The base current is `100 muA` and collector current is 3 mA
(a) Calculate the values of `beta,I_(e)` and `alpha`
(b) A change of `20 muA` in the base current produces a charge of 0.5 mA in the collector current. Caculate `beta_(ac)`

Explanation

`I_(b)=100muA=0.100mA_(1)I_(c)=3mA`
(a) `beta=(I_(c))/(I_(b))=(3)/(0.100)=30,alpha=(beta)/(1+beta)=(30)/(1+30)=(30)/(31)=0.97 and I"_(e) = (I_(c))/(alpha)=(3xx31)/(30)=3.1 mA`
(b) `DeltaI_(b)=20muA = 0.02 mA, DeltaI_(c)=0.5 mA :. beta_(ac)=(DeltaI_(c))/(DeltaI_(b))=(0.5)/(0.02)=25`.