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

Options

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

Answer

A

Explanation

Current amplification factor `beta=(DeltaI_(C))/(DeltaI_(B))` Collector resistance
`DeltaI_(C)=(2V)/(2xx10^(3)Omega)=1xx10^(-3)A`
Base current `DeltaI_(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=10mV`