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

Answer

D

Explanation

Here, `R_(C )=2kOmega=2xx10^(3)Omega`
`V_(o)=2V`
`R_(B)=1kOmega =1xx10^(3)Omega`
`beta=100`
Output voltage, `V_(o)=I_(C )R_(C )`
`or I_(C )=(V_(o))/(R_(C ))=(2V)/(2xx10^(3)Omega)=10^(-3)A=1mA`
As `beta=(I_(C ))/(I_(B))or I_(B)=(I_(C ))/(beta)`
`I_(B)=(10^(-3)A)/(100)=10^(-5)A`
Input voltage,
`V_(i)=I_(B)R_(B)=(10^(-5)A)(1xx10^(3)Omega)=10^(-2)V=10mV`