A) 16.36
B) 28
C) 78
D) 108
Correct Answer: A
Solution :
Using voltage gain \[{{A}_{v}}=\frac{\mu }{1+\frac{{{r}_{p}}}{{{R}_{L}}}}\] also \[\mu ={{r}_{p}}\times {{g}_{m}}\] \[\Rightarrow {{r}_{p}}=\frac{\mu }{{{g}_{m}}}=\frac{20}{3\times {{10}^{-3}}}\] \[\therefore {{A}_{v}}=\frac{20}{1+\frac{20}{3\times {{10}^{-3}}\times 3\times {{10}^{4}}}}\]=\[\frac{180}{11}\]=16.36.You need to login to perform this action.
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