For a transistor amplifier in common emitter configuration for load impedance of 1kΩ ($h_{fe} = 50$ and $h_{oe}= 25mAV^{-1}$), the current gain is:
Answer & explanation
Correct answer: option 4
For a transistor amplifier in common emitter configuration, current gain
$A_i=\frac{h_{fe}}{1+h_{oe}R_L}$
Where $h_{fe}$ and $h_{oe}$ are hybrid parameters of a transistor.
$∴A_i=\frac{50}{1+25×10^{-6}×1×10^3}=48.78$