In a NPN transistor,the collector current is | vedclass

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(D) Given that the collector current $i_{c} = 24 \, mA$.Since $80\%$ of the electrons emitted from the emitter reach the collector,the collector curre

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$6$

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(D) Given that the collector current $i_{c} = 24 \, mA$.Since $80\%$ of the electrons emitted from the emitter reach the collector,the collector current is $80\%$ of the emitter current $(i_{e})$.So,$i_{c} = 0.80 	imes i_{e}$.Substituting the value of $i_{c}$,we get $24 = 0.80 	imes i_{e}$.Therefore,$i_{e} = \frac{24}{0.80} = 30 \, mA$.Using the relation for transistor currents,$i_{e} = i_{b} + i_{c}$.Substituting the known values,$30 = i_{b} + 24$.Thus,the base current $i_{b} = 30 - 24 = 6 \, mA$.

In a $NPN$ transistor,the collector current is $24 \, mA$. If $80\%$ of electrons reach the collector,what is the base current in $mA$?

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