updated the code

1 files changed, 5 insertions, 22 deletions

diff --git a/1445/CH8/EX8.33/Ex8_33.sce b/1445/CH8/EX8.33/Ex8_33.sce
index aaf8cdd40..9e16a1081 100644
--- a/1445/CH8/EX8.33/Ex8_33.sce
+++ b/1445/CH8/EX8.33/Ex8_33.sce
@@ -1,10 +1,10 @@
 //CHAPTER 8- DIRECT CURRENT MACHINES
 //Example 33
 
+clc;
 disp("CHAPTER 8");
 disp("EXAMPLE 33");
 
-//230 V DC series motor
 //VARIABLE INITIALIZATION
 v_t=230;                       //in Volts
 N1=1500;                       //in rpm
@@ -15,34 +15,17 @@ r_se=0.2;                      //series field resistance in Ohms
 //SOLUTION
 
 //solution (a)
-//for series motors, phi dir prop Ia
-// therefore, Te dir prop Ia^2
-// at starting Eb=0 and Vt= Ia1.(r_a+r_se+r_ext)
-//rearranging for r_ext, we get
-// r_ext = (Vt-Ia1.(r_a+r_se))/ Ia1
-E_b=0;                         //back emf at starting
-nr1=v_t-I_a1*(r_a+r_se);       //value of numerator in the expression for r_ext
+E_b=0;                         //at starting
+nr1=v_t-I_a1*(r_a+r_se);       //value of numerator
 r_ext=nr1/I_a1;
-disp(sprintf("(a) At starting, the resistance that must be added is %.0f Ω",r_ext));
+disp(sprintf("(a) At starting, the resistance that must be added is %f Ω",r_ext));
 
 //solution (b)
-//Ia2=Ia1=20 A
-//as phi dir prop Ia, we get 
-//Eb2/Eb1 = phi2.n2/ phi1. N1 = Ia2.N2/Ia1.N1
-//=> Eb2/Eb1=N2/N1  as Ia2=Ia1               (eq 1)
 I_a2=I_a1;
 N2=1000;
 ratio=N2/N1;
-// now, we know that Eb1=Vt-Ia1.(r_a+r_se) and 
-// Eb2 = Vt - Ia2.(r_a+r_se+r_ext)
-//substituting values of Eb1 and Eb2 in eq 1 above, we get
-//n2/n1 = (Vt - Ia2.(r_a+r_se+r_ext))/ (Vt-Ia1.(r_a+r_se))
-//since ia1=Ia2 (rated torque)
-//we get 
-//r_ext = (N2/N1).(v_t-I_a1*(r_a+r_se))/Ia2 -(v_t-I_a2*(r_a+r_se))/Ia2
-//
 nr2=v_t-I_a2*(r_a+r_se);
 r_ext=((ratio*nr1)-nr2)/(-I_a2);
-disp(sprintf("(b) At 1000 rpm, the resistance that must be added is %.3f Ω",r_ext));
+disp(sprintf("(b) At 1000 rpm, the resistance that must be added is %f Ω",r_ext));
 
 //END