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Topic: JonF's "Simple" Circuit (Read 1277 times) previous topic - next topic

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Re: JonF's "Simple" Circuit
Reply #175
I gotta say, this doesn't seem to be all that simple of a problem.
Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know?  And he wrote 10 equations with 10 unknowns ... blindfolded with his hands tied behind his back! Honest. Just ask him. And his simple solution is comin' right up.
  • Last Edit: January 17, 2017, 04:30:49 AM by Dave Hawkins

Re: JonF's "Simple" Circuit
Reply #176
All the R values are knowns.  That is, the goal was to write the voltages in terms of R's only which would be given.
I'm with Dave. That it could be interpreted in another way never occurred to me.
Really doesn't work. Defining an unknown in terms of other unknowns doesn't change anything to a known
:facepalm:
What specifically are you taking issue with? As a reminder, I try to be completely civil with you, so I'd appreciate you actually trying to engage with me rather than posting silly pictures.

  • JonF
Re: JonF's "Simple" Circuit
Reply #177
I gotta say, this doesn't seem to be all that simple of a problem.
True.

But it is a simple circuit, which is a technical term for a particular class of circuits.  I posted definitions.
"I would never consider my evaluation of his work to be fair minded unless I had actually read his own words." - Dave Hawkins

  • JonF
Re: JonF's "Simple" Circuit
Reply #178
I was just a lowly radio tech, but isn't that just a DC version of a Wien Bridge?
Same topology, but whether or not it's a Wien bridge depends on the resistor values.
"I would never consider my evaluation of his work to be fair minded unless I had actually read his own words." - Dave Hawkins

  • JonF
Re: JonF's "Simple" Circuit
Reply #179
I gotta say, this doesn't seem to be all that simple of a problem.
Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know?
Yep, indeed I did.

Quote
And he wrote 10 equations with 10 unknowns
Yep. And I acknowledged that those were wrong.

Quote
... blindfolded with his hands tied behind his back! Honest. Just ask him.
Dishonest.  Nothing like that.

Quote
And his simple solution is comin' right up.
Try to keep up, Davie.  I never said I would solve the equations.  The challenge was for you.

 I also acknowledged you've met the challenge by my standards.  And that you've proved that when I claimed you couldn't solve a simple resistive circuit I was right at the time. Your flailing around proves that.
  • Last Edit: January 17, 2017, 06:14:54 AM by JonF
"I would never consider my evaluation of his work to be fair minded unless I had actually read his own words." - Dave Hawkins

Re: JonF's "Simple" Circuit
Reply #180
Jesus Christ.

Re: JonF's "Simple" Circuit
Reply #181
Jesus Christ.

Isn't that taking your lords name in vain?
"That which can be asserted with evidence can also be dismissed without evidence." (Dave Hawkins)

  • JonF
Re: JonF's "Simple" Circuit
Reply #182
Jesus Christ.
Do you disagree with anything I wrote?

Will you apologize for the lies in message 175?  I count three.

For one thing, I never posted a definition of "simple".
  • Last Edit: January 17, 2017, 06:16:56 AM by JonF
"I would never consider my evaluation of his work to be fair minded unless I had actually read his own words." - Dave Hawkins

Re: JonF's "Simple" Circuit
Reply #183


This is your KCL for nodes 2 and 5, with current directions per my diagram.  I don't think they are independent because of the shared (V5 - V2)/R3 term.

1: i0 + i3 = i2,4  (1 - V2)/R0 + (V5 - V2)/R3 = V2/R2,4
2: i1,5 = i3 + i6  (1 - V5)/R1,5 = (V5 - V2)/R3 + V5/R6

Rearrange equation 2.

1: (1 - V2)/R0 + (V5 - V2)/R3 = V2/R2,4
2: (1 - V5)/R1,5 - (V5 - V2)/R3  = V5/R6

Add them and the (V5 - V2)/R3 term drops out.  Then you are left with:

(1 - V2)/R0 + (1 - V5)/R1,5 = V2/R2,4 + V5/R6

That's one equation with two unknowns.  At best you can get a proportional relationship between V2 and V5, but I don't think you can get absolute values for them without another independent equation.  Again, circuit shit is not my forte and I acknowledge that I could be missing something here.

The two equations you have are independent. The fact that both have a (V2 - V5) term grouped together is irrelevant; all that matters is that they don't fully cancel for any nontrivial manipulation.

Try simply writing the equations you have using the resistances to make coefficients for V2 and V5, using distribution to separate the V2 - V5 term.
Pulling this forward

  • JonF
Re: JonF's "Simple" Circuit
Reply #184
Why?
"I would never consider my evaluation of his work to be fair minded unless I had actually read his own words." - Dave Hawkins

Re: JonF's "Simple" Circuit
Reply #185
Spoiler (click to show/hide)

This is your KCL for nodes 2 and 5, with current directions per my diagram.  I don't think they are independent because of the shared (V5 - V2)/R3 term.

1: i0 + i3 = i2,4  (1 - V2)/R0 + (V5 - V2)/R3 = V2/R2,4
2: i1,5 = i3 + i6  (1 - V5)/R1,5 = (V5 - V2)/R3 + V5/R6

Rearrange equation 2.

1: (1 - V2)/R0 + (V5 - V2)/R3 = V2/R2,4
2: (1 - V5)/R1,5 - (V5 - V2)/R3  = V5/R6

Add them and the (V5 - V2)/R3 term drops out.  Then you are left with:

(1 - V2)/R0 + (1 - V5)/R1,5 = V2/R2,4 + V5/R6

That's one equation with two unknowns.  At best you can get a proportional relationship between V2 and V5, but I don't think you can get absolute values for them without another independent equation.  Again, circuit shit is not my forte and I acknowledge that I could be missing something here.

The two equations you have are independent. The fact that both have a (V2 - V5) term grouped together is irrelevant; all that matters is that they don't fully cancel for any nontrivial manipulation.

Try simply writing the equations you have using the resistances to make coefficients for V2 and V5, using distribution to separate the V2 - V5 term.
You're right.  I think I have it.  When I punch in the R values V5 = 0.4 V

ETA: the vertical line to the right of the 0 three equations up from the bottom is the cursor.
  • Last Edit: January 17, 2017, 12:25:19 PM by Dean W

Re: JonF's "Simple" Circuit
Reply #186
Spoiler (click to show/hide)
Please go ahead and write out the solution.
Rewriting your equations:

V2 (-1/R0 - 1/R3 - 1/R2,4) + V5 (1/R3) = 1/R0
V2 (1/R3) + V5 (-1/R1,5 - 1/R3 - 1/R6) = 1/R1,5

V2 = ((1/R0 + 1/R3 + 1/R2,4)(1/R1, 5 + 1/R3 + 1/R6) - 1/R3^2)((-1/R1,5 - 1/R3 - 1/R6) 1/R0 - 1/R3 1/R1,5)
V5 = ((1/R0 + 1/R3 + 1/R2,4)(1/R1, 5 + 1/R3 + 1/R6) - 1/R3^2)(-1/R3 1/R0 - (1/R0 + 1/R3 + 1/R2,4) 1/R1,5)

Re: JonF's "Simple" Circuit
Reply #187
Spoiler (click to show/hide)

Corrected Cramer's rule answer:

(R1 R3 + R1 R4 + R2 R4 + R3 R4)/(R1 R2 R3 + R1 R2 R4 + R1 R2 R5 + R1 R3 R5 + R1 R4 R5 + R2 R3 R4 + R2 R4 R5 + R3 R4 R5)

Re: JonF's "Simple" Circuit
Reply #188
It checks out.

  • Pingu
Re: JonF's "Simple" Circuit
Reply #189
I gotta say, this doesn't seem to be all that simple of a problem.
Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know?  And he wrote 10 equations with 10 unknowns ... blindfolded with his hands tied behind his back! Honest. Just ask him. And his simple solution is comin' right up.

Dave, even I know that a simple phenomenon isn't necessarily a "simple problem" to solve.

I mean, the four colour map problem hasn't been solved yet.  Or has it?

Re: JonF's "Simple" Circuit
Reply #190
I gotta say, this doesn't seem to be all that simple of a problem.
Oh it's simple. Just ask Jon. He posted the definition of "simple" doncha know?  And he wrote 10 equations with 10 unknowns ... blindfolded with his hands tied behind his back! Honest. Just ask him. And his simple solution is comin' right up.

Dave, even I know that a simple phenomenon isn't necessarily a "simple problem" to solve.

I mean, the four colour map problem hasn't been solved yet.  Or has it?
Define "solved". (Also, "solved" is a bit of an odd term here; "proven" is probably more accurate)

It was originally reduced to a finite (but large) number of cases, and those cases were checked by computer - so if we believe in the integrity of the computer (and the program), then it's been solved (i.e. the 4-color theorem has been proven). Apparently since then, it's been reduced to proof-checking software, so if you trust Coq (a theorem-prover), then this theorem has been proven. It's proven enough that the mathematical community generally accepts it.

Damnit quick edit, stop being awful.
  • Last Edit: January 17, 2017, 02:23:56 PM by uncool