In an 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,we have the relation

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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,we have the relation: $i_c = 0.80 	imes i_e$. Substituting the value of $i_c$: $24 = 0.80 	imes i_e$. Solving for the emitter current: $i_e = \frac{24}{0.80} = 30 \, mA$. Using the Kirchhoff's current law for a transistor: $i_e = i_b + i_c$. Rearranging to find the base current: $i_b = i_e - i_c$. Substituting the known values: $i_b = 30 \, mA - 24 \, mA = 6 \, mA$. Thus,the base current is $6 \, mA$.

In an $NPN$ transistor,the collector current is $24 \, mA$. If $80\%$ of electrons reach the collector,its base current in $mA$ is:

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