Question

For a transistor ${I_c}$ = $25$ mA, and ${I_b}$ = $1$ mA. What is the value of $\alpha $?
A. \[\dfrac{{25}}{{26}}\]
B. \[\dfrac{{26}}{{25}}\]
C. \[\dfrac{{24}}{{25}}\]
D. \[\dfrac{{25}}{{24}}\]

Answer 1

Hint: We know that in a transistor there are three types of current flowing: a. Collector current b. Base current and c. Emitter current. For solving this question, we would require the relationship between collector current ( ${I_c}$ ) and the base current ( ${I_b}$ ). We got to understand what the term $\alpha $ means.


Complete step by step solution:
The term $\alpha $ is defined as the current gain in a common base configuration of a transistor and also is defined as the change in collector current to the change in emitter current.
Thus $\alpha $ = $\dfrac{{{I_c}}}{{{I_e}}}$ . Here we are provided with the values of collector current and base current. We have to find the value of emitter current to get the required answer.
We know that Emitter current is the sum of the other two currents in a transistor, thus
${I_e} = {I_c} + {I_b}$ . Now we can calculate emitter current using the formula.\[\]
Thus $\alpha $ = $\dfrac{{{I_c}}}{{{I_e}}}$ = $\dfrac{{{I_c}}}{{{I_b} + {I_c}}}$ = $\dfrac{{25}}{{1 + 25}}$ = $\dfrac{{25}}{{26}}$ .
Hence, we get the required answer as option A.

Additional information: Note that the value of alpha is always less than 1. The value of Beta or current gain due to a common emitter is always greater than one and is related to alpha according to the relation: \[\beta = \dfrac{\alpha }{{1 - \alpha }}\] . Thus, it is always better to use a common emitter amplifier as it always has a very high output.

Note: It is important that we know the formula given in the additional information section, linking the two transistor setups. These questions generally ask about the values of current gains in different combinations. The definitions of alpha and Beta are to be kept in mind.