blob: 60903ad05a55b13d49d021aedcd8e6641fd342d2 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
|
//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 33
disp("CHAPTER 1");
disp("EXAMPLE 33");
//VARIABLE INITIALIZATION
I=10; //current source in Amperes
v=10; //voltage source in Volts
r1=4; //top resistance in Ohms
r1=4; //right resistance in Ohms
r3=4; //bottom resistance in Ohms
r4=6; //left resistance in Ohms
r5=1; //in Ohms
//SOLUTION
//by applying nodal analysis at node 1, the following equations are obtained:
//(17)v1+(-12)v2=150.......eq (1)
//(-4)v1+(6)v2=10..........eq (2)
//solving the equations by matrix method
A=[17 -12;-4 6];
b=[150;10];
x=inv(A)*b;
v1=x(1,:); //to access the 1st element of 2X1 matrix
v2=x(2,:); //to access the 1st element of 2X1 matrix
if(v1>v2) then
I=(v1-v2)/r5;
disp(sprintf("By nodal analysis, the current through 1Ω resistor is %.3f A",I));
else
I=(v2-v1)/r5;
disp(sprintf("By nodal analysis, the current through 1Ω resistor is %.3f A",I));
end;
//END
|
