The correct relation between alpha (ratio of | vedclass

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(D) The current gain $\alpha$ is defined as the ratio of collector current $(I_{C})$ to emitter current $(I_{E})$: $\alpha = \frac{I_{C}}{I_{E}}$.The

Vedclass · Accepted Answer

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

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(D) The current gain $\alpha$ is defined as the ratio of collector current $(I_{C})$ to emitter current $(I_{E})$: $\alpha = \frac{I_{C}}{I_{E}}$.The current gain $\beta$ is defined as the ratio of collector current $(I_{C})$ to base current $(I_{B})$: $\beta = \frac{I_{C}}{I_{B}}$.From Kirchhoff's current law for a transistor,the emitter current is the sum of base and collector currents: $I_{E} = I_{B} + I_{C}$.Substituting $I_{E}$ in the expression for $\alpha$: $\alpha = \frac{I_{C}}{I_{B} + I_{C}}$.Dividing the numerator and denominator by $I_{C}$: $\alpha = \frac{1}{\frac{I_{B}}{I_{C}} + 1}$.Since $\frac{1}{\beta} = \frac{I_{B}}{I_{C}}$,we substitute this into the equation: $\alpha = \frac{1}{\frac{1}{\beta} + 1} = \frac{1}{\frac{1+\beta}{\beta}} = \frac{\beta}{1+\beta}$.Therefore,the correct relation is $\alpha = \frac{\beta}{1+\beta}$.

The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is :

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