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

Answer

A

Explanation

Current amplification factor `beta=(DeltaI_(C))/(DeltaI_(B))`
Collector resistance `DeltaI_(C)=(2V)/(2xx10^(3) Omega)=1xx10^(-3) A`
Base current `Delta I_(B)=(V_(B))/(R_(B))=(V_(B))/(R_(B))=(V_(B))/(1xx10^(3))=V_(B)xx10^(-3)`
Given, `beta=100`
Now, `100=(10^(-3))/(V_(B)xx10^(-3))`
`V_(B)=1/100 V=10 mV`