A) 0.8 mA
B) 0.6 mA
C) 0.4 mA
D) 1 mA
Correct Answer: B
Solution :
By using \[{{g}_{m}}=\frac{\Delta {{i}_{p}}}{\Delta {{v}_{g}}}\]Þ \[3\times {{10}^{-4}}=\frac{\Delta {{i}_{p}}}{-1-(-3)}\] Þ \[\Delta {{i}_{p}}=6\times {{10}^{-4}}A=0.6\,mA\]
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