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CIRCUIT ANALYSIS: 2 resistors, Indep. Current Source, V.C.C.S - find v0

  • Engineering
  • Thread starter VinnyCee
  • Start date
489
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1. Homework Statement

Find [itex]v_0[/itex] in the circuit below and the power dissipated by the controlled source.

http://img291.imageshack.us/img291/5479/chapter2problem22wy5.jpg [Broken]


2. Homework Equations

KCL

[tex]v\,=\,i\,R[/tex]


3. The Attempt at a Solution

KCL: [tex]i\,=\,2\,v_0\,+\,10\,A[/tex]

[tex]v_o\,=\,i\,R[/tex]

[tex]v_o\,=\,\left(2\,v_0\,+\,10\right)\,(4\Omega)[/tex]

[tex]v_0\,=\,8\,v_0\,+\,40[/tex]

[tex]v_0\,=\,-\frac{40}{7}\,V[/tex]

Am I on the right track here?
 
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Answers and Replies

berkeman
Mentor
56,086
6,136
Looks good so far. BTW, I don't like the wording of the other part of the question:

"and the power dissipated by the controlled source."

The power *dissipated* depends on the efficiency of the power supply. So it will be the power delivered to the circuit, divided by the efficiency of the power supply, which is not given. So if that really is the wording of the question, you might calculate the power delivered to the rest of the circuit, and point out that you need the efficiency of the power supply to know what its power "dissipation" is.
 
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If [itex]v_0[/itex] looks correct then I can calculate the rest.

[tex]i\,=\,\left[2\,\left(-\frac{40}{7}\right)\,+\,10\right]\,A\,=\,-\frac{10}{7}\,A[/tex]

[tex]p\,=\,v\,i\,=\,\left(-\frac{40}{7}\,V\right)\,\left(-\frac{10}{7}\,A\right)[/tex]

[tex]p\,=\,\frac{400}{49}\,W\,\approx\,8.16\,W[/tex]

Does this look right?
 
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berkeman
Mentor
56,086
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But only part of the total power delivered to the resistive load(s) comes from the controlled current source, right? I missed that my first time looking at the circuit as well.
 
489
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I guess, the resistors are the other part of the resistive load?

How do I figure for that then?

[tex]P\,=\,v\,i\,=\,\left(v_0\right)\,\left(2\,v_0\right)\,W[/tex]

[tex]P\,=\,2\,v_0^2\,=\,2\,\left(-\frac{40}{7}\right)^2\,W[/tex]

[tex]P\,=\,2\,\left(\frac{1600}{49}\right)\,W\,=\,\frac{3200}{49}\,W[/tex]

[tex]P\,\approx\,32.7\,W[/tex]

Correct?
 
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berkeman
Mentor
56,086
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I guess, the resistors are the other part of the resistive load?

How do I figure for that then?
I don't understand your reply. Calculate the total load resistor voltage and then the current contribution from the controlled source.
 
489
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[tex]2\,v_0[/tex] is the current contribution of the controlled source.

[tex]2\,v_0\,=\,-\frac{80}{7}\,A\,\approx\,-11.43\,A[/tex]

Do I then use the p = v i ?

[tex]P\,=\,\left(-\frac{40}{7}\right)\,\left(-11.43\,A\right)\,=\,65.3\,W[/tex]
 
berkeman
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Nice job, Vinny. Keep it up and you will go far.
 
489
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Thank you!

Just to clarify, [tex]v_0\,=\,-5.71\,V[/tex] and [tex]P_{V.C.C.S}\,=\,65.3\,W[/tex]. These are correct?
 
161
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I believe the question wants you to determine the P = VI for the dependent source. Since you have the I as 2v0, all you need next is to calculate the V across the dependent source and then multiplying the two terms for the power.
 
489
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Would that V be half of the [itex]v_0[/itex]?

P is 32.7W and not 65.3W?
 
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161
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No Vinny, that V would be the voltage across the VCCS. Equivalently, that is also the voltage across the two resistors, since the (4+6)ohm, 10A source, and VCCS are all in parallel.
 
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The voltage across the VCCS is what I need for the P = v i equation right? It is the v and the i is the current contribution of the VCCS? That would mean [tex]p\,=\,v\,i\,=\,\left(-\frac{20}{7}\right)\,\left[2\,\left(-\frac{40}{7}\right)\right]\,=\,32.7\,W[/tex]

If that is not it, how is it calculated?
 
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As I have earlier said, it will be the V across the (6+4)ohm. Since you have already found the current across the resistors, this V shouldn't be too difficult to find I am sure.
 
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The current across the resistors is [tex]i\,=\,2\,v_0\,+\,10\,=\,-1.43\,A[/tex], right?

Now to get the V across both resistors.

[tex]V\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V[/tex]

I then use the P = vi.

[tex]P\,=\,v\,i\,=\,(-14.3\,V)\,(-1.429\,A)\,=\,20.4\,W[/tex]

I hope thats right:)
 
157
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The current across the resistors is [tex]i\,=\,2\,v_0\,+\,10\,=\,-1.43\,A[/tex], right?

Now to get the V across both resistors.

[tex]V\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V[/tex]

I then use the P = vi.

[tex]P\,=\,v\,i\,=\,(-14.3\,V)\,(-1.429\,A)\,=\,20.4\,W[/tex]

I hope thats right:)
yea the current right. I was looking at your other post and this one, and it seems you have the same question a current through a series resistor. Remember that the current is the same through each resistor when they are in series. THe method for getting P is right to me. Do you know why finding V through the resistors allowed you to get P for the controlled source?
 
Do you know why finding V through the resistors allowed you to get P for the controlled source?
no! why? (sorry) I thought the power dissipated by the resistors were a result of BOTH sources, not just one.

edit: Oh I meant using the same current through the resistors to find P of D.S
 
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157
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no! why? (sorry) I thought the power dissipated by the resistors were a result of BOTH sources, not just one.

edit: Oh I meant using the same current through the resistors to find P of D.S
Don't really understand your post...

The question is asking for "power dissipated by the controlled source"
NOT the resistors.

But looking at the work again, he found the P for the series resistors and not the controlled source. Have to use the 2(V0)*V = P(controlled souce)
 
Yes, What you said just now was what I was asking.
 
157
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Oh alright. So yea, he's just not finished with the problem yet. Hope he checks back. :uhh:
 
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Yes Vinny, like Number2Pencil and teknodude said, please check that when applying P = VI, it is the V and I of the VCCS: V = -14.3V is correct, but I = -1.429A is not.
 
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SO the V is for the VCCS only and that is right. So I need to recalculate the i for ONLY the vccs.

[tex]i_{VCCS}\,=\,2\,v_0\,=\,2\,\left(-\frac{40}{7}\right)\,=\,-\frac{80}{7}\,A\,\approx\,-11.43\,A[/tex]

[tex]v_{VCCS}\,=\,-14.3\,V[/tex]

[tex]p\,=\,v\,i\,=\,(-14.3\,V)\,(-11.43\,A)\,=\,163.4\,W[/tex]

NOTE: But doesn't that change the calculation of [tex]v\,=\,i\,R[/tex]?

[tex]v\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V[/tex]

VS.

[tex]v\,=\,i\,R\,=\,(-11.43\,A)\,(10\Omega)\,=\,-114.3\,V[/tex]

I'm thoroughly confused!
 
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161
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To write V = IR for the pair of resistors, you have to take into account both the current from the VCCS and the 10A, or I = -11.43+10 = -1.43A, since both the current sources contribute to the current across the resistors. Plugging that into V = IR will give you the same V = -14.3V. There is no confusion, is there?
 
489
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Like this?

[tex]P_{VCCS}\,=\,v\,i\,=\,i^2\,R\,=\,(-1.43\,A)^2\,(4\Omega)\,\approx\,8.18\,W[/tex]

or

[tex]P_{VCCS}\,=\,v\,i\,=\,\left(-\frac{40}{7}\,V\right)\,(-1.43\,A)\,\approx\,8.17\,W[/tex]

So...

[tex]v_0\,=\,-\frac{40}{7}\,A\,\approx\,-5.71\,A[/tex]

and

[tex]P_{VCCS}\,\approx\,8.17\,W[/tex]

Right?
 
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I'm confused Vinny. How did the above equations arise?

What you did earlier in Post 22 is correct, at least as far as P (VCCS) = 163.4W is correct.
 

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