The current transfer ratio `beta` of a transistor is 50. The input resistance of the transistor when used in common emitter mode is 1 kilo ohm. The peak value of the collector alternating current for an input peak voltage of 0.01 volt is

Options

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

Answer

D

Explanation

Here `R_(i)=10^(3) Omega, beta=50, V_(i)=0.01 V`,
`I_(b)=V_(i)/R_(i)=0.01/10^(3)=10^(-5)A`
As `I_(C)=betaxxI_(b)`
`=50xx10^(-5)=500xx10^(-6)A=500 muA`