1 files changed, 43 insertions, 0 deletions
diff --git a/1370/CH3/EX3.4/example3_4.sce b/1370/CH3/EX3.4/example3_4.sce
new file mode 100755
index 000000000..08b6ea65c
--- /dev/null
+++ b/1370/CH3/EX3.4/example3_4.sce
@@ -0,0 +1,43 @@
+//example3.4
+clc
+disp("The given values are,")
+disp("(I_0)=1 A, cos(psi_0)=0.4, I2=50 A, and cos(psi_2)=0.8")
+k=200/400
+disp(k,"K=E2/E1=")
+k=0.5*50
+disp(k,"Therefore, I2''(in A)=K*I2=")
+disp("The angle of (I2'') is to be decided from cos(psi_2)=0.8")
+disp("Now, cos(psi_2)=0.8")
+p=acosd(0.8)
+format(6)
+disp(p,"Therefore, (psi_2)[in degree]=")
+disp("I2'' is in antiphase with I2 which lags E2 by 36.86 degree")
+disp("Consider the phasor diagram shown in the fig 3.18. The fluz (psi) is the reference")
+disp("Now cos(psi_0)=0.4")
+c=acosd(0.4)
+disp(c,"Therefore, psi_0(in degree)")
+disp("vector(I1)=vector(I2'')+vector(I_0)")
+disp("Resolve (I_0) and (I2'') into two components, along reference (psi) and in quadrature with (psi) in phase with V1.")
+x=1*sind(66.42)
+format(7)
+disp(x,"x component of (I_0)=(I_0)*sin(psi_0)=")
+x=1*cosd(66.42)
+format(4)
+disp(x,"y component of (I_0)=(I_0)*(psi_0)=")
+disp("Therefore, vector(I_0)=0.9165+j(0.4) A")
+i=25*sind(36.86)
+format(7)
+disp(i,"x component of I2''(in A)=I2''*sin(psi_2)=")
+i=25*0.8
+disp(i,"y component of I2''(in A)=I2''*cos(psi_2)=")
+disp("Thus the two component of I1 are as shown in the fig 3.18(c)")
+io=sqrt((15.9165^2)+(20.4^2))
+disp(io,"I1(in A)=sqrt[((15.9165)^2)+(20.4)^2]=")
+disp("This is the primary current magnitude.")
+disp("while tan(phi_1)=15.9165/20.4")
+t=atand(15.9165/20.4)
+disp(t,"Therefore, (psi_1)[in degree]=")
+disp("Hence the primary power factor is,")
+i=cosd(37.96)
+disp(i,"cos(psi_1)[lagging]=cos(37.96)=")
+disp("Key point: Remember that (psi_1) is angle between V1 and I1 and as V1 is vertical, (psi_1) is measured with respect V1. So do not convert rectangular to polar as it goes angle with respect to x-axis and we want it with respect to y-axis.")