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clear
//Given
c1=0.5 //micro F
c2=0.3 //micro F
c3=0.2 //micro F
//Calculation
Cp=c1+c2+c3
Cs=(1/c1)+(1/c2)+(1/c3)
//Result
printf("\n The ratio ofmaximum capacitance to minimum capacitance is %0.1f ",Cs)
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