What is the maximum power transfer for a load in a circuit using nodal analysis?

[CaptProton987]

Circuit Analysis--Nodal error?

Homework Statement

Find R_Load for maximum power that can be transferred to the load.

Homework Equations

KCL@V1: (V1-6v)/3kΩ + (V1-V3)/6kΩ + (V1 - V2)/2kΩ - 2mA = 0

KCL@V2: 2mA + (V2 - V1)/2kΩ + Isc = 0

KCL@V3: 2mA + (V3-V1)/6kΩ - Isc = 0

The Attempt at a Solution

NOTE: The above equations are currents for KCL that I came up with while attempting Nodal Analysis so they may be incorrect...

Okay basically I'm trying to do two things here:

(1) reduce the circuit to one voltage source and one resistor for Thevenin's theorem.
I've started by removing the load resistor, and trying to find the value for Isc [short circuit current] via nodal analysis... but I'm stuck I don't believe the above nodal equations are correct and even if they are I really! screwed up trying to solve for V1, V2, and V3.

(2) Once I have the Thev. equivalent circuit I want to plug that load resistor back in and find its value that will "Maximize power" -- I have absolutely no idea how to do this.

I've uploaded some images of the circuit I drew [software isn't working for me but at least it makes nice pictures] to give you an idea of what I'm doing...

 

[jhicks]

There's a problem that I see with KCL@V1 (and V2 and V3). It's a subtle point but (V1-6V)/(3k) suggests that positive current flows from V1 to 6V - away from node V1. However, you use -2mA to indicate current flowing away as well on the same side of the equation so your signs are all mixed up. It doesn't matter whether you originally assume the current direction correctly as long as you are consistent.

Maximum power occurs when the resistance is equal to Voc / Isc.

 

[CEL]

Substituting I1,R1 by their Thevenin equivalent will simplify your calculations.
Once you have your Thevenin equivalent and plug R_load in., write the equation for the power, with Vth and Rth as known values and R_load as a variable. Differentiate the power equation wrt R_load and set it to zero.
You have the value of R_load that maximizes power transfer.
Incidentally, this value is Rth, that is what jhicks said.

 

[najnaj224]

The correct equations for this poblem are:

V1: (6-V1)/3k + (V1-V2)/6k + (V1-V2)/2k - 2mA = 0

V2: (V2-V3)/Rload + (V1-V2)/2k + 2mA = 0

V3: (V2-V3)/Rload + (V1-V3)/6k = 0

(V2-V3)/Rload = Isc

Right? (I asking rather than telling)

 

[meatspin]

@V1: (V1 - 6V)/3k + (V1-V3)/6k + 2mA + (V1-V2)/2k = 0V

@V2: -2mA + (V2-V1)/2k = 0V


V2 = 0V