The coIIector current in an n-p-n transistor is 10 mA. If 99. 5% of the emitted eIectrons reach the coIIector, determine the emitter current, base current and ampIification factor of the transistor.

Explanation

In an n-p-n transistor we know collector current =`I_(C)`=10mA emitter current =`I_(E)`, base current =`I_(B)`and amplification factor , `beta=I_(C)/(I_(B))`.
According to the problem `I_(C)=99.5% "of" I_(E)`
`:." "I_(C)=(995)/(1000)I_(E) " ""or",I_(E)=(1000)/(995)xx10=10.05 `mA
`:." "I_(B)=I_(E)-I_(C)=(10.05-10)`mA=0.05 mA
`:." "beta=(I_(C))/(I_(B))=(10)/(0.05)=200`