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

  • Engineering
  • Thread starter VinnyCee
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  • #26
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Start Again!!!

Anyways, you are saying that the [itex]P_{VCCS}\,=\,163.4\,W[/itex] is correct?

Let's just start from the beginning!

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

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

Right?

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

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

Right?

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

So...

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

Right?

I need to find the power dissipated by the controlled source. [itex]P_{VCCS}\,=\,?[/itex]

So, I need to use the EQs [itex]P\,=\,v\,i[/itex] or [itex]P\,=\,\frac{v^2}{R}[/itex], right?

The i should only be for the contribution by the VCCS. So, [itex]i_{VCCS}\,=\,2\,V_0[/itex], right?

Now what do I do, assuming the above is all right?



[tex]P\,=\,I\,V\,=\,I_{VCCS}\,V_{VCCS}[/tex]

[tex]I_{VCCS}\,=\,2\,V_0\,=\,2\,\left(-\frac{40}{7}\right)\,=\,-\frac{80}{7}\,\approx\,-11.43\,A[/tex]

Does this seem right?
 
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  • #27
P=IV

you know what the current I is, so multiply it by the voltage ACROSS THE SOURCE. if you need a hint on how to get this, just know that if two or more branches are connected to two of the same nodes (the top node and the bottom node in this case), they have the same voltage drop across them. Look at this diagram if you don't understand what I'm saying

parallel-1.jpg
 
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  • #28
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voltage ACROSS THE SOURCE
That would be [tex]V_{VCCS}[/tex], right?

Is it just divided equally between the two sources so the voltage across the source is just [tex]\frac{1}{2}\,V_0\,=\,-2.86\,V[/tex]
 
  • #29
yes on the first question, no on the second.

look at that diagram again, all the voltage drops/rises are the same...you can apply this to your circuit too...the voltage across the two sources are the same...and infact, the voltage across the two resistors (combined) is the same as these two as well. this fact could prove useful.....especially about the part where:

Voltage across the resistors = Voltage across the V.C.C.S
 
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  • #30
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Are you are saying that

[tex]V_0\,=\,V_{VCCS}\,=\,V_{4\,\Omega}\,=\,V_{6\,\Omega}\,=\,V_{10\,A}\,=\,-\frac{40}{7}\,V[/tex]
 
  • #31
almost

[tex] V_4ohm + V_6ohm = V_C.S(10amp) = V_V.C.C.S [/tex]

Ok so I'm not good with the fancy looking formulas. lol. if you can solve for one of these...you know the other two as well
 
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  • #32
161
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Okay, let's take it from the start. :)

Anyways, you are saying that the [itex]P_{VCCS}\,=\,163.4\,W[/itex] is correct?

Let's just start from the beginning!

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

Right?

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

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

Right?

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

So...

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

Right?
Yes.

I need to find the power dissipated by the controlled source. [itex]P_{VCCS}\,=\,?[/itex]

So, I need to use the EQs [itex]P\,=\,v\,i[/itex] or [itex]P\,=\,\frac{v^2}{R}[/itex], right?
Yes, you use P = VI. But no, you can't use P = V^2/R since there is no R for the VCCS. To find P (VCCS), obtain the V and I across the VCCS, and then combine these terms as a product. Please note that you can never use P = V^2/R or P = RI^2 to calculate the power supplied/dissipated by a source, never ever.

The i should only be for the contribution by the VCCS. So, [itex]i_{VCCS}\,=\,2\,V_0[/itex], right?
Okay... so we now know the I across the VCCS.

Now what do I do, assuming the above is all right?

[tex]P\,=\,I\,V\,=\,I_{VCCS}\,V_{VCCS}[/tex]
We know the I (VCCS) but what about the V (VCCS)?

[tex]I_{VCCS}\,=\,2\,V_0\,=\,2\,\left(-\frac{40}{7}\right)\,=\,-\frac{80}{7}\,\approx\,-11.43\,A[/tex]

Does this seem right?
Yes, I (VCCS) = -11.43A is correct. But again, what about the V (VCCS)?
 
  • #33
489
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lol - so we have this

[tex]V_{4\Omega}\,=\,V_0\,=\,-\frac{40}{7}[/tex]

and

[tex]\left(-\frac{40}{7}\right)\,+\,V_{6\Omega}\,=\,V_{V.C.C.S.}\,=\,V_{10\,A}[/tex]

[tex]V_{V.C.C.S.}\,=\,V_{6\Omega}\,-\,\frac{40}{7}[/tex]

How do I get [tex]V_{6\Omega}[/tex] though?
 
  • #34
you get the voltage across the 6 ohm resistor with non other than ohm's law:

[tex] V_6 = I_6 R_6 [/tex]
 
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  • #35
489
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Cool!

[tex]i\,=\,2\,V_0\,+\,10\,=\,2\,\left(-\frac{40}{7}\right)\,+\,10\,=\,-\frac{10}{7}\,\approx\,-1.43\,A[/tex]

[tex]V_{6\Omega}\,=\,(-1.43\,A)\,(6\Omega)\,=\,-8.57\,V[/tex]

[tex]P_{V.C.C.S.}\,=\,I_{V.C.C.S.}\,V_{V.C.C.S.}\,=\,(-11.43\,A)\,(-8.57\,V)\,=\,97.96\,W[/tex]

What you did earlier in Post 22 is correct, at least as far as P (VCCS) = 163.4W is correct.
But I thought that [tex]P_{V.C.C.S}\,=\,163.4\,W[/tex] ???
 
  • #36
the problem is you set V6 = V of the V.C.C.S. remember it's both resistors combined

I updated that diagram to hopefully make it more obvious
 
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  • #37
489
0
[tex]V_{V.C.C.S.}\,=\,V_{6\Omega}\,+\,V_{4\Omega}\,=\,(-8.57\,V)\,+\,(-5.71\,V)\,=\,-14.28\,V[/tex]

[tex]P_{V.C.C.S.}\,=\,V_{V.C.C.S.}\,I_{V.C.C.S.}\,=\,(-14.28\,V)\,(-11.43\,A)\,=\,163.2\,W[/tex]

Finally!
 
  • #38
Good job. Practice with these circuits and you'll be able to do them in your sleep in no time
 
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