A) \[320\Omega \] and 10 mA
B) \[70\Omega \] and 10 mA
C) \[820\Omega \] and 10 mA
D) \[820\Omega \] and 8 mA
Correct Answer: D
Solution :
\[\frac{20}{Ig}-1680={{R}_{G}};\] |
\[\frac{30}{Ig}-2930={{R}_{G}}\] |
\[\therefore {{I}_{g}}=8mA\] and \[{{R}_{G}}=820\Omega \] |
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