Is this superposition example wrong?

Discussion in 'General Electronics Chat' started by hunterage2000, May 2, 2010.

  1. hunterage2000

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    May 2, 2010
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  2. Ghar

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    Mar 8, 2010
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    The higher resistance will always have less current than the lower resistance when in parallel.
    In this case you have R1 = 4 ohms and R2 = 2 ohms.

    Since they're in parallel in that 7 V circuit, R2 must have twice as much current as R1.
    R3 will get the sum of those two currents.

    In the 28 V circuit R2 and R3 are in parallel, and R3 is 1 ohm. Therefore the situation is swapped, with R3 having twice the current of R2.
    R1 will get the sum of those two currents.
     
  3. hunterage2000

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    May 2, 2010
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    I used for I1 2/2+4 x 3 = 1 and I2 4/2+4 x 3 = 2, cant see where im going wrong
     
  4. Ghar

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    Mar 8, 2010
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    You really need to use brackets.

    I1 = 2/(2+4) * 3 = 1

    So I'm not sure what the issue is. I1 is the current through R1, yes?
    Then it agrees... 4 ohm R1 gets 1 A, 2 ohm R2 gets 2 A.
    Where's the disagreement?
     
  5. hunterage2000

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    May 2, 2010
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    Just I thought as R1 = 4 R2 is 2 then I1 would be I1 = 4/(2+4) * 3 = 2 not 1 which it says in the tutorial. Is this the correct way of doing this because the 28v circuit was correct.
     
  6. Ghar

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    Mar 8, 2010
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    With current division the numerator is the resistance in the other branch.

    So:
    I1 = Itotal * R2 / (R1 + R2)
    I2 = Itotal * R1 / (R1 + R2)

    The 28V circuit can do the same thing:
    Itotal = 6 A.
    R2 = 2 ohms
    R3 = 1 ohm

    I2 = 6 * 1 / (2 + 1) = 2 A
    I3 = 6 * 2 / (2 + 1) = 4 A
     
  7. hunterage2000

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    May 2, 2010
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    cheers mate thanks for clearing that up :)
     
  8. Ghar

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    Mar 8, 2010
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    You're welcome.
     
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