blob: 378b29151d0b7b0507293bdda756d8aad80c1ff3 (
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
|
//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 6
disp("CHAPTER 2");
disp("EXAMPLE 6");
//VARIABLE INITIALIZATION
f=50; //in Hertz
I1=20; //in Amperes
pf1=0.75; //power factor
v=230; //in Volts
pf2=0.9; //power factor(lagging)
//SOLUTION
//V.I1.cos(Φ1) = P
phi1=acos(pf1);
res1=tan(phi1); //result1 = tan(Φ1)
phi2=acos(pf2);
res2=tan(phi2); //result2 = tan(Φ2)
Ic=I1*pf1*(res1-res2);
w=2*%pi*f; //w=2.pi.f
c=Ic/(v*w);
disp(sprintf("The value of capacitance is %5.2f μF",c*(10^6)));//text book answer is 82.53 mF
Qc=v*Ic; // reactive power in kVAr
disp(sprintf("The reactive power is %6.4f kVAR",Qc/(10^3)));//text book answer is 1.3716
I2=I1*(pf1/pf2); //I1.cos(Φ1) = I2.cos(Φ2)
disp(sprintf("The new supply current is %5.2f A",I2));
//END
|
