Adding Apparent Load

Discussion in 'Homework Help' started by DYLH, Nov 25, 2013.

  1. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    Appologies, this isn't actually a homework problem, but I thought that this may be the best place to get academic input.

    Which is the correct way to solve this?

    Find the total VA and PF for a circuit with the two loads:
    Apparent Load 1 = 250 VA w/ 0.75 PF
    Apparent Load 2 = 150 VA w/ 0.90 PF

    Method 1
    Real Power 1 = 0.75 * 250 = 187.5 W
    Real Power 2 = 0.90 * 150 = 135 W

    Total Apparent Load = 250 VA + 150 VA = 400 VA
    Total Real Power = 187.5 + 135 = 322.5 W

    Resulting Power Factor = 322.5 / 400 = 0.80625

    Method 2
    Real Power 1 = 0.75 * 250 = 187.5 W
    Real Power 2 = 0.90 * 150 = 135 W
    Total Real Power = 187.5 + 135 = 322.5 W

    Reactive Power 1 = sqrt(250^2 – 187.5^2) = 165.36 VAR
    Reactive Power 2 = sqrt(150^2 – 135^2) = 65.38 VAR
    Total Reactive Power = 230.7 VAR

    Total Apparent Power = sqrt(322.5^2 + 230.74^2) = 396.55 VA
    Resulting Power Factor = 322.5 / 396.55 = 0.813
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Method 2 is correct.
     
  3. DYLH

    Thread Starter New Member

    Aug 13, 2013
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  4. WBahn

    Moderator

    Mar 31, 2012
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    How about just doing the math from the definitions.

    Do you know about complex power? If so, things are very easy. If not, then things are a bit harder, but not too much.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You are neglecting to take into account whether the power factor is leading or lagging. If both are leading or both are lagging, then you will get the results in Method 2. But if one is leading and the other is lagging, which is precisely the case when you are doing power factor correction, then you have to take that into account.
     
    Last edited: Nov 25, 2013
    DYLH likes this.
  6. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    Good Point.. Thanks!
     
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