In a common emitter transistor amplifier circuit, the current gain $\alpha=0.8$. If the base current changes by $3 \mu \mathrm{A}$ then the collector current changes by

The common emitter current gain $\beta$ relates to the common base current gain $\alpha = 0.8$ through the formula $\beta = \frac{\alpha}{1 - \alpha} = \frac{0.8}{0.2} = 4$.

For a transistor in common emitter configuration, the change in collector current relates to the base current change as $\Delta I_C = \beta \Delta I_B$.

With $\Delta I_B = 3 \mu A$, the collector current changes by $\Delta I_C = 4 \times 3 \mu A = 12 \mu A$.

Final answer: $12 \mu A$