In a CE transistor amplifier, the audio signal voltage across the...

Current amplification factor,

$\beta = \dfrac{\Delta{I}_{C}}{\Delta{I}_{B}}$

Collector current,

$\Delta{I}_{C} = \dfrac{2V}{2 \times {10}^{3}Ω} = 1 \times {10}^{- 3}A$

Base current,

$\Delta{I}_{B} = \dfrac{{V}_{B}}{{R}_{B}} = \dfrac{{V}_{B}}{1 \times {10}^{3}} = {V}_{B} \times {10}^{- 3}$

Given, $\beta = 100$

Now, $100 = \dfrac{{10}^{- 3}}{{V}_{B} \times {10}^{- 3}}$

$⇒ {V}_{B} = \dfrac{1}{100}V = 10mV$