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Q.
For an $n-p-n$, transistor circuit, the collector current is $10\, mA$. If $90 \%$ of the electrons emitted reach the collector then :
AFMCAFMC 2002
Solution:
Emitter current is sum of base current and collector current.
We know that amplification factor $\alpha=\frac{\Delta I_{C}}{\Delta I_{E}}$
$\therefore \Delta I_{E}=\frac{\Delta I_{C}}{\alpha}=\frac{10}{0.9}=11\, mA$
Also, $\Delta I_{E}=\Delta I_{B}+\Delta I_{C}$
$\Rightarrow \Delta I_{B} =\Delta I_{E}-\Delta I_{C}$
$=11-10$
$=1\, mA$ Alternative : The collector current
$I_{C}=10\, mA$
As given,
$90 \%$ of emitter current = collector current
or $\frac{90}{100} I_{E} =I_{C}$
$I_{E} =\frac{100}{90} I_{C}$
$=\frac{100}{90} \times 10$
$\simeq 11\,mA$
Thus, $I_{B} =I_{E}-I_{C}$
$=11-10=1\, mA$