A) \[1\,k\Omega \]
B) \[5\,k\Omega \]
C) \[10\,k\Omega \]
D) \[20\,k\Omega \]
Correct Answer: B
Solution :
Mutual conductance \[{{g}_{m}}=25m\,A/V=2.5\times {{10}^{-3}}A/V\] Plate resistance \[{{r}_{p}}=20k\Omega =20\times {{10}^{3}}\Omega \] Amplification \[A=100,\,\,{{R}_{L}}=?\] \[\upsilon ={{r}_{p}}\times {{g}_{m}}\] \[=20\times {{10}^{3}}\times 25\times {{10}^{-3}}=500\] \[A=\frac{\mu {{R}_{L}}}{{{R}_{P}}+{{R}_{L}}}\] \[100=\frac{500{{R}_{L}}}{20\times {{10}^{3}}+{{R}_{L}}}\] \[\therefore \] \[{{R}_{L}}=5k\Omega \]You need to login to perform this action.
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