Q149 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
In common emitter configuration of transistor amplifier, $r_i$, $R_L$ and $\beta$ represent the input resistance, load resistance and the a.c. current gain respectively. The voltage gain $A_V$ and power gain $A_P$ are represented in magnitude respectively by
- A$\beta\left(\dfrac{r_i}{R_L}\right) , \beta^2\left(\dfrac{r_i}{R_L}\right)$
- B$\beta\left(\dfrac{R_L}{r_i}\right) , \beta\left(\dfrac{R_L}{r_i}\right)^2$
- Correct answerC$\beta\left(\dfrac{R_L}{r_i}\right) , \beta^2\left(\dfrac{R_L}{r_i}\right)$
- D$\beta\left(\dfrac{r_i}{R_L}\right) , \beta\left(\dfrac{r_i}{R_L}\right)^2$
Solution
Voltage gain $A_V$ is the ratio of output voltage to input voltage.
$A_V = \dfrac{V_{out}}{V_{in}} = \dfrac{I_{out} R_L}{I_{in} r_i}$
Since the a.c. current gain is $\beta = \dfrac{I_{out}}{I_{in}}$,
$A_V = \beta \left(\dfrac{R_L}{r_i}\right)$
Power gain $A_P$ is the ratio of output power to input power.
$A_P = \dfrac{P_{out}}{P_{in}} = \dfrac{I_{out}^2 R_L}{I_{in}^2 r_i}$
$A_P = \left(\dfrac{I_{out}}{I_{in}}\right)^2 \left(\dfrac{R_L}{r_i}\right) = \beta^2 \left(\dfrac{R_L}{r_i}\right)$
Answer: $\beta\left(\dfrac{R_L}{r_i}\right) , \beta^2\left(\dfrac{R_L}{r_i}\right)$
$A_V = \dfrac{V_{out}}{V_{in}} = \dfrac{I_{out} R_L}{I_{in} r_i}$
Since the a.c. current gain is $\beta = \dfrac{I_{out}}{I_{in}}$,
$A_V = \beta \left(\dfrac{R_L}{r_i}\right)$
Power gain $A_P$ is the ratio of output power to input power.
$A_P = \dfrac{P_{out}}{P_{in}} = \dfrac{I_{out}^2 R_L}{I_{in}^2 r_i}$
$A_P = \left(\dfrac{I_{out}}{I_{in}}\right)^2 \left(\dfrac{R_L}{r_i}\right) = \beta^2 \left(\dfrac{R_L}{r_i}\right)$
Answer: $\beta\left(\dfrac{R_L}{r_i}\right) , \beta^2\left(\dfrac{R_L}{r_i}\right)$
Hence, the correct answer is option: 'C'.