| Node Voltage Method |
Node Voltage Method
The node voltage method is deduced from the Kirchhoff�s current law (KCL) and
Ohm�s laws
The following formulas are used to solve circuit.
where
 Node
i conductivity is the sum of conductivities of all branches which connected with
Node i.
Node i to node j conductivity is the negative sum of conductivities of all
branches which connect Node i with Node j.
 .Node
i - current
Sample: Circuit Solution By Node Voltage
Step 1. Construct following circuit using circuit editor.
Step 2. Run solution by node voltage method.
Step 3. Correct following solution.
Electrical scheme
Initial variables
R1=10 ohms; R2=20
ohms; R3=25 ohms; R4=10
ohms; R5=10 ohms;
E1=10V; E2=15V;
J1=10A;
Solution
V1�G11+V2�G12+V3�G13=I11
V1�G21+V2�G22+V3�G23=I22
V1�G31+V2�G32+V3�G33=I33
Node 1 conductivity
G11=1/R2+1/R1+1/R4=0,25
Node 1 to Node 2 conductivity
G12=-1/R2=-0,05
G13=-1/R1=-0,1
G22=1/R2+1/R5=0,15
G23=-1/R5=-0,1
G33=1/R5+1/R1+1/R3=0,24
I11=-E2/R1=-1,5
I22=-J1=-10
I33=E2/R1-E1/R3=1,1
Linear equations
0,25V1-0,05V2-0,1V3=-1,5
-0,05V1+0,15V2-0,1V3=-10
-0,1V1-0,1V2+0,24V3=1,1
V1=-69,706
V2=-147,06
V3=-85,735
V4=0
Branch currents calculation using Ohm�s Law
I1=(V1-V2)/R2=3,86765
I2=(V2-V3)/R5=-6,13235
I3=(V1-V3+E2)/R1=3,10294
I4=J1=10
I5=(V3-V4+E1)/R3=-3,02941
I6=(V1-V4)/R4=-6,97059
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