The question is as shown below:
The answer from the book is as follows:
However, below is my attempt, is my approach accurate?
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1\$\begingroup\$ Thank you for posting the full question and your fully-documented attempt at an answer! That's a great example that everyone should follow when posting this type of question. But please do add the 'homework' tag to your question as well. \$\endgroup\$– Mr. SnrubCommented Jan 19, 2021 at 8:32
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1\$\begingroup\$ You were sublime until the KVL equation. You messed the signs. Since you enter the resistor R2, it should be like: R2Id - 6 + Vgs = 0. The rest follows and if you correct that bit, all will be fine. \$\endgroup\$– a360pilotCommented Jan 19, 2021 at 8:33
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1 Answer
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1. Your calculation for R2 contains a sign error
Double-check your equation $$ -v_{gs} + i_dR_2 - 6 = 0 $$ There is a sign error in there. It looks like you're using KVL; double-check and be very detail-oriented in how you apply KVL. Correct the sign error and you will get the correct answer.
2. Your calculation for R1 is correct -- actually more correct than than the "solution"!
By your calculations you end up with a value of 4.8kΩ. In the answer key they just round this to 5kΩ: $$ R_1=\frac{6V-1.2V}{1mA}=4.8k\Omega \approx 5k\Omega $$
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2\$\begingroup\$ You swapped the resistor tags. Replace R2 and R1, please. In the titles, I mean. \$\endgroup\$ Commented Jan 19, 2021 at 8:42
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\$\begingroup\$ @BenFM D'oh! Thanks. Fixed. \$\endgroup\$ Commented Jan 20, 2021 at 2:51