blob: 86044a5a53976cb517fbed0ffa338e100f45673b (
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
|
//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 36
disp("CHAPTER 1");
disp("EXAMPLE 36");
//VARIABLE INITIALIZATION
I=10; //current source in Amperes
v=10; //voltage source in Volts
r1=4; //top resistance in Ohms
r2=4; //right resistance in Ohms
r3=4; //bottom resistance in Ohms
r4=6; //left resistance in Ohms
r5=1; //in Ohms
//SOLUTION
//(1)v1+(12/5)In=30........eq (1)
//(2)v1+(-4)In=10..........eq (2)
A=[1 12/5;2 -4];
b=[30;10];
x=inv(A)*b;
v1=x(1,:); //to access the 1st element of 2X1 matrix
In=x(2,:); //to access the 2nd element of 2X1 matrix
req1=(r1*r4)/(r1+r4);
req2=(r2*r3)/(r2+r3);
rn=req1+req2;
I1=(rn*In)/(rn+r5);
disp(sprintf("By Norton Theorem, the current through 1Ω resistor is %.3f A",I1));
//END
|
