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The transfer ration of a transistor is ...

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

A

`100 mu A`

B

`0.01 mA`

C

`0.25 mA`

D

`500 mu A`

Text Solution

Verified by Experts

The correct Answer is:
D
`i_(b)=(V_(i))/(R_(i))=(0.01)/(10^(3))=10^(-5)A=10muA`
`i_(C)=betai_(p)=50xx10xx10^(-6)=500muA`
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