A) 5\[\text{k}\]
B) 25\[\text{k}\]
C) 20 \[\text{k}\]
D) 50\[\text{k}\]
Correct Answer: C
Solution :
\[{{i}_{P}}=2.5\,mA\,=2.5\times {{10}^{-3}}\,A,\] \[{{r}_{P}}=20\times {{10}^{3}}\Omega \] \[A=10,\,{{R}_{L}}=?\] \[{{V}_{P}}={{i}_{P}}\times {{r}_{P}}=50\] \[\begin{align} & {{g}_{m}}=\frac{{{i}_{p}}}{{{V}_{p}}}=\frac{2.5\times {{10}^{3}}}{50}=5\times {{10}^{-5}} \\ & \mu ={{r}_{P}}\times {{g}_{m}}=20\times {{10}^{3}}\times 5\times {{10}^{-5}}=1 \\ \end{align}\] \[A=\frac{\mu {{R}_{L}}}{{{R}_{P}}+{{R}_{L}}}\,\text{solving}\,{{R}_{L}}=20\,\text{k}\]
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