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In an n-p-n transistor only $3 \%$ of the charge carriers are allerved to recombine in the base region. The value of $ \alpha$ is

$(a)\;0.03 \\ (b)\;93 \\ (c)\;0.97 \\ (d)\;0.07 $
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Let $I_E$ be emitter current.
$I_B= 3 \% $ of $I_E$
$\qquad= 0.03\;I_E$
$I_C= I_E -I_B$
$\qquad= I_E - 0.03\; I_E$
$\qquad= 0.97\;IE$
$\alpha =\large\frac{I_c}{I_E}$
$\qquad= \large\frac {0.97 I_E}{I_E}$
$\qquad= 0.97$
hence c is the correct answer.
answered Jan 28, 2014 by meena.p

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