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For a transistor amplifier in common emi...

For a transistor amplifier in common emitter configuration for load impedance of `1 k Omega. (h_(fe) = 50 and h_(oe) = 25 xx 10^(-6))` the current gain is

A

`-5.2`

B

`-15.7`

C

`-28.8`

D

`-48.76`

Text Solution

Verified by Experts

The correct Answer is:
D
In common emitter configuration current gain
`A_(i)=(-h_(fe))/(1+h_(ie)R_(L))=(-50)/(1+25xx10^(-6)xx10^(3))=-48.78`
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