A) 50
B) 25
C) 100
D) 12.5
Correct Answer: B
Solution :
Using \[{{A}_{V}}=\frac{\mu }{1+\frac{{{r}_{p}}}{{{R}_{L}}}}\]and \[\mu ={{r}_{p}}\times {{g}_{m}}\] \[\Rightarrow {{r}_{p}}=\frac{\mu }{{{g}_{m}}}=\frac{50}{2\times {{10}^{-3}}}=25\times {{10}^{3}}\Omega \] \[\therefore {{A}_{v}}=\frac{50}{1+\frac{25\times {{10}^{3}}}{25\times {{10}^{3}}}}=25.\]
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