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+//Tested on Windows 7 Ultimate 32-bit
+//Chapter 8 Power Amplifiers Pg no. 277,278 and 279
+clear;
+clc;
+
+//Given Data
+//Figure 8.13
+
+VCC=20;//collector supply voltage in volts
+RC=270;//collector resistance in ohms
+RE=150;//emitter resistance in ohms
+R1=3.3D3;//divider network resistance R1 in ohms
+R2=1.5D3;//divider network resistance R2 in ohms
+VBE=0.7;//forward voltage drop of emitter diode in volts
+B=100;//DC CE current gain beta
+RL=470;//load resistance in ohms
+C1=15D-6;//input coupling capacitance in farads
+C2=15D-6;//output coupling capacitance in farads
+
+//Solution
+
+VB=VCC*R2/(R1+R2);//base to ground voltage in volts
+VE=VB-VBE;//emitter to ground voltage in volts
+IE=VE/RE;//emitter current in amperes
+ICQ=IE;//neglecting base current, collector current is equal to emitter current in amperes
+VC=VCC-ICQ*RC;//collector to ground voltage in volts
+VCEQ=VC-VE;//collector to emitter quiscent voltage in volts
+PD=VCEQ*ICQ;//power dissipation in watts
+RL_dash=RC*RL/(RC+RL);//equivalent a.c. load resistance in ohms
+IC_sat=ICQ+VCEQ/RL_dash;//saturation collector current in amperes
+VCE_cutoff=VCEQ+ICQ*RL_dash;//cutoff collector to emitter voltage in volts
+Pout=0.5*ICQ^2*RL_dash;//output a.c. power in watts
+e=Pout/VCC/ICQ;//efficiency of circuit = Pout/Pin(dc)
+printf("(a) The minimum transistor power rating required PD = %.3f Watts\n ",PD);
+printf("(b) AC output power Pout = %d milli-Watts\n ",Pout*10^3);
+printf("(c) Efficiency of the amplifier η = %.2f\n ",e);
+
+//decimal approximation taken here in efficiency