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+//example6.27
+clc
+disp("From the fig 6.50 we can write,")
+disp("V_CC=20V and R_L=12 ohm")
+disp("i) The maximum ac power that can be delivered to the load is,")
+p=(20^2)/24
+disp(p,"(P_ac)_max[in W]= ")
+disp("Let new power delivered to load be (P_ac)''.")
+disp("The corresponding new supply voltage be (V''_cc)")
+disp("(P_ac)''[in W]=1.36(P_ac)_max ..36% more")
+p=1.36*16.67
+disp(p,"= 1.36*16.67=")
+disp("And (P_ac)''=(V''_cc^2)/R_L")
+disp("Therefore 22.67=(V''_cc^2)/(2*12)")
+v=sqrt(544.1088)
+disp(v,"V''_cc(in V)=")
+disp("Hence the percentage increase in supply voltage is,")
+p=(23.326-20)/0.2
+disp(p,"= ((V''_cc-V_cc)*100)/V_cc= ")
+disp("The mimimum breakdown voltage per transistor this condition is,")
+v=2*23.326
+disp(v,"=2*V''_cc=2*23.326=")