Question

In a transistor amplifier, the two a.c current gains $\alpha$ and $\beta$ are defined as $\alpha = \delta \dfrac{{{I_C}}}{{{I_E}}}$ and $\beta = \delta \dfrac{{{I_C}}}{{{I_B}}}$, the relation between $\alpha$ and $\beta$ is
(A) $\beta = \dfrac{{1 + \alpha }}{\alpha }$
(B) $\beta = \dfrac{{1 - \alpha }}{\alpha }$
(C) $\beta = \dfrac{\alpha }{{1 - \alpha }}$
(D) $\beta = \dfrac{\alpha }{{1 + \alpha }}$

Answer 1

In this question it is given that the transistor amplifier is used and there is a current flowing in it, so by using the transistor current formula and then by differentiating the transistor current formula, the relation between the $\alpha$ and $\beta$ is determined.

Formula used:
The transistor current formula is given by,
${I_E} = {I_B} + {I_C}$
Where, ${I_E}$ is the current of the transistor in the emitter terminal, ${I_B}$ is the current of the transistor in the base terminal, and ${I_C}$ is the current of the transistor in the collector terminal.

Complete step by step answer:
Given that,
The current gain $\alpha$ is, $\alpha = \delta \dfrac{{{I_C}}}{{{I_E}}}$
The current gain $\beta$ is, $\beta = \delta \dfrac{{{I_C}}}{{{I_B}}}$
Now,
The transistor current formula is given by,
${I_E} = {I_B} + {I_C}\,...................\left( 1 \right)$
By differentiating the above equation (1) with respect to the current in the collector terminal $\Rightarrow {I_C}$, then the above equation is written as,
$\Rightarrow \delta \dfrac{{{I_E}}}{{{I_C}}} = \delta \dfrac{{{I_B}}}{{{I_C}}} + 1$
By substituting the terms which are given in the question, then the above equation is written as,
$\Rightarrow \dfrac{1}{\alpha } = \dfrac{1}{\beta } + 1$
By taking the term $\dfrac{1}{\beta }$ in one side and the other terms in the other side, then the above equation is written as,
$\Rightarrow \dfrac{1}{\beta } = \dfrac{1}{\alpha } - 1$
By cross multiplying the terms in the terms in the RHS, then the above equation is written as,
$\Rightarrow \dfrac{1}{\beta } = \dfrac{{1 - \alpha }}{\alpha }$
By rearranging the terms in the above equation, then the above equation is written as,
$\Rightarrow \beta = \dfrac{\alpha }{{1 - \alpha }}$

The relation between current gains is $\beta = \dfrac{\alpha }{{1 - \alpha }}$. Hence, option (C) is the correct answer.

Note:
The transistor current formula is ${I_E} = {I_B} + {I_C}$ . In general, the transistors are widely used to amplify current and so undergo an examination. It is asked to fill the current or collector current on the gain. Therefore, the transistor current formula can be examined by the base current and switch a resistive load.