A) \[1.6\times {{10}^{-3}}A\]
B) \[2.0\times {{10}^{-4}}A\]
C) \[1.6\times {{10}^{-3}}A\]
D) \[1.6\times {{10}^{-4}}A\]
Correct Answer: A
Solution :
\[\Delta {{V}_{g}}=-4.5-(-2.5)=2.0V\] \[{{g}_{m}}=8\times {{10}^{4}}mho\] Change in plate current\[\Delta {{I}_{p}}=\Delta {{V}_{g}}\times {{g}_{m}}\] \[=2.0\times 8\times {{10}^{-4}}\] \[=1.6\times {{10}^{-3}}A\]
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