Correct Answer – A
`R_(0)=5000 Omega , R_(i)=2000, V_(i)=10 mV, beta=50, V_(0)=?`
`A_(v)=(V_(0))/(V_(i))=beta*(R_(0))/(R_(i))`
`V_(0)=betaV_(i)(R_(0))/(R_(i))`
`=50xx10xx10^(-3)xx(5000)/(2000)`
`=50xx10xx2.5xx10^(-3)`
`=1250xx10^(-3)V`
`=1.25 V`

Correct Answer – B
In common emitter mode, the transistor is current amplifier
`DeltaI_(B)=(10xx10^(-3))/2000=5xx10^(-6) A`
Again, `beta=(DeltaI_(C))/(DeltaI_(B))` or `DeltaI_(C)=betaxxDeltaI_(B)`
`=50xx5xx10^(-6)A =250xx10^(-6)A`
`=2.5xx10^(-4) A`
Peak value of output voltage
`=2.5xx10^(-4)xx5000` volt=`1.25` volt

Correct Answer – C
`A_(v)=(Delta V_(0))/(DeltaV_(i))=beta(R_(0))/(R_(i)) and DeltaV_(0)=DeltaV_(i)xxbeta(R_(0))/(R_(i))`
`therefore DeltaV_(0)= 10xx50 xx(500)/(2000)`
`DeltaV_(0)=1250mV=1.25V`

Correct Answer – A
Voltage gain `=(Output Vol tag e)/(Input vol tag e)=1.25/(10xx10^(-3))=125`
Power gain =voltage gainxcurrent gain
`125xx50`