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+// Example 12.4
+// Computation of (a) Required resistance of a noninductive diverter that will 
+// bypass 27 percent of the total armature current(b) Power rating of the 
+// diverter
+// Page No. 494
+
+clc;
+clear;
+close;
+
+// Given data
+Rs=0.00306;               // Shunt generator resistance rating
+Is=0.73;                  // Shunt generator current rating
+Id1=0.27;                 // Armature winding resistance
+Pload=170000;             // Load of power
+VT=250;                   // Shunt generator voltage rating
+Id2=680;                  // No load voltage
+Rd=0.27;                  // Resistance drop
+
+// (a) Required resistance of a noninductive diverter that will bypass 
+// 27 percent of the total armature current
+Rd=Rs*Is/Id1;
+
+
+// (b) Power rating of the diverter
+Ia=Pload/VT; 
+Pd=((Id1*Id2)^2)*Rd;
+
+
+
+//Display result on command window
+printf("\n Required resistance of a noninductive diverter = %0.5f Ohm ",Rd);
+printf("\n Power rating of the diverter = %0.0f W ",Pd);