A `npn` transistor is connected in common emitter configuration in a given amplifier. A load resistance of `800 Omega`is connected in the collector circuit and the voltage drop across `0.96` and the input resistance of the circuit is `192 Omega`, the voltage gain and the power gain of the amplifier will respectively be :

Options

A. 4, 3.84
B. 3.69, 3.84
C. 4, 4
D. 4, 3.69

Answer

A

Explanation

Here, `R_(L)=800Omega,V_(L)=0.8 V`,
`R_(i)=192Omega, beta=0.96`.
`I_(C)=V_(L)/R_(L)=0.8/800=10^(-3)A`
Current amplication
`beta=(I_(c))/(I_(b))=0.96` or `I_B=I_c/0.96=10^(-3)/0.96`
Voltage gain,
`A_(V)=V_(L)/V_(i)=V_(L)/(I_(R_(i)))=0.8/((10^(-3)//0.96)xx192)=4`
Power gain,
`A_(P)=(I_c^(2)R_(L))/(I_B^(2)R_(i))=((10^(-3))^(2)xx800)/((10^(-3)//0.96)^(2)xx192)=3.84`