A) 1 kW
B) 5 kW
C) 10 kW
D) 20 kW
Correct Answer: B
Solution :
\[\mu ={{r}_{P}}\times {{g}_{m}}=20\times 2.5=50\] From \[A=\frac{\mu \,{{R}_{L}}}{{{r}_{P}}+{{R}_{L}}}\]Þ \[{{r}_{P}}+{{R}_{L}}=\frac{\mu \,{{R}_{L}}}{A}=\frac{50{{R}_{L}}}{10}=5{{R}_{L}}\] Þ \[4{{R}_{L}}={{r}_{p}}\Rightarrow {{R}_{L}}=\frac{{{r}_{p}}}{4}=\frac{20}{4}=5k\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
