For a transistor,alpha_{dc} and beta_{dc} are | vedclass

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(A) We know the relationship between the current gain parameters $\alpha_{dc}$ and $\beta_{dc}$ is given by $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alph

Vedclass · Accepted Answer

$1$

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(A) We know the relationship between the current gain parameters $\alpha_{dc}$ and $\beta_{dc}$ is given by $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$.Rearranging this equation,we get $1 - \alpha_{dc} = \frac{\alpha_{dc}}{\beta_{dc}}$.Now,we need to evaluate the expression $\frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.This can be written as $\frac{\beta_{dc}}{\alpha_{dc} \cdot \beta_{dc}} - \frac{\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}} = \frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}}$.Substituting $\frac{1}{\alpha_{dc}} = \frac{1 + \beta_{dc}}{\beta_{dc}}$ is not necessary here; instead,use $\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = \frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.From the relation $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$,we have $\frac{1}{\beta_{dc}} = \frac{1 - \alpha_{dc}}{\alpha_{dc}} = \frac{1}{\alpha_{dc}} - 1$.Therefore,$\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = 1$.Thus,the value of the expression is $1$.

For a transistor,$\alpha_{dc}$ and $\beta_{dc}$ are the current ratios. Then the value of $\frac{\beta_{dc}-\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$ is

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