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

[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