JEE Class main Answered
Urgent! please solve the problem in the given image with steps.
![question image](https://images.topperlearning.com/topper/new-ate/topr_11643710532850296Screenshot20190812011708.jpeg)
Asked by miaowmira1900 | 12 Aug, 2019, 01:21: AM
Voltage between Gate and Source, vGS = 5× R2 /( R1 + R2 ) = 5× 100/ 675.7 = 0.74 V
Drain current iD = K( vGS - vT )2 = 1.16×10-3 ( 0.74 - 0.5 )2 = 67 μA
if vDS ≥ ( vGS - vT ) or vDS ≥ (0.74 - 0.5 ) or vDS ≥ 0.24 V , then voltage drop across RD ≤ ( 5-0.24) V
Hence maximum voltage drop across RD = 4.76 V
since drain current is 67 μA , resistance RD = 4.76 V / 67 μA = 71.04 kΩ
Answered by Thiyagarajan K | 12 Aug, 2019, 07:29: AM
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