Solving a Puzzling Circuit: Find V2

[gneill]

I get n=13.232. What now?

That's a bit off from what I'm seeing. But you've made excellent progress!

Myself, I didn't solve for n first, but rather substituted $n = \frac{8}{v2 - 8}$ from the voltage divider part into the node equation which then reduced to a pretty nice quadratic for v2.

 

[kaspis245]

Sorry, I've recalculated it and got this equation:

$n^3-11n^2-16n-4=0$

Then let wolfram alpha do the work and got n=12.325

This way V₂=9.94V

Is it correct?

 

[gneill]

The value for n looks good. The value of V2 does not. What expression did you use?

 

[kaspis245]

I used this expression:

$V_2= \frac{10n+10n^2}{1+3n+n^2}$

which is derived from:

$\frac{10-V_2}{V_2}-\frac{V_2}{nR}-\frac{V_2}{R+nR}=0$

 

[gneill]

I used this expression:

$V_2= \frac{10n+10n^2}{1+3n+n^2}$

which is derived from:

$\frac{10-V_2}{V_2}-\frac{V_2}{nR}-\frac{V_2}{R+nR}=0$

I presume that first term should be over R, not $V_2$.
Okay, that should work. But I'm seeing a different value for $V_2$ using that equation.

I would have used the simpler voltage divider derived equation myself.

 

[kaspis245]

I am such an idiot, I've made another mistake in my calculations.

Here's my final answer:

$V_2=8.65 V$

 

[NascentOxygen]

Sorry, I've recalculated it and got this equation:

$n^3-11n^2-16n-4=0$

Then let wolfram alpha do the work and got n=12.325

This way V₂=9.94V

Is it correct?

A cubic?? You have probably overlooked a term in the numerator which would have canceled a factor of this, leaving you with just a quadratic to solve.

n=12.325 is my answer, too, from the quadratic solution

 

[kaspis245]

As long as we are getting the same answer I am ok with that