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+//Chapter-3, Example 3.6, Page 111
+//=============================================================================
+clc
+clear
+//INPUT DATA
+N1=250;//no of turns in a coil
+I1=2;//current in coil in A
+phi1=0.3;//flux in coil in wb
+dt=2//time in  millisec
+Em2=63.75//induced voltage in V
+K=0.75
+//CALCULATIONS
+L1=N1*(phi1/I1);//self inductance of first coil in H
+M=Em2*(dt/I1);//mutual inductance of two coils in H
+L2=((Em2/K)^2)/(L1);//self inductance of second coil in H
+phi2=K*phi1;//flux in second coil in wb
+N2=(Em2*dt)/phi2;//no of turns in second coil
+//OUTPUT
+mprintf("Thus the self inductance of first coil is %2.1f mH \n",L1);
+mprintf("mutual inductance of two coils %2.2f mH \n",M);
+mprintf("self inductance of second coil %4.0f mH \n",L2);
+mprintf("no of turns in second coil %3.0f turns \n",N2);
+//note:the answer given  for N2 in textbook is wrong .please check the calculations
+
+//=================================END OF PROGRAM==============================
+
+