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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.3
+
+clc;clear; // clears the console and command history 
+
+// Given data
+V_t = 230    // supply voltage series motor in V
+R_a = 0.2    // armature resistance in ohm
+R_f = 0.2    // field resistance in ohm
+I_a1 = 10    // dc series motor armature current in A
+N = 1000     // speed in rpm
+I_a2 = 8     // dc series motor armature current in A when R=2 ohm
+R_2 = 2      //  series resistance in ohm
+R_1 = 0      // series resistance in ohm
+// phi2 = 0.8*phi1 
+// phi1/phi2 = 1.25
+
+// caclulations 
+R_a1 = R_a+R_1            // new armature resistance in ohm
+E_b1 = V_t-I_a1*R_a1      // back emf in V
+R_a2 = R_a+R_2            // new armature resistance in ohm
+E_b2 = V_t-I_a2*R_a2      // back emf in V
+N_2 = (E_b2/E_b1)*1.25*N  // new speed in rpm 
+
+// display the result 
+disp("Example 6.3 solution");
+printf(" \n New speed \n N_2 = %.1f rpm \n", N_2);