If alpha and beta are the current gains in th | vedclass

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(B) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.First,calculate the numerator

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$1$

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(B) We know the relationship between current gains $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.First,calculate the numerator $(\beta - \alpha)$:$\beta - \alpha = \frac{\alpha}{1 - \alpha} - \alpha = \frac{\alpha - \alpha(1 - \alpha)}{1 - \alpha} = \frac{\alpha - \alpha + \alpha^2}{1 - \alpha} = \frac{\alpha^2}{1 - \alpha}$.Next,calculate the denominator $(\alpha \beta)$:$\alpha \beta = \alpha 	imes \left( \frac{\alpha}{1 - \alpha} ight) = \frac{\alpha^2}{1 - \alpha}$.Finally,divide the numerator by the denominator:$\frac{\beta - \alpha}{\alpha \beta} = \frac{\frac{\alpha^2}{1 - \alpha}}{\frac{\alpha^2}{1 - \alpha}} = 1$.

If $\alpha$ and $\beta$ are the current gains in the $CB$ and $CE$ configurations respectively of the transistor circuit,then $\frac{\beta - \alpha}{\alpha \beta} = $

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