In an n-p-n transistor, the base and collector currents are $100 m u A$ and $9 m A$ respectively. Then the emitter current will be

9.1 m A

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18.2 m A

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3.91 m u A

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18.2 m u A

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Solution

The correct option is A $9.1 m A$
Given that:
$I_{b} = 100 μ A = 0.1 m A$
$I_{c} = 9 m A$
We know,
$I_{e} = I_{b} + I_{c} = 0.1 + 9 = 9.1 m A$

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