Node Voltage Practice Problem

[sshoptaugh1991]

This could be another issue of not being able to look at the circuit in an easier way, but I am a little confused at the answer here.  Maybe there is a step in my work that I did not take, or maybe I took too many steps.

The problem says to find the value of Vo, and it is a node voltage practice exercise.  When running the circuit through LTSpice, I got the same answer as the book, Vo = 2.79V, but I just do not see what I did wrong.  Here is the circuit, and my work:

Node 1 Equations:
-8mA + V1/3k + (V1-V2)/6k + 2mA = 0
V1(1/2k) - V2(1/6k) = 6mA

Node 2 Equations:
-(V1-V2)/6k + V2/6k + (V2-V3)/2k = 0
-V1(1/6k) + V2(5/6k) - V3(1/2k) = 0

Node 3 Equations:
-(V2-V3)/2k - 2mA + V3/1k = 0
-V2(1/2k) + V3(3/2k) = 2mA

All equations to form system of linear equations:
V1(1/2k) - V2(1/6k) + 0 = 6mA
-V1(1/6k) + V2(5/6k) - V3(1/2k) = 0
0 -V2(1/2k) + V3(3/2k) = 2mA

Solving the system with my calculator yields the results of:
V1 = 10.7692V
V2 = 3.6923V
V3 = 2.5641V

I assumed that V3 should equal Vo, but my numbers are off.  Where did I go wrong?  Am I over thinking something?

[Ian.M]

An alternative and possibly  easier way to look at that circuit that avoids nodal or mesh analysis is to apply a wye-delta transformation to R2, R3 & R4. then you can simplify it. R1||R2_3' and R3_4' || R5.  Apply a Norton to Thévenin transformation to the sources and their parallel resistors and solve it as a simple loop by K.V.L then use the known node voltages to back calculate the R3,R4,R5 junction node voltage using the original circuit.

[BigBoss]

Use always matrix notation

[sshoptaugh1991]

I feel stupid now.  I did not double-check my calculator input, and the second term in the first equation was put in as a positive coefficient instead of negative.  I entered the numbers again, being careful of the sign, and I got he right answer.

[MrAl]

Hi,


A sneaky way to do it is to use superposition.

To do that with this circuit split up the current source I2 into two separate current sources, which then gives you three current sources in the whole circuit where one end of each current source is connected to ground:
I1, I2a, I2b

You can then use superposition for all three current sources to arrive at the output voltage.

Just to note, it is almost never good enough to have just ONE method of doing the circuit because then we have no way to check the result.  If we have two ways to do it we can then make sure that we get the very same answer from both methods.  That is ALWAYS better than just doing it one way as you probably found out already.

[Ian.M]

Although my suggestion above is amenable to 'back of the envelope' calculation, I've put together a parameterised LTspice .op simulation showing the original circuit and the three easy steps to solve it as I described, with .op node voltage labels, formulae and resulting resistor values captured by .measure. (see Spice error log)