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In an NPN transistor the collector current is 24 mA. If $80 \%$ of electrons reach collector its base current in mA is
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6
Given $i_c=\frac{80}{100} \times i_e \Rightarrow 24=\frac{80}{100} \times i_e \Rightarrow i_e=30 \mathrm{~mA}$
By using $i_e=i_b+i_c \Rightarrow i_b=30-24=6 \mathrm{~mA}$.
By using $i_e=i_b+i_c \Rightarrow i_b=30-24=6 \mathrm{~mA}$.
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