Question

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)
(A) $\alpha = \dfrac{\beta }{{1 + \beta }}$
(B) $\alpha = \dfrac{\beta }{{1 - \alpha }}$
(C) $\beta = \dfrac{1}{{1 - \alpha }}$
(D) $\beta = \dfrac{\alpha }{{1 + \alpha }}$

Answer 1

Hint: In order to solve this question, we will first express the $\alpha $ and $\beta $ in their respective currents ratio, and then using the general equation of emitter, collector, and base current for a transistor, we will derive the relation between $\alpha $ and $\beta $.

Complete step by step solution:
As we know, the ratio of collector current to emitter current of the transistor is denoted by
$\alpha = \dfrac{{{I_C}}}{{{I_E}}}$ where ${I_C},{I_E}$ are collector and emitter current whereas the ratio of collector current to base current is denoted by $\beta = \dfrac{{{I_C}}}{{{I_B}}}$ where ${I_B}$ is the base current and In transistor circuit we know the relation between collector, emitter and base current given by,
${I_E} = {I_C} + {I_B}$

Now, divide this whole equation by ${I_C}$ we get,
$\dfrac{{{I_E}}}{{{I_C}}} = \dfrac{{{I_C}}}{{{I_C}}} + \dfrac{{{I_B}}}{{{I_C}}}$
on putting the values, $\beta = \dfrac{{{I_C}}}{{{I_B}}}$ and $\alpha = \dfrac{{{I_C}}}{{{I_E}}}$ we get
$\dfrac{1}{\alpha } = 1 + \dfrac{1}{\beta }$
on simplifying above equation, we get
$\therefore \alpha = \dfrac{\beta }{{1 + \beta }}$

Hence, the correct option is (A) $\alpha = \dfrac{\beta }{{1 + \beta }}$.

Note: It should be noted that, the relation between $\alpha $ and $\beta $ other than $\alpha = \dfrac{\beta }{{1 + \beta }}$ is also possible in terms of rearranging the coefficients and it became like $\beta = \dfrac{\alpha }{{1 + \alpha }}$ hence, always check all the options carefully to ensure the correct answer.