## prateekbansal 2 years ago i hv jst started wid 6.002 as a self learner and want to confirm my answers for hw1...

1. prateekbansal

2. prateekbansal

just to check where mine answers are correct or not

3. 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.

4. prateekbansal

oooky got is so i need answer for the 1.2

5. 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

6. KenLJW

You should eventually give Best Response and/medal if answer is concise and accurate.

7. 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