In n-p-n transistor circuit the collector current is $10 m A$ . If $90 %$ of the electron reach the collector, then emitter current will be

1 m A

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0.1 m A

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2 m A

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nearly

11 m A

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Solution

The correct option is D nearly $11 m A$
$I_{c} = 10 m A$ ; $I_{e} = I_{c} + I_{b}$
If $I_{c} = 90 %$ of $I_{e}$
then $I_{c} = \frac{90}{100} \times I_{e}$
$∴$ $I_{e} = \frac{100}{90} I_{c}$

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