question_answer

A NPN transistor is connected in common emitter configuration in which collector supply is 8V and the voltage drop across the load resistance of 800 ohm connected in the collector circuit is 0.8  V.  If current 25 amplification factor is \[\frac{25}{26}\] , then power gain is - (given \[\to \] input resistance = \[200\,\Omega \])

A) 6.93     

B)   3.69

C) 9.63                             

D) None of these

Correct Answer: B

Solution :

\[\beta =\frac{25}{26}\] \[{{R}_{in}}\,\,=\,\,200\,\Omega \] \[\beta =\frac{{{i}_{C}}}{{{i}_{b}}}\,\,=\,\,\frac{25}{26}\] Here, \[{{I}_{C}}{{R}_{L}}\,\,=\,\,0.8\,V\] \[\therefore \,\,\,{{I}_{C}}\,=\,\,\frac{0.8}{{{R}_{L}}}=\frac{0.8}{800}\,\,=\,\,{{10}^{-3}}\,A\,\,=\,\,1\,\,mA\] Power gain \[=\,\,{{\beta }^{2}}\,\,=\,\,\frac{{{R}_{L}}}{{{R}_{in}}}\,\,=\,\,\left( \frac{25}{26} \right){{\,}^{2}}\,\times \,\frac{800}{200}\,\,=\,\,3.69\]