Let i_ E , i_ C and i_ B represent the emitter current, the colle…

MCQ

Let $\mathrm{i}_{\mathrm{E}}$, $\mathrm{i}_{\mathrm{C}}$ and $\mathrm{i}_{\mathrm{B}}$ represent the emitter current, the collector current and the base current respectively in a transistor. Then,

✓
$\mathrm{i}_{\mathrm{C}}$ is slightly smaller than $\mathrm{i}_{\mathrm{E}}$.
B
$\mathrm{i}_{\mathrm{C}}$ is slightly greater than $\mathrm{i}_{\mathrm{E}}$.
C
$i_B$ is much smaller than $i_E$.
D
$\mathrm{i}_{\mathrm{B}}$ is much greater than $\mathrm{i}_{\mathrm{E}}$.

✓

Answer

Correct option: A.

$\mathrm{i}_{\mathrm{C}}$ is slightly smaller than $\mathrm{i}_{\mathrm{E}}$.

The highlighted parts could not be edited, as the meaning could not be understood. Also, please check the last line for logical accuracy. Only one option is given as correct, while the solution gives two correct options.
We know that in the transistor base is slightly doped, therefore when the majority carriers due to forward biasing of emitter base junction, feel the repulsive force from the battery and pass over to the base region. This gives the emitter current $i_E$.
As the base is thin and lightly doped, only few majority carriers of the emitter are neutralised at the base. This gives the base current.
Hence, base current $(i_B)$ is low.
The remaining majority carriers of the emitter pass to the collector and give collector current $i_C$.
Thus, we get the relation given below:
$i_E= i_B+ i_C$
Thus, because of the base, current $\mathrm{i}_{\mathrm{C}}$ is slightly smaller than $\mathrm{i}_{\mathrm{E}}$.