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+clear;
+clc;
+//Example 3.14
+R1=56;
+R2=12.2;
+Rc=2;
+Re=.4;
+Vcc=10;
+Vbe=0.7;
+b=100;
+//fig.3.53(b)
+Rth=R2*R1/(R1+R2);
+printf('\nThevenin rquivalent resistance=%0.1f KOhm\n',Rth)
+Vth=(R2/(R1+R2))*Vcc;
+printf('\nThevenin equivalent voltage=%0.2f V\n',Vth)
+Ibq=(Vth-Vbe)/(Rth+(1+b)*Re);
+printf('\nbase current=%f mA\n',Ibq)
+Icq=b*Ibq;
+printf('\ncollector current=%.3f mA\n',Icq)
+Ieq=(1+b)*Ibq;
+printf('\nemitter current=%.2f mA\n',Ieq)
+Vceq=Vcc-Icq*Rc-Ieq*Re;
+printf('\ncollector emitter voltage=%.3f V\n',Vceq)
+b=[50,100,150]
+for x=b
+Ibq=(Vth-Vbe)/(Rth+(1+x)*Re);
+disp("Ibeq,Iceq,Ieq,Vceq")
+disp(Ibq)
+Icq=x*Ibq;
+disp(Icq)
+Ieq=(1+x)*Ibq;
+disp(Ieq)
+Vceq=Vcc-Icq*Rc-Ieq*Re;
+disp(Vceq)
+disp("")
+end