A) 10 W
B) 20 W
C) 30 W
D) 40 W
Correct Answer: A
Solution :
\[\frac{i}{{{i}_{g}}}=1+\frac{G}{S}\]\[\Rightarrow \frac{i.G}{{{V}_{g}}}=1+\frac{G}{S}\Rightarrow \frac{100\times {{10}^{-3}}\times 40}{800\times {{10}^{-3}}}=1+\frac{40}{S}\] Þ \[S=10\Omega \].
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