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+// Exa 13.6
+
+clc;
+clear all;
+
+// Given data
+
+K=0.32; // Coupling co efficient
+Op=1;// Output in oz.in.
+
+// Solution
+
+// 1 oz.in.= 1 oz.in. * (1 ft/12 in.) * (1 lb/16 oz) * (1.3561/1 ft lb) = 7.06*10^-3 J ;
+
+Elec_mech= 7.06*10^-3; // Electrical energy converted to mechanical energy(J)
+Ee=Elec_mech/K; // Applied Electrical energy
+printf(' The electrical energy of %.2f mJ must be applied \n',Ee*10^3);
+// The answer mentioned in the book is incorrect