In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`, then the peak value of output voltage is

Options

A. `5xx10^(-6)` volt
B. `2.5xx10^(-4)`
C. `1.25` volt
D. `125` volt

Answer

B

Explanation

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