In the study of a transistor as an amplifier, | vedclass

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(B) We know that the emitter current is the sum of the collector current and the base current:$I_{E} = I_{C} + I_{B}$Dividing both sides by $I_{C}$:$\

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$\beta = \frac{\alpha}{1 - \alpha}$

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(B) We know that the emitter current is the sum of the collector current and the base current:$I_{E} = I_{C} + I_{B}$Dividing both sides by $I_{C}$:$\frac{I_{E}}{I_{C}} = 1 + \frac{I_{B}}{I_{C}}$Since $\alpha = \frac{I_{C}}{I_{E}}$,it follows that $\frac{1}{\alpha} = \frac{I_{E}}{I_{C}}$.Since $\beta = \frac{I_{C}}{I_{B}}$,it follows that $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$.Substituting these into the equation:$\frac{1}{\alpha} = 1 + \frac{1}{\beta}$Rearranging to solve for $\frac{1}{\beta}$:$\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$,where $I_{C}$,$I_{B}$,and $I_{E}$ are the collector,base,and emitter currents respectively,then:

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