Q 12-14-131JEE MainJEE Main 2021 (24 Feb, Shift 1)Easy
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of $\beta$ will be:
Answer: (D) 7
$\Delta I_B = \Delta I_E - \Delta I_C = 4 - 3.5 = 0.5$ mA
$$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3.5}{0.5} = 7$$
Solution by Sreeraj P, M.Sc Physics