The $β$ of a transistor is $74$ . It is connected in common base configuration. If the emitter current is $5 m A$ , the collector current is (approximately)

0.45 m A

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4.93 m A

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370 m A

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5 m A

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Solution

The correct option is B $4.93 m A$
Given that:
$β = 74$ ,
$I_{e} = 5 m A$
we have
$I_{e} = I_{b} + I_{c}$ .............(1)
$\frac{I_{e}}{I_{b}} = 1 + \frac{I_{c}}{I_{b}}$
$\frac{I_{e}}{I_{b}} = 1 + β$ ...........( $∵ β = \frac{I_{c}}{I_{b}}$ )
$\frac{5}{I_{b}} = 1 + 74 = 75$
$I_{b} = \frac{5}{75} = 0.066$
Hence, from (1)
$I_{c} = I_{e} - I_{b} = 5 - 0.066 = 4.934 m A$

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