| 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
0,3V1=-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
0,3V1=-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
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