In a junction transistor, the collector curre | vedclass

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(B) Given: Change in collector current $\Delta I_C = 6.8 \,mA$ and change in emitter current $\Delta I_E = 7 \,mA$. Since $\Delta I_E = \Delta I_C + \

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$34$

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(B) Given: Change in collector current $\Delta I_C = 6.8 \,mA$ and change in emitter current $\Delta I_E = 7 \,mA$. Since $\Delta I_E = \Delta I_C + \Delta I_B$, the change in base current is $\Delta I_B = \Delta I_E - \Delta I_C = 7 \,mA - 6.8 \,mA = 0.2 \,mA$. The current amplification factor $\beta$ for a common-emitter configuration is defined as $\beta = \frac{\Delta I_C}{\Delta I_B}$. Substituting the values: $\beta = \frac{6.8 \,mA}{0.2 \,mA} = 34$. Thus, the current amplification factor is $34$.

In a junction transistor, the collector current changes by $6.8 \,mA$, if the emitter current is changed by $7 \,mA$. For such a transistor, the current amplification factor is

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