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prateekbansal

sum 1 plz help me with hw1 exercise 1-2

  • 11 months ago
  • 11 months ago

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

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

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

    • 11 months ago
  4. prateekbansal
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    tankzz helped a lot

    • 11 months ago
  5. DEVESHkishore
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    welcome :-)

    • 11 months ago
  6. KenLJW
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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)

    • 11 months ago
  7. KenLJW
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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

    • 11 months ago
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