[sdiy] A little OT, Begging a hand

[Ian Fritz]

I disagree with this. V0 is a measured voltage.  It is assumed to be 
measured with a high impedance meter.  Thus it has no effect on the 
currents and voltages in the circuit.  Therefore it shouldn't appear in the 
equations.  It should only appear after the calculation is finished as the 
voltage divider equation involving the already determined current through R8.


At 04:03 PM 2/8/2011, David G. Dixon wrote:
> > But in the end R1+R2+R3 = K  *must* be satisfied. So all three are not
> > independent.
>
>Yes.  Actually, that constraint adds the fifth equation we need to solve the
>system.  It's not linear, but it doesn't need to be.
>
>The problem is soluble.  We know that G8 = 1/2k2, and we have six equations
>for six conductances:
>
>G1 - G6 = 0
>
>G1*(VA - V1) + G2*(VA - VB) + G4*(VA - VC) = 0
>
>G2*(VB - VA) + G3*(VB - V2) + G57*(VB - VC) = 0
>
>G6*(Vout - VC) + G8*(Vout - V3) = 0
>
>G57*(VC - VB) + G4*(VC - VA) + G6*(VC - Vout) = 0
>
>1/G1 + 1/G2 + 1/G3 = 1/K
>
>This will give us G1, G2, G3, G4, G57, G6 and G8.  G5 and G7 can then be
>determined from the following equations:
>
>G2 = G7
>
>1/G57 = 1/G5 + 1/G7
>
>This is not a linear system, but it is soluble.

Show me the solution, then. :-)