A) 23.5 mW
B) 48.7 mW
C) 25.6 mW
D) None of these
Correct Answer: C
Solution :
\[{{r}_{p}}=\frac{\mu }{{{g}_{m}}}=\frac{64}{1600\times {{10}^{-6}}}=4\times {{10}^{4}}\Omega \] Voltage gain \[{{A}_{v}}=\frac{\mu }{1+\frac{{{r}_{p}}}{{{R}_{L}}}}=\frac{64}{1+\frac{4\times {{10}^{4}}}{40\times {{10}^{3}}}}=32\] \ Output signal voltage \[{{V}_{0}}={{A}_{v}}\times {{V}_{i}}=32\times 1=32\,V\,(r.m.s.)\] Signal power in load \[=\frac{V_{0}^{2}}{{{R}_{L}}}=\frac{{{(32)}^{2}}}{40\times {{10}^{3}}}=25.6\,mW\]
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