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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 : 6.6
+
+clc;clear; // clears the console and command history 
+
+// Given data
+V_t = 220      // shunt motor supply voltage in V
+P_0 = 3550     // o/p power in W 
+n = 0.85       // efficiency
+// condition given : starting current of the motor should not exceed twice the load current
+
+// caclulations 
+P_in = P_0/n          // i/p power in W
+P_tloss = P_in-P_0    // total loss in W
+I_a = P_in/V_t        // neglet the field current then armature current in A
+P_cu = P_tloss/2      // copper loss in W
+R_a = P_cu/I_a^2      // armature resistance in ohm
+I_1 = 2*I_a           // maximum starting current in A
+R_1 = V_t/I_1         // maximum resistance in ohm
+k = (R_1/R_a)^(1/4)   // constant
+R_2 = R_1/k           // other value of resistance in ohm
+R_3 = R_2/k           // other value of resistance in ohm
+R_4 = R_3/k           // other value of resistance in ohm
+R_5 = R_4/k           // other value of resistance in ohm
+R_1step = R_1-R_2     // resistance of the 1st step in ohm
+R_2step = R_2-R_3     // resistance of the 1st step in ohm
+R_3step = R_3-R_4     // resistance of the 1st step in ohm
+R_4step = R_4-R_5     // resistance of the 1st step in ohm
+
+// display the result 
+disp("Example 6.6 solution");
+printf(" \n resistance of the 1st step in ohm \n R_1step = %.1f ohm \n", R_1step);
+printf(" \n resistance of the 2nd step in ohm \n R_2step = %.2f ohm \n", R_2step);
+printf(" \n resistance of the 3rd step in ohm \n R_3step = %.2f ohm \n", R_3step);
+printf(" \n resistance of the 4th step in ohm \n R_3step = %.2f ohm \n", R_4step);