A) 10
B) 20
C) 15
D) 30
Correct Answer: B
Solution :
Voltage gain \[{{A}_{V}}=\frac{\mu }{1+\frac{{{r}_{p}}}{{{R}_{L}}}}\]and \[\mu ={{r}_{p}}\times {{g}_{m}}\] \[\Rightarrow \mu =10\times {{10}^{3}}\times 3\times {{10}^{-3}}=30\] \[\therefore {{A}_{v}}=\frac{\mu }{1+\frac{{{r}_{p}}}{2{{r}_{p}}}}=\frac{2}{3}\mu =\frac{2}{3}\times 30=20.\]
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