A) 1.69 mA
B) 3.95 mA
C) 2.87
D) 7.02 mA
Correct Answer: A
Solution :
\[{{i}_{p}}=k{{({{V}_{p}}+\mu {{V}_{g}})}^{3/2}}mA\] Þ 4 = k(200 ? 10 ´ 4)3/2 = k ´ (160)3/2 ?.(i) and \[{{i}_{p}}=k{{(160-10\times 7)}^{3/2}}=k\times {{(90)}^{3/2}}\] ?.(ii) From equation (i) and (ii) we get \[{{i}_{p}}=4\times {{\left( \frac{90}{160} \right)}^{3/2}}=4\times {{\left( \frac{3}{4} \right)}^{3}}=1.69\,mA\]
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