A transistor is operated in common emitter configuration at \( V_{\mathrm{CE}} = 3\,\mathrm{V} \) such that a change in the base current from 50 µA to 250 µA produces a change in the collector current from 5 mA to 20 mA. The current gain is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 75
The current gain of a transistor in the common emitter configuration (the a.c. gain) is the ratio of the change in collector current to the change in base current at constant collector voltage.
\( \displaystyle \beta_{\mathrm{ac}} = \frac{\Delta I_C}{\Delta I_B} \)
\( \Delta I_C = 20 - 5 = 15\,\mathrm{mA} = 15\times10^{-3}\,\mathrm{A} \)
\( \Delta I_B = 250 - 50 = 200\,\mu\mathrm{A} = 200\times10^{-6}\,\mathrm{A} \)
\( \displaystyle \beta_{\mathrm{ac}} = \frac{15\times10^{-3}}{200\times10^{-6}} = 75 \)
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):
Basic Transistor Circuit Configurations and Transistor Characteristics – “… (iii) Current amplification factor (β): This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage (VCE) when the transistor is in active state.
\( \displaystyle \beta_{ac} = \left(\frac{\Delta I_C}{\Delta I_B}\right)_{V_{CE}} \qquad (14.10) \)”