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

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(D) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.Alternatively,we can use the f

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(D) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.Alternatively,we can use the fundamental current relation $I_E = I_C + I_B$.Dividing by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.This implies $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.Rearranging the terms,we get $\frac{1}{\alpha} - \frac{1}{\beta} = 1$.Substituting the given values: $\frac{1}{0.98} - \frac{1}{49} \approx 1.0204 - 0.0204 = 1$.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$,where $I_C, I_B,$ and $I_E$ are collector,base,and emitter currents respectively,then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to:

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