Let i_E, i_C and i_B represent the emitter current, collector current…

The correct answer is Option (4) → \( i_E \) is equal to the sum of \( i_C \) and \( i_B \)

In a transistor the emitter injects charge carriers into a thin, lightly doped base. A few of them recombine there and form the small base current; most of them cross into the collector and form the collector current. By charge conservation (Kirchhoff's current law),

    \( i_E = i_C + i_B \)

with \( i_B \) only a few per cent of \( i_E \), so \( i_C \) is slightly less than \( i_E \).

Option (1) \( i_C \) is slightly greater than \( i_E \) — Incorrect. \( i_C = i_E - i_B \), so the collector current is slightly less than the emitter current, never greater.

Option (2) \( i_B \) is much greater than \( i_E \) — Incorrect. The base is thin and lightly doped, so \( i_B \) is much smaller than \( i_E \).

Option (3) \( i_C \) is equal to \( i_B \) — Incorrect. \( i_C \) is many times \( i_B \); their ratio is the current gain \( \beta \), typically 50 to 200.

Option (4) \( i_E \) is equal to the sum of \( i_C \) and \( i_B \) — Correct. The emitter current divides into the base and collector currents.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Transistor: Structure and Action – “… The base current is thus a small fraction of the emitter current. … It is obvious from the above description and also from a straight forward application of Kirchhoff’s law to Fig. 14.28(a) that the emitter current is the sum of collector current and base current:

    \( \displaystyle I_E = I_C + I_B \qquad (14.7) \)”