A non transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is `100 Omega` and the output load resistance is `10 k Omega`. The common emitter current gain `beta` is:

Options

A. `10^(4)`
B. 60
C. `10^(2)`
D. `6 xx 10^(2)`

Answer

C

Explanation

`10 log_(10)(P_("out"))/(P_(in)) = 60 rArr P_("out")/P_("in") =10^(6)`.
`P_("out")= Delta V_("out") .DeltaI_(C), P_("in") =DeltaV_(in). DeltaI_(g)`
`therefore V_("out") = I_(C). R_("out") rArr DeltaV_("out")=R_("out") .DeltaI_( C) therefore V_("in")= I_(B).R_("in") rArr Delta V_(in) =R_(in). DeltaI_(s)`
`R_("out")/R_("in") .B^(2) =10^(6), beta =100`