Discussion
For that device to carry \( 5 \ \mathrm{mA} \), the gate voltage must be about
Answer: B
Rearranging the square law for the gate voltage:
\[ \begin{aligned} V_{GS} &= V_{P}\left(1 - \sqrt{\tfrac{5}{10}}\right) \\ &= -4(1-0.707) \approx -1.17 \ \mathrm{V} \end{aligned} \]
so about -1.17 volts. Halving the current needs only about a quarter of the way to pinch-off, which is the square law working in reverse. Taking the square root rather than halving is the step this calculation turns on.
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