iC is slightly smaller than iE.
iB is much smaller than iE.
Explanation:
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 iE.
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 (iB) is low.
The remaining majority carriers of the emitter pass to the collector and give collector current iC.
Thus, we get the relation given below:
iE = iB + iC
Thus, because of the base, current iC is slightly smaller than iE.
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