Q 12-14-179JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
An NPN transistor operates as a common emitter amplifier, with a power gain of $60\ \text{dB}$. The input circuit resistance is $100\ \Omega$ and the output load resistance is $10\ \text{k}\Omega$. The common emitter current gain $\beta$ is:
Answer: (B) $10^2$
$60\ \text{dB}$ means a power gain of $10^6$. For a CE amplifier
$$\text{Power gain} = \beta^2\frac{R_L}{R_{in}} = \beta^2\times\frac{10^4}{100} = 100\,\beta^2$$
So $\beta^2 = 10^4$ and $\beta = 10^2$.
Solution by Sreeraj P, M.Sc Physics