anonymous
  • anonymous
i hv jst started wid 6.002 as a self learner and want to confirm my answers for hw1...
MIT 6.002 Circuits and Electronics, Spring 2007
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
i just want the answers....
anonymous
  • anonymous
just to check where mine answers are correct or not
KenLJW
  • KenLJW
The answer to question come with procedure used, ask individual question and someone will answer. Example Problem 1.1a DC supply in series with R!,R2,R3 Find I, P for each KVL, V = V1+V2+V3 V=IR, V = IR1+IR2+IR3 = I(R1+R2+R3), I = V/(R1+R2+R3) Power in equals power out P = IV, P = I^2R IV= V^2/(R1+R2+R3) I^2 R1+I^2 R2+I^ R3 = V^2 R1/(R1+R2+R3)^2+V^2 R2/(R1+R2+R3)^2+V^2 R3/(R1+R2+R3)^2 V^2/(R1+R2+R3) The answers correlate, there's different way's of coming to a solution, its good to know all of them so you can pick the basic and simplest. In asking only for answers assume your why is the best way.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oooky got is so i need answer for the 1.2
KenLJW
  • KenLJW
You should ask how to do Problem 1.1b, DC current supply in parallel with R1,R2,R2 Find Voltage across each and power across each resister KCL I = I1+I2+I3 = V/R1+V/R2+V/R3 V = I/{1/R!1 +1/R2+1/R3} the simple way to express this is R1||R2||R3 = 1/{1/R!1 +1/R2+1/R3} power in = IV= I^ 2R1||R2||R3 Power loss across R1 V^2/R1 = I^2 (R1||R2||R3)^2/R1 continue for R2 and R3 and prove power in equals power out
KenLJW
  • KenLJW
You should eventually give Best Response and/medal if answer is concise and accurate.
KenLJW
  • KenLJW
1K||1K = 1K (1||1); 1K + 1K =1K (1+1) (1+1)||1+1= 5/3, realize > 1 therefore must add one (1+1||1)||1=3/5 realize < 1 therefore must have || with one end

Looking for something else?

Not the answer you are looking for? Search for more explanations.