A) 16.36
B) 28
C) 78
D) 108
Correct Answer: A
Solution :
\[{{r}_{p}}=\frac{\mu }{{{g}_{m}}}=\frac{20}{3\times {{10}^{-3}}}\] \[=\frac{2}{3}\times {{10}^{4}}\Omega \] \[\therefore \]Voltage gain\[(A)=\frac{\mu }{\frac{{{r}_{p}}}{R}+1}=\frac{20}{\frac{2\times {{10}^{4}}}{3\times 3\times {{10}^{4}}}+1}\] \[=\frac{20\times 9}{11}=16.39\]
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