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

[VinnyCee]

Homework Statement

Find v_0 in the circuit below and the power dissipated by the controlled source.

http://img291.imageshack.us/img291/5479/chapter2problem22wy5.jpg

Homework Equations

KCL

v\,=\,i\,R

The Attempt at a Solution

KCL: i\,=\,2\,v_0\,+\,10\,A

v_o\,=\,i\,R

v_o\,=\,\left(2\,v_0\,+\,10\right)\,(4\Omega)

v_0\,=\,8\,v_0\,+\,40

v_0\,=\,-\frac{40}{7}\,V

Am I on the right track here?

 

[berkeman]

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.

 

[VinnyCee]

If v_0 looks correct then I can calculate the rest.

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

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

p\,=\,\frac{400}{49}\,W\,\approx\,8.16\,W

Does this look right?

 

[berkeman]

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.

 

[VinnyCee]

I guess, the resistors are the other part of the resistive load?

How do I figure for that then?

P\,=\,v\,i\,=\,\left(v_0\right)\,\left(2\,v_0\right)\,W

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

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

P\,\approx\,32.7\,W

Correct?

 

[berkeman]

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.

 

[VinnyCee]

2\,v_0 is the current contribution of the controlled source.

2\,v_0\,=\,-\frac{80}{7}\,A\,\approx\,-11.43\,A

Do I then use the p = v i ?

P\,=\,\left(-\frac{40}{7}\right)\,\left(-11.43\,A\right)\,=\,65.3\,W

 

[berkeman]

Nice job, Vinny. Keep it up and you will go far.

 

[VinnyCee]

Thank you!

Just to clarify, v_0\,=\,-5.71\,V and P_{V.C.C.S}\,=\,65.3\,W. These are correct?

 

[doodle]

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.

 

[VinnyCee]

Would that V be half of the v_0?

P is 32.7W and not 65.3W?

 

[doodle]

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.

 

[VinnyCee]

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 p\,=\,v\,i\,=\,\left(-\frac{20}{7}\right)\,\left[2\,\left(-\frac{40}{7}\right)\right]\,=\,32.7\,W

If that is not it, how is it calculated?

 

[doodle]

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.

 

[VinnyCee]

The current across the resistors is i\,=\,2\,v_0\,+\,10\,=\,-1.43\,A, right?

Now to get the V across both resistors.

V\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V

I then use the P = vi.

P\,=\,v\,i\,=\,(-14.3\,V)\,(-1.429\,A)\,=\,20.4\,W

I hope that's right:)

 

[teknodude]

The current across the resistors is i\,=\,2\,v_0\,+\,10\,=\,-1.43\,A, right?

Now to get the V across both resistors.

V\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V

I then use the P = vi.

P\,=\,v\,i\,=\,(-14.3\,V)\,(-1.429\,A)\,=\,20.4\,W

I hope that's 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?

 

[Number2Pencil]

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

 

[teknodude]

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)

 

[Number2Pencil]

Yes, What you said just now was what I was asking.

 

[teknodude]

Oh alright. So yea, he's just not finished with the problem yet. Hope he checks back.

 

[doodle]

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.

 

[VinnyCee]

SO the V is for the VCCS only and that is right. So I need to recalculate the i for ONLY the vccs.

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

v_{VCCS}\,=\,-14.3\,V

p\,=\,v\,i\,=\,(-14.3\,V)\,(-11.43\,A)\,=\,163.4\,W

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

v\,=\,i\,R\,=\,(-1.429\,A)\,(10\Omega)\,=\,-14.3\,V

VS.

v\,=\,i\,R\,=\,(-11.43\,A)\,(10\Omega)\,=\,-114.3\,V

I'm thoroughly confused!

 

[doodle]

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?

 

[VinnyCee]

Like this?

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

or

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

So...

v_0\,=\,-\frac{40}{7}\,A\,\approx\,-5.71\,A

and

P_{VCCS}\,\approx\,8.17\,W

Right?

 

[doodle]

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.

 

[VinnyCee]

Start Again!

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

Let's just start from the beginning!

http://img291.imageshack.us/img291/5479/chapter2problem22wy5.jpg

i\,=\,2\,v_0\,+\,10\,A

Right?

v_o\,=\,i\,R

v_o\,=\,\left(2\,v_0\,+\,10\right)\,(4\Omega)

Right?

v_0\,=\,8\,v_0\,+\,40

So...

v_0\,=\,-\frac{40}{7}\,V

Right?

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

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

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

Now what do I do, assuming the above is all right?
P\,=\,I\,V\,=\,I_{VCCS}\,V_{VCCS}

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

Does this seem right?

 

[Number2Pencil]

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

 

[VinnyCee]

voltage ACROSS THE SOURCE

That would be V_{VCCS}, right?

Is it just divided equally between the two sources so the voltage across the source is just \frac{1}{2}\,V_0\,=\,-2.86\,V

 

[Number2Pencil]

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

 

[VinnyCee]

Are you are saying that

V_0\,=\,V_{VCCS}\,=\,V_{4\,\Omega}\,=\,V_{6\,\Omega}\,=\,V_{10\,A}\,=\,-\frac{40}{7}\,V

 

[Number2Pencil]

almost

V_4ohm + V_6ohm = V_C.S(10amp) = V_V.C.C.S

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

 

[doodle]

Okay, let's take it from the start. :)

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

Let's just start from the beginning!

i\,=\,2\,v_0\,+\,10\,A

Right?

v_o\,=\,i\,R

v_o\,=\,\left(2\,v_0\,+\,10\right)\,(4\Omega)

Right?

v_0\,=\,8\,v_0\,+\,40

So...

v_0\,=\,-\frac{40}{7}\,V

Right?

Yes.

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

So, I need to use the EQs P\,=\,v\,i or P\,=\,\frac{v^2}{R}, 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, i_{VCCS}\,=\,2\,V_0, right?

Okay... so we now know the I across the VCCS.

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

P\,=\,I\,V\,=\,I_{VCCS}\,V_{VCCS}

We know the I (VCCS) but what about the V (VCCS)?

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

Does this seem right?

Yes, I (VCCS) = -11.43A is correct. But again, what about the V (VCCS)?

 

[VinnyCee]

lol - so we have this

V_{4\Omega}\,=\,V_0\,=\,-\frac{40}{7}

and

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

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

How do I get V_{6\Omega} though?

 

[Number2Pencil]

you get the voltage across the 6 ohm resistor with non other than ohm's law:

V_6 = I_6 R_6

 

[VinnyCee]

Cool!

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

V_{6\Omega}\,=\,(-1.43\,A)\,(6\Omega)\,=\,-8.57\,V

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

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

But I thought that P_{V.C.C.S}\,=\,163.4\,W ?

 

[Number2Pencil]

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

 

[VinnyCee]

V_{V.C.C.S.}\,=\,V_{6\Omega}\,+\,V_{4\Omega}\,=\,(-8.57\,V)\,+\,(-5.71\,V)\,=\,-14.28\,V

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

Finally!

 

[Number2Pencil]

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