Vth using Node

opening load, finding V_th using Node .. code

${\displaystyle V_{th}=6.4516}$

In using Node

shorting load, finding I_n using Node .. code

${\displaystyle I_{N}=1.064773736}$

Rth or Rn

${\displaystyle V_{th}/I_{N}={\frac {6.4516}{1.064773736}}=6.0591ohms}$

Finding Rth using source injection and node

• 
  zero sources to find Rth
• 
  add source injection since parallel/series combinations do not exist
• 
  redraw up and down before solving

Here is the mupad/matlab code that generates the answer R_th = 6.0591 ohms.

Comparing Node with Thevenin Equivalent

solving using Node .. code

solving using Thevenin equivalent

Solving the node equations yields:

${\displaystyle V_{a}=5.393}$

${\displaystyle V_{b}=1.1673}$

${\displaystyle V_{c}=1.107}$

${\displaystyle i_{12}=0.3571}$

${\displaystyle v_{12}=4.286}$

Using the Thevenin equivalent (and voltage divider) to compute voltage across the 12 ohm resistor:

${\displaystyle v_{12}=V_{s}*{\frac {12}{R{total}}}}$

${\displaystyle v_{12}=6.4516*{\frac {12}{6.0591+12}}=4.287}$

So they match ...

Thevenin voltage and resistance can not be computed from a node analysis of the entire circuit, but the node analysis of the entire circuit can be used to check if the thevenin equivalent produces the same numbers.