Question
For a transistor, the current amplification factor is $0.8.$ The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by $6\, mA$ is .....$mA$
✓
Answer
$\alpha=0.8 \Rightarrow \beta=\frac{0.8}{(1-0.8)}=4$
Also $\beta=\frac{\Delta \mathrm{i}_{\mathrm{c}}}{\Delta \mathrm{i}_{\mathrm{b}}} $
$\Rightarrow \Delta \mathrm{i}_{\mathrm{c}}=\beta \times \Delta \mathrm{i}_{\mathrm{b}}=4 \times 6=24 \mathrm{\,mA}$
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