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

Answer

D

Explanation

`beta=50, R_(i)=1000 Omega, V_(i)=0.01 V`
`beta=(i_(c))/(i_(b))` and `ib=(V_(i))/(R_(i))=0.01/(10^(3)) =10^(-5) A`
Hence `i_(c)=50xx10^(-5) A=500 muA`