For a transistor,alpha and beta are given as | vedclass

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(B) Given: $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$.We know that the emitter current is the sum of collector current and base

Vedclass · Accepted Answer

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

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(B) Given: $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$.We know that the emitter current is the sum of collector current and base current: $I_{E} = I_{C} + I_{B}$.Substituting $I_{E}$ in the expression for $\alpha$:$\alpha = \frac{I_{C}}{I_{C} + I_{B}}$.Dividing the numerator and denominator by $I_{B}$:$\alpha = \frac{I_{C}/I_{B}}{(I_{C}/I_{B}) + (I_{B}/I_{B})} = \frac{\beta}{\beta + 1}$.Now,rearrange to solve for $\beta$:$\alpha(\beta + 1) = \beta$$\alpha \beta + \alpha = \beta$$\alpha = \beta - \alpha \beta$$\alpha = \beta(1 - \alpha)$$\beta = \frac{\alpha}{1 - \alpha}$.

For a transistor,$\alpha$ and $\beta$ are given as $\alpha = \frac{I_{C}}{I_{E}}$ and $\beta = \frac{I_{C}}{I_{B}}$. Then the correct relation between $\alpha$ and $\beta$ is:

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