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prateekbansal

  • 2 years ago

i hv jst started wid 6.002 as a self learner and want to confirm my answers for hw1...

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  1. prateekbansal
    • 2 years ago
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    i just want the answers....

  2. prateekbansal
    • 2 years ago
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    just to check where mine answers are correct or not

  3. KenLJW
    • 2 years ago
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    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.

  4. prateekbansal
    • 2 years ago
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    oooky got is so i need answer for the 1.2

  5. KenLJW
    • 2 years ago
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    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

  6. KenLJW
    • 2 years ago
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    You should eventually give Best Response and/medal if answer is concise and accurate.

  7. KenLJW
    • 2 years ago
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    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

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