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prateekbansal

  • one year ago

sum 1 plz help me with hw1 exercise 1-2

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  1. DEVESHkishore
    • one year ago
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    i have given a possible solution of ur problem to Keni_jk plz look at it

  2. DEVESHkishore
    • one year ago
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    |dw:1369544893178:dw| solve this u will get solution of 1st part

  3. DEVESHkishore
    • one year ago
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    |dw:1369544973454:dw| solve box . get answere. now solve with resistor in series

  4. prateekbansal
    • one year ago
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    tankzz helped a lot

  5. DEVESHkishore
    • one year ago
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    welcome :-)

  6. KenLJW
    • one year ago
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    |1K|1K| = 1K(1||1|. 1K + 1K = 1K(1+1) So I'll only use 1's (1+1)||1 +1 = 5/2 (1+1||1)||1 = 3/5 I had to try different arrangements until I came upon this. 1||1 = 1/2 2||1 = 2/3 general rule 1||a = a/(1+a)

  7. KenLJW
    • one year ago
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    should be 5/3 This principle of factoring resistance can be used on all passive networks aR--a, L--L/R, C--RC You need to make appropriate change in driving function, whether you preserving loop currents or nodal voltages This is used in filter design

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