For a Common-Emitter amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100. The base resistance is 1 kΩ. The input signal voltage will be:

0.01 V

The correct answer is Option (4) → 0.01 V

Given,

$R_C$ Resistance = $2kΩ=2×10^3Ω$ ($1kΩ=1000Ω$)

$V_o$, Audio signal voltage = $2V$

$β$ (current amplification factor) = 100

$R_b$ (base resistance) = $1kΩ=1×10^3Ω$

$I_B=\frac{V_o}{β×R_c}$

where, $I_B$ = Base current

$I_B=\frac{V_o}{β×R_c}=\frac{2}{100×2000}=10^{-5}A$

$V_i=I_B×R_b=10^{-5}×10^3$

$=0.01V$