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+//Example 1_2 page no:23
+clc;
+//given
+//solving a sub part
+voltage = 500;//in v
+current = 32;//in A
+arm_res = 0.4;//in ohm
+fl_win_res = 250;//in ohm
+rpm = 450;
+field_current = 2;
+input_pow = (voltage*current)/1000;
+arm_current = current - field_current;
+//when running at 600rpm
+rpm1 = 600;
+k_phi = (voltage - 12)/rpm1;
+//when running at 450rpm
+R = -(k_phi*rpm-voltage)/arm_current;
+R = R - arm_res;
+disp("To decrease the speed to 450 rev/min");
+disp(R,"the resistance added with the armature is (in ohm)");
+disp(current,"the current is (in A)");
+disp(input_pow,"the kw-input taken from the supply is(in kW)");
+//solving b sub part
+disp("To increase the speed to 700 rev/min");
+flux_ratio = 600/700;
+res_added = (fl_win_res/flux_ratio)- fl_win_res;
+disp(res_added,"the resistance to be added is (in ohm)");
+arm_current = arm_current*(1/flux_ratio);
+fld_current = 1.25;
+tot_current = arm_current + fld_current;
+pow = tot_current * voltage/1000;
+disp(arm_current,"the armature current is (in A)");
+disp(fld_current,"the field current is (in A)");
+disp(tot_current,"the total current is (in A)");
+disp(pow,"the kw-input taken from the supply is(in kW)");
+//the resistance value is rounded off in text book so armature current, total current , input power vary slightly with text book