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A transistor having a $\beta$ equal to 80 has a change in base current of $250 \mu \mathrm{A}$, then the change in collector current is
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$20 \mathrm{~mA}$
$\beta=\left(\frac{\Delta i_C}{\Delta i_B}\right)$
$80=\frac{\Delta i_C}{250 \times 10^{-6}}$
$\Delta i_C=80 \times 250 \times 10^{-6}$
$\Delta i_C=20000 \times 10^{-6}$
$=20 \times 10^{-3}=20 \mathrm{~mA}$
$80=\frac{\Delta i_C}{250 \times 10^{-6}}$
$\Delta i_C=80 \times 250 \times 10^{-6}$
$\Delta i_C=20000 \times 10^{-6}$
$=20 \times 10^{-3}=20 \mathrm{~mA}$
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