In a transistor circuit,the collector current | vedclass

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(A) Given:Change in collector current,$\Delta I_C = 8.9 \ mA$Change in emitter current,$\Delta I_E = 9.0 \ mA$We know that the emitter current is the

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$89$

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(A) Given:Change in collector current,$\Delta I_C = 8.9 \ mA$Change in emitter current,$\Delta I_E = 9.0 \ mA$We know that the emitter current is the sum of collector current and base current: $\Delta I_E = \Delta I_C + \Delta I_B$Therefore,the change in base current is: $\Delta I_B = \Delta I_E - \Delta I_C = 9.0 \ mA - 8.9 \ mA = 0.1 \ mA$The current amplification factor $\beta$ is defined as the ratio of change in collector current to the change in base current:$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{8.9 \ mA}{0.1 \ mA} = 89$

In a transistor circuit,the collector current is changed by $8.9 \ mA$,if the emitter current is changed by $9.0 \ mA$. The value of current amplification factor,$\beta$ is

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