In a `CE` transistor amplifier, the audio signal voltage across the collector resistance of `2 k Omega` is `2 V`. If the base resistance is `1 k Omega` and the current amplification of the transistor is `100`, the input signal voltage is

Options

A. `0.1V`
B. `1.0V`
C. `1mV`
D. `10mV`

Answer

D

Explanation

`V_(o)=DeltaI_(c )R_(L)`
`2=DeltaI_(c )xx2xx10^(3)impliesDeltaI_(c )=10^(-3)A`
`beta=(DeltaI_(c ))/(DeltaI_(B))`
`100=(10^(-3))/(deltaI_(B))impliesDeltaI_(B)=10^(-5)A`
`v_(i)deltai_(B)R_(B)=10^(-5)xx10^930=0.01V=10mV`