Modern Physics 6 Question 32
37. In an $n-p-n$ transistor circuit, the collector current is $10 mA$. If $90 %$ of the electrons emitted reach the collector
$(1992,2 M)$
(a) the emitter current will be $9 mA$
(b) the emitter current will be $11 mA$
(c) the base current will be $1 mA$
(d) the base current will be $-1 mA$
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Solution:
- Given : $i _c=10 mA=(0.9) i _e \quad$ [Given that $i _c$ is $90 %$ of $i _e$ ]
$$ \begin{array}{ll} \therefore & i _e=\frac{10}{0.9} mA \approx 11 mA \\ \text { and } & i _b=i _e-i _c=(11-10) mA=1 mA \end{array} $$
