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//chapter 3
//example 3.13
//page 98
printf("\n")
printf("given")
C1=150*10^-6;C2=C1;vi=4;vo=1;f=120;
Xc2=8.84;//FROM EXAMPLE 3.12
Xl=Xc2*((vi/vo)+1)
L1=Xl/(2*3.14*f);
printf(" suitable value of L1 is %3.3fH\n",L1)
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