A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800Ω is connected in the collector circuit and the voltage drop across it is 0.8V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192Ω, the voltage gain and the power gain of the amplifier will respectively be?

Transistor is working in common emitter configuration. Hence, input terminal is at base and output terminal is at collector.
Given:Rc=800Ω,VL=0.8V,RB=192Ω
Current amplification factor =0.96
Hence,Ic/IB=0.96
β=0.96
Assumption:Transistor is ideal and working in forward active region (Junction BE is forward biased and Junction BC is reverse biased).
→VBE=0
VE=0
→VB=0
Collector current is given by
IC=VL/Rc=0.8/800=10⁻³A
Base current is given by,
IB=IC/β=10⁻³/0.96A
Applying KVL on Base circuit, input voltage
VBB=IBRB
VBB=10⁻³/0.96×192=0.2V
Voltage Gain,Av=Vout/Vin=VL/VBB=0.8/0.2=4
Current Gain,AI=Iout/Iin=IC/IB=β=0.96
Power Gain,AP=AVAI=4×0.96=3.84