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In an n-p-n t...

Question

In an $n-p-n$ transistor circuit, the collector current is $10 mA$ . If $90 %$ of the electrons emitted from the emitter reach the collector, then the emitter current and base currents are

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Solution

Given:

$I_{C} = 10 mA$ and

$\frac{90}{100} I_{E} = I_{C}$

$I_{E} = \frac{10}{0.9} m A = 11.11 mA$

$I_{B} = I_{E} - I_{C} = 11.11 - 10 = 1.11 mA$

Hence, Option $(C)$ is the correct answer.

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