A change of `20 muA` in the base current produces a change of 0.5mA in the collector current. Calculate `beta_(ac)`

Explanation

`I_(b)= 100 mu A= 100 xx 10^(-6)A, I_(c)= 3mA = 3 xx 10^(-3)A, beta`= ?, `Delta I_(b)= 20 mu A, I_(c)= ?, Delta I_(c) = 0.5mA, alpha=`?
`beta= (I_(c))/(I_(b))= (3 xx 10^(-3))/(100 xx 10^(-6))=30`
`I_(e) = I_(b) + I_(c)`
`100 xx 10^(-6) +3 xx 10^(-3)`
`=0.1 xx 10^(-3) + 3 xx 10^(-3)`
`I_(e)= (0.1 +3) 10^(-3)= 3.1 xx 10^(-3)A`
`I_(e)= 3.1 mA`
`alpha= (beta)/(1+ beta)= (30)/(1+30)`
`=(30)/(31)= 0.968`
`alpha= 0.97`
`beta_(ac)= (Delta I_(c))/(Delta I_(b))= (0.5 xx 10^(-3))/(20 xx 10^(-6))=25`