Question

If an emitter current is changed by $4mA$, the collector current changes by $3.5mA$. The value of $\beta $ will be:
(A) $7$
(B) $0.875$
(C) $0.5$
(D) $3.5$

Answer 1

Hint: In order to solve this question, we will first find the value of $\alpha $ which is the ratio of change in collector current to the change in emitter current and then using general relation between $\alpha ,\beta $ we will solve to find the value of $\beta $.

Formula used:
$\alpha = \dfrac{{\vartriangle {I_C}}}{{\vartriangle {I_E}}}$
Where, $\vartriangle {I_C}$ is the change in collector current and $\vartriangle {I_E}$ is the change in emitter current.
The relation between $\alpha ,\beta $ is
$\beta = \dfrac{\alpha }{{1 - \alpha }}$
where $\beta = \dfrac{{\vartriangle {I_C}}}{{\vartriangle {I_B}}}$ which is the ratio of change in collector current to the change in base current.

Complete answer:
We have given that the emitter current is changed by the value of $4mA$ which can be written as $\vartriangle {I_E} = 4mA$ and the collector current is changed by the value of $3.5mA$ which can be written as $\vartriangle {I_C} = 3.5mA$

So, The value of $\alpha $ can be calculated using the formula $\alpha = \dfrac{{\vartriangle {I_C}}}{{\vartriangle {I_E}}}$ on putting the values we get,
$
  \alpha = \dfrac{{3.5}}{4} \\
  \alpha = \dfrac{7}{8} \\
 $
Now, In order to find the value of $\beta $ we can use the relation between $\alpha ,\beta $ as $\beta = \dfrac{\alpha }{{1 - \alpha }}$

so, on putting the required value and solving for $\beta $ we get,
$
  \beta = \dfrac{{\dfrac{7}{8}}}{{1 - \dfrac{7}{8}}} \\
  \beta = 7 \\
 $
So, The value of $\beta $ is $7$

Hence, the correct answer is option (A) $7$.

Note: It should be remembered that the emitter current, collector current, and base current are in the context of a PNP or NPN three-pin transistors and transistors are the electronic components that are widely used in almost all electrical devices because transistors can be used as an amplifier, rectifier or switch and PNP and NPN transistor mean n-type semiconductor is sandwiched between p-types and vice-versa.