[help] two balanced star-connected

Discussion in 'Homework Help' started by cupcake, Sep 20, 2010.

  1. cupcake

    Thread Starter Member

    Sep 20, 2010
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    Hi, please help me with this question..

    Two balanced star-connected loads of (8+j5) ohm/phase and (6-j2) ohm/phase are supplied by a three-phase source at a line-to-line voltage of 440 V. Find
    a. the line current drawn by each load
    b. total line current supplied by the source
    c. real, reactive, and complex power absorbed by each load
    d. real, reactive and complex power delivered by the source.

    Ok, I did the part a, and got the answer for line current by each load are, 26.92\angle -32 for (8+j5) and 40.16 \angle18.45 for (6-j2).. and now, I'm stuck with part b onwards.. do I just add up the current from each load to get the total current supplied by the source?? can anyone enlighten me how to approach part b onwards? thanks
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Part (b)

    Just add the two current having regard to their phase relationship - i.e. a complex number addition.

    Note that this is just the current in one phase [reference 0°] and while the magnitudes are the same in all phases of the balanced load, the individual line currents are phase displaced by 120°.

    Parts (c) & (d) can be done in various ways.

    For instance as a start one approach might be ...

    Find find the complex or apparent power in a given phase by multiplying the complex {phasor} line current you calculated {in part (a)} by the line-to-neutral voltage.

    So if I take 26.92[​IMG] and multiply this by 440/√2 [angle 0°] I will have the phase complex (apparent) power for the (8+j5)Ω branch.

    Similar reasoning can be applied to the source complex power. For the same phase, you would use the total line current for that phase you will have already calculated in part (b) multiplied by the same voltage of 440/√2 [angle 0°].

    Use the relationship Q=P+jS to derive other values as needed.
     
  3. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    ok, I got it already..

    now, supposed, when the load impedances of (6-j2) are delta-connected, how to find the total line current supplied by the source..?

    it seems than the other impedances (8+j5) are still star-connected.. I did try to find current by each loads and add them together.. but the answer is not correct.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One approach would be to convert the re-jigged (??) load of (6-j2) in delta to a star equivalent. Which would be (6-j2)/3 per phase leg.

    You seem a little unsure as to what is connected in star and what is connected in delta. Is this the original problem or a proposal to change the load arrangement (in part) from star to delta to see what happens?
     
  5. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    yes..this is the continuation of the previous question, it asks to repeat all problems when the load impedances of (6-j2) ohm/phase are delta-connected. so meant, the other impedance is still star-connected right?

    I think it just wants to show us that the delta-connected loads take 3-times the line currents and complex power of star-connected.

    I have done like what you did, converting the 6-j2 in delta to star equivalent but somehow the answer is off.. for part a, is ok. but, for part b and d I'm not quiet sure how to approach the questions. thanks
     
  6. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    I SOLVED IT..

    thanks :D
     
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