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

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prateekbansalBest ResponseYou've already chosen the best response.0
i just want the answers....
 10 months ago

prateekbansalBest ResponseYou've already chosen the best response.0
just to check where mine answers are correct or not
 10 months ago

KenLJWBest ResponseYou've already chosen the best response.1
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.
 10 months ago

prateekbansalBest ResponseYou've already chosen the best response.0
oooky got is so i need answer for the 1.2
 10 months ago

KenLJWBest ResponseYou've already chosen the best response.1
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 R1R2R3 = 1/{1/R!1 +1/R2+1/R3} power in = IV= I^ 2R1R2R3 Power loss across R1 V^2/R1 = I^2 (R1R2R3)^2/R1 continue for R2 and R3 and prove power in equals power out
 10 months ago

KenLJWBest ResponseYou've already chosen the best response.1
You should eventually give Best Response and/medal if answer is concise and accurate.
 10 months ago

KenLJWBest ResponseYou've already chosen the best response.1
1K1K = 1K (11); 1K + 1K =1K (1+1) (1+1)1+1= 5/3, realize > 1 therefore must add one (1+11)1=3/5 realize < 1 therefore must have  with one end
 10 months ago
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