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What is the emitter current $I_E$ in a transfer where base current $I_B$ is $20 \mu A$ and $\beta =40$

$(a)\;1\;mA \\ (b)\;0.52\;mA \\ (c)\;1.82\;mA \\ (d)\;0.82\; mA $
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We know
$\beta =\large\frac{I_c}{I_B}$
$I_c= 40 \times 20 \mu A$
$\qquad= 800 \mu A$
$\qquad= 0.8 mA$
$I_B= 0.02 mA$
$\therefore I_E =I_B+I_C$
$\qquad= 0.02 mA+ 0.8 mA$
$\qquad= 0.82 mA$
Hence d is the correct answer.
answered Jan 28, 2014 by meena.p

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