A change of 8.0 ,mA in the emitter current br | vedclass

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(D) Given that, the change in emitter current is $\Delta I_{E} = 8.0 \,mA$. The change in collector current is $\Delta I_{C} = 7.9 \,mA$. We know that

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$0.99, 79$

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(D) Given that, the change in emitter current is $\Delta I_{E} = 8.0 \,mA$. The change in collector current is $\Delta I_{C} = 7.9 \,mA$. We know that the current gain $\alpha$ is defined as the ratio of change in collector current to the change in emitter current: $\alpha = \frac{\Delta I_{C}}{\Delta I_{E}} = \frac{7.9}{8.0} = 0.9875 \approx 0.99$. We also know the relationship between $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$. Alternatively, using the currents directly: $\beta = \frac{\Delta I_{C}}{\Delta I_{B}}$. Since $\Delta I_{E} = \Delta I_{C} + \Delta I_{B}$, we have $\Delta I_{B} = \Delta I_{E} - \Delta I_{C} = 8.0 \,mA - 7.9 \,mA = 0.1 \,mA$. Therefore, $\beta = \frac{7.9 \,mA}{0.1 \,mA} = 79$. Thus, the values are $\alpha = 0.99$ and $\beta = 79$.

$A$ change of $8.0 \,mA$ in the emitter current brings a change of $7.9 \,mA$ in the collector current. The values of $\alpha$ and $\beta$ are

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