I was too late in posting it here.

Ratch

careless25,

Sorry I am late is answering.

Let's call the right current side the "A" side and the left voltage side the "B" side. Then ωA = 5000, ωB = 10000, XLA = 5j, XLB = 10j, XCA = -10j, XCB = -5j, I= 0.5, V = 5j . Notice that the voltage source is 90 degrees ahead of the current source because it is a sin instead of a cos.

Setting up the equations for the "A" side gives vxA = -V2A, and -I+V2A/XCA+V2A/R+(V2A-(2vxA))/XLA=0 .

Solving gives vxA = -0.344828-0.862069 and V2A = 0.344828+0.862069

Setting up the equatons for the "B" side gives VxB = V-V2B, and V2B/XCB+(V2B-V)/R+(V2b-(V+2vxB))/XLB=0

Solving gives VxB = -4-2j and V2B = 4+7j

So V2 = V2A+V2B = 4.34483+7.86207j

The power from just the current side "A" is (|vxA|^2)/R = 0.172414

Hope I didn't make any mistakes. If you have any questions, just ask.

Ratch