Q 12-14-191JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
The transfer characteristic curve of a transistor, having input and output resistance $100\ \Omega$ and $100$ kΩ respectively, is shown in the figure. The voltage and power gain, are respectively:
Answer: (A) $5\times10^4$, $2.5\times10^6$
Current gain from the slope:
$$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{5\ \text{mA}}{100\ \mu\text{A}} = 50$$
$$A_V = \beta\frac{R_{\text{out}}}{R_{\text{in}}} = 50\times\frac{10^5}{100} = 5\times10^4,\qquad A_P = \beta A_V = 2.5\times10^6$$
Solution by Sreeraj P, M.Sc Physics