Physics - Semiconductors Question with Solution | TestHub

Given current gain $\beta =10$, input impedance $R_{in}=20k\Omega$, and load resistance $R_L=100k\Omega$.

The power gain $A_p$ is given by:

$A_p={\beta }^2\times \frac{R_L}{R_{in}}$

$A_p=(10{)}^2\times \frac{100}{20}$

$A_p=100\times 5=500$

Answer: $500$