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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Exa 6.14
+clc;
+clear;
+close;
+// Given data
+L1= 2;// in mH
+L1= L1*10^-3;// in H
+L2= 1.5;// in mH
+L2= L2*10^-3;// in H
+// Formula f= 1/(2*%pi*sqrt((L1+L2)*C)
+// For f= 1000 kHz, C will be maximum
+f=1000;// in kHz
+f=f*10^3;// in Hz
+Cmax= 1/((2*%pi*f)^2*(L1+L2));// in F
+// For f= 2000 kHz, C will be maximum
+f=2000;// in kHz
+f=f*10^3;// in Hz
+Cmin= 1/((2*%pi*f)^2*(L1+L2));// in F
+disp(Cmin*10^12,"Minimum Capacitance in pF is : ")
+disp(Cmax*10^12,"Maximum Capacitance in pF is : ")