The transfer ration of a transistor is `50`. The input resistance of the transistor when used in the common -emitter configuration is `1 kOmega`. The peak value for an `A.C.` input voltage of `0.01 V` peak is

Options

A. `100muA`
B. `0.01mA`
C. `0.25mA`
D. `500muA`

Answer

D

Explanation

`beta=50,R_(i)=1000 Omega,v_(i)=0.01V`
`v_(i)=DeltaI_(B)R_(i)impliesDeltaI_(B)=(0.01)/(1000)=10^(-5)A`
`beta=(DeltaI_(C ))/(DeltaI_(B))impliesDeltaI_(C )=betaDeltaI_(B)=50xx10^(-5)A=500muA`