A) 16
B) 12
C) 10
D) None of the above
Correct Answer: B
Solution :
Voltage gain \[{{A}_{v}}=\frac{\mu }{1+\frac{{{r}_{p}}}{{{R}_{L}}}}\] \[\because {{R}_{L}}=1.5\,{{r}_{p}}\]\[\Rightarrow {{A}_{v}}=\frac{\mu }{1+\frac{{{r}_{p}}}{1.5\,{{r}_{p}}}}=\frac{3}{5}\mu \]\[=\frac{3}{5}\times 20=12\].
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