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

Answer

D

Explanation

`V_(0)=I_(C)R_(C)=2`
`I_(C)=(2)/(2xx10^(3))=10^(-3)AmP`
Current gain `=(I_(C))/(I_(B))=100`
`I_(B)=(I_(C))/(100)=(10^(-3))/(100)=10^(-5)AmP`
`V_(i)=R_(B)I_(B)=1xx10^(3)xx10^(-5)=10^(-2)`volt
`=10mV`