Norton & Thevenin theorem

Thé venin's Theorem

Any voltage network which may be
viewed from two terminals can be
replaced by a voltage-source
equivalent circuit comprising a single
voltage source E and a single series
resistance R.. The voltage V is the
open-circuit voltage between the two
terminals and the resistance Z is the
resistance of the network viewed from
the terminals with all voltage sources
removed from circuit.

Sample

All circuits are equivalent. Resistors
R₁,R₂, R₃ and voltage source are
transformed into Required Eq,

see parallel,
series simplifications.

To determine Eequ we shall break off
branch connecting node 1 and node
2

Norton's Theorem

Any current network which may be
viewed from two terminals can be
replaced by a current-source
equivalent circuit comprising a single
current source I and a single shunt
conductance G. The current I is the
short-circuit current between the two
terminals and the conductance G is
the conductance of the network
viewed from the terminals with all
branches containing current sources
are broken off.

Superposition theorem

In a network with multiple voltage
sources, the current in any branch is
the sum of the currents which would
flow in that branch due to each
voltage source acting alone with all
other voltage sources replaced by
their internal impedances.
The goal of folowing text is to check
superposition theorem.

Step 1. Construct following circuit
using Circuit Magic then run Node
Voltage Analysis. (popular circuits
analysis technique). You can
alsocalculate currents using other
techniques.


Electrical scheme
Inital variables
R₂=10Ohms; R₁ =10Ohms;
R₃ =10Ohms;
E₁ =3V; E₃ =4V;
Solution
V₁ ·G₁₁=I₁₁
G₁₁=1/R₁ +1/R₂+1/R₃=0,3
I₁₁ =-E₁/R₁ -E₃ /R₃ =-0,7
0,3V 1 =-0,7
V₁ =-2,3333
V₂ =0
I₁ =(V₁ -V₂ +E₁)/R₁ =0,0666667
I₂=(V₁-V₂)/R₂ =-0,233333
I₃=(V₁-V₂+E₃ )/R₃=0,166667
These values are used to check
currents determined from
superposition theorem
Step 2. Remove a voltage source
from the third branch then run
Node Voltage Analysis.


Electrical scheme
Inital variables
R₂=10Ohms; R₁ =10Ohms;
R₃ =10Ohms;
E₁=3V;
Solution
V₁ ·G₁₁=I₁₁
G₁₁=1/R₁ +1/R₂ +1/R₃=0,3
I₁₁ =-E₁ /R₁ =-0,3
0,3V 1 =-0,3
V₁ =-1
V₂ =0
I₁(1) =(V₁ -V₂ +E₁)/R₁ =0,2
I₂(1) =(V₁ -V₂ )/R₂ =-0,1
I₃(1) =(V₁ -V₂ )/R₃ =-0,1
These values are used to
determine current from
superposition theorem.
Step 3. Remove a voltage source
from the first branch then run
Node Voltage Analysis.


Electrical scheme
Inital variables
R₂=10Ohms; R₁ =10Ohms;
R₃ =10Ohms;
E₃=4V;
Solution
V₁ ·G₁₁=I₁₁
G₁₁=1/R₁ +1/R₂ +1/R₃=0,3
I₁₁ =-E₃ /R₃ =-0,4
0,3V₁ =-0,4
V₁ =-1,3333
V₂ =0
I₁(2) =(V₁ -V₂ )/R₁ =-0,133333
I₂(2) =(V₁ -V₂ )/R₂ =-0,133333
I₃(2) =(V₁ -V₂ +E₃)/R₃ =0,266667
Superposition theorem checking
I₁ =I₁(1) +I₁(2) =0,2-0,133333=0,0666666
I₂ =I₂(1) +I₂(2) =-0,1-0,133333=-0,233333
I₃ =I₃(1) +I₃(2) ==-0,1+0,266667=0,166667