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Diffstat (limited to '3311/CH6/EX6.28/Ex6_28.sce')
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diff --git a/3311/CH6/EX6.28/Ex6_28.sce b/3311/CH6/EX6.28/Ex6_28.sce new file mode 100755 index 000000000..8285703d0 --- /dev/null +++ b/3311/CH6/EX6.28/Ex6_28.sce @@ -0,0 +1,24 @@ +// chapter 6 +// example 6.28 +// Compute source inductance, load resistance and overlap angle +// page-371-372 +clear; +clc; +// given +E=400; // in V (supply voltage) +f=50; // in Hz (supply frequency) +alpha=%pi/4; // in radian (firing angle) +Id=10; // in A (load current) +Edc=360; // in V (load voltage) +// calculate +alpha=alpha*(180/%pi); +disp(alpha); +Emph=E*sqrt(2/3);// calculation of peak voltage +// Since Edc=(3*sqrt(3)*Emph/%pi)*cosd(alpha)-(3*2*%pi*f*Ls/%pi)*Id, we get +Ls=(%pi/(3*2*%pi*f*Id))*((3*sqrt(3)*Emph/%pi)*cosd(alpha)-Edc);// calculation of source inductance +R=Edc/Id;// calculation of load resistance +// since Edc=(3*sqrt(3)*Emph/%pi)*cosd(alpha+u)-(3*2*%pi*f*Ls/%pi)*Id, we get +u=acosd((%pi/(3*sqrt(3)*Emph))*(Edc-(3*2*%pi*f*Ls/%pi)*Id))-alpha;// calculation of overlap angle +printf("\nThe source inductance is \t Ls=%.1f mH",Ls*1E3); +printf("\nThe load resistance is \t\t R=%.1f ohm",R); +printf("\nThe overlap angle is \t\t u=%.f degree",u);
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