1026/CH6/EX6.18/Example6_18.sce


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//chapter6,Example6_18,pg 148

delIc=2.5*10^-3

delIb=40*10^-6

B=delIc/delIb

alpha=B/(1+B)

printf("alpha=%.5f\n",alpha)

printf("B=%.2f",B)