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(C) Given,current amplification factor $\alpha = 0.8$.For a common emitter configuration,the current gain $\beta$ is related to $\alpha$ by the formul

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$24$

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(C) Given,current amplification factor $\alpha = 0.8$.For a common emitter configuration,the current gain $\beta$ is related to $\alpha$ by the formula $\beta = \frac{\alpha}{1 - \alpha}$.Substituting the value of $\alpha$:$\beta = \frac{0.8}{1 - 0.8} = \frac{0.8}{0.2} = 4$.We know that the current gain $\beta$ is defined as the ratio of the change in collector current $\Delta I_c$ to the change in base current $\Delta I_b$:$\beta = \frac{\Delta I_c}{\Delta I_b}$.Given $\Delta I_b = 6 \, mA$,we can calculate $\Delta I_c$ as:$\Delta I_c = \beta 	imes \Delta I_b = 4 	imes 6 \, mA = 24 \, mA$.Therefore,the change in collector current is $24 \, mA$.

For a transistor,the current amplification factor is $\alpha = 0.8$. The transistor is connected in a common emitter configuration. The change in the collector current when the base current changes by $6 \, mA$ is.....$mA$.

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