Q 12-14-194JEE MainJEE Main 2018 (15 Apr, Shift 1)Easy
In a common emitter configuration with suitable bias, it is given that $R_L$ is the load resistance and $R_{BE}$ is small-signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by: ($\beta$ is current gain, $I_B$, $I_C$ and $I_E$ are respectively base, collector, and emitter currents.)
Answer: (A) $\beta\dfrac{R_L}{R_{BE}},\ \dfrac{\Delta I_C}{\Delta I_B},\ \beta^2\dfrac{R_L}{R_{BE}}$
In the CE configuration:
- Current gain: $\beta = \dfrac{\Delta I_C}{\Delta I_B}$.
- Voltage gain: $A_V = \dfrac{\Delta I_C R_L}{\Delta I_B R_{BE}} = \beta\dfrac{R_L}{R_{BE}}$.
- Power gain: $A_P = A_V\times\beta = \beta^2\dfrac{R_L}{R_{BE}}$.
This is option (1).
Solution by Sreeraj P, M.Sc Physics