Nodal Analysis of a Voltage Dependant voltage source, no pretty answer

[cybhunter]

Homework Statement

five branches

left most branch: 50 Volt source
second left most: 10 ohm resistor
shared branch between the second left most branch and center: 10 ohms (Voltage drop from left to right= V delta)

center branch: 30 ohm resistor

between center and second right most: dependent voltage source (V delta/ 5)
second right most branch: 39 ohms
right most branch: 78 ohms

the second and right most branch have an unknown voltage 'vo'


find the vo voltage

Homework Equations

due to KCL, the current entering the voltage dependent source must equal the current leaving the dependent source:

vo=v2-(V delta/5) and v delta= 50 volts - v2

equating the currents:

(v2-50 Volts)/(10 ohms) +(v2)/(30 ohms) =(v2-(50 volts -v2)/5)/(39 ohms) + (v2-(50 volts -v2)/5)/(78 ohms)

rearranging the equation to equate to zero amperes:

(v2-50 Volts)/(10 ohms) +(v2)/(30 ohms) -(v2-(50 volts -v2)/5)/(39 ohms) - (v2-(50 volts -v2)/5)/(78 ohms)= 0 Amperes

using a Ti-89 to slove this (using the above equation as 'n1' and the output as 'v2')
solve(n1,{v2}) results in v2 being equal to 900/17 volts (~52.9412 Volts) meaning a V delta value of -2.9412 volts.

Checking the currents, the left side of the dependent source (not including the current across the 10 ohm resistor) is equal to 2.0589 Amperes, and the right side is equal to 2.0583 Amperes. Considering the ugly numbers I ended up mentioning it to my professor and he note that since it is a textbook example, the numbers should be integers and that the answer I got is wrong.

The Attempt at a Solution

(see attached)


Considering that the currents are very very close, I can assume to an extant that I am setting up the equations properly. What am I missing though?

 

[gneill]

due to KCL, the current entering the voltage dependent source must equal the current leaving the dependent source:

vo=v2-(V delta/5) and v delta= 50 volts - v2

equating the currents:

(v2-50 Volts)/(10 ohms) +(v2)/(30 ohms) =(v2-(50 volts -v2)/5)/(39 ohms) + (v2-(50 volts -v2)/5)/(78 ohms)

If you look carefully at the terms of your equation, each one describes a current flowing out of the supernode:

As such, when you wrote the terms on the RHS of the equals sign they should have taken the opposite sign so that they would represent currents flowing into the supernode. As it stands you've equated one set of currents flowing out of the supernode with another set of currents also flowing out.

 

[cybhunter]

Thanks