In a single state transistor amplifier,When the signal changes by 0.02 V,the base current change by 10 `muA` and collector current by 1mA.If collector load `R_(C)=2 kOmega` and `R_(L)=10kOmega` ,Calculate ,(i)Current Gain (ii)Input impedance ,(iii)Effective a.c load (iv)Voltage gain and (v)Power gain.

Explanation

(i)Current Gain `beta=(deltai_(c))/(deltai_(h))=(1mA)/(10muA)=100`
(ii)Input impedance
`R_(i)=(deltaV_(BE))/(deltai_(b))=(0.02)/(10muA)=2000Omega=2kOmega`
(iii)Effctive (a.c)load
`R_(AC)=R_(C)||R_(L)=(2xx10)/(2+10)=1.66 kOmega`
(iv)Voltage gain `A_(v)=betaxx(R_(AC))/(R_(in))=(100xx1.66)/(2)=83`
(v)Power gain ,Ap=Current gain `xx` Voltage gain
`=100x83=8300`