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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 8.17
+clc;
+clear;
+close;
+//given data
+C=0.1;// in nF
+C=C*10^-9;//in F
+V=5;//in V
+t=1;// in micro S
+t=t*10^-6;// in sec
+// v= V*(1-%e^(-t/(R*C)))
+// Since hold value does not drop by more than 0.5% or by 0.005 V, hold value is 0.995 V, Thus
+// 0.995*V= V*(1-%e^(-t/(R*C)))
+// or %e^(-t/(R*C))= 1-0.995 = 0.005
+R= t/(C*log(1/0.005));// in ohm
+I= V/R*(1-%e^(-t/(R*C)));// Maximum currnet through R in amphere
+disp(I*10^3,"Maximum permissible leakage current through the hold capacitor in mA")