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Answers Please. Electrical Technology.

1. A load of 2.5kW takes a current of 12 Amp from a 230 Volt supply. Calculate the power factor.


2. Three single phase loads are to be connected to a three phase supply. All at unity power factor. The loads are,


i, 60 Amp.


ii, 50 Amp.


iii, 40 Amp.


By any method determine the neutral current.


Z.
  • 1. about 0.906

    2. 17.32A


      - Andy.
  • 90.57971 by plastic brain - a.k.a a calculator..


    15.5A by a rule and pen on the back of an envelope. (Guessed the angle by the "Rack of the eye" method)


    Next calculation I might use a "Guessing Stick" Well that`s what our tutors used to call a slide rule as calculators (plastic Brain) were just starting to appear on the scene. 

    I can`t remember a slang name for those adding machines that were like a mini cash register with levers, a handle and moveable carriage. We thought they were brill though, 4 in the whole school and was a treat to get to use one for a few minutes , I guess it was the second half of the 60s and I think they were about £40 each which was a lorra lorra money back then


    Mind you, our school had a maths teacher (not my teacher, I had one of the others) that actually gave one girl 101% in a maths test - 101% ? yes 101%, she answered all the questions correctly plus her work was so neat and so well set out that he gave her an extra one per cent on top. 101% and that was at a Grammar type school !
  • ebee:

    90.57971 by plastic brain - a.k.a a calculator..


    15.5A by a rule and pen on the back of an envelope. (Guessed the angle by the "Rack of the eye" method)


    Next calculation I might use a "Guessing Stick" Well that`s what our tutors used to call a slide rule as calculators (plastic Brain) were just starting to appear on the scene. 

    I can`t remember a slang name for those adding machines that were like a mini cash register with levers, a handle and moveable carriage. We thought they were brill though, 4 in the whole school and was a treat to get to use one for a few minutes , I guess it was the second half of the 60s and I think they were about £40 each which was a lorra lorra money back then


    Mind you, our school had a maths teacher (not my teacher, I had one of the others) that actually gave one girl 101% in a maths test - 101% ? yes 101%, she answered all the questions correctly plus her work was so neat and so well set out that he gave her an extra one per cent on top. 101% and that was at a Grammar type school !


    Who threw that? Stop talking at the back. Put that gum in the bin Don't forget that the power factor has to be less than one. Put that away John.


    Z.


  • Sorry Sir, I was expressing the power factor as a percentage. I shoulda/menta said 0.905791 which agrees with AJJs answer
  • Very good so far.


    Any more answers please?


    Z.
  • In = +17.32A >/ 64deg

    edit: whoops wrong angle and direction....didn't realise that editing can take place well after posting...

    Legh
  • showing the (known to be erroneous - see comments below) workings.


    12A 230V is 2760VA

    PF is 2500/2760 = 250/276  = 0.90597 

    So 25 degree phase angle between P and Q


    2) Subtract 40A from each phase as that has no effect on neutral current.

    Problem now  reduces to completing a parallelogram  with one side length 20 units, and the other side length 10 units and calculating the diagonal


    resolve parallel and orthogonal to the phase carrying the 20A excess.

    Orthogonal to the  dominant phase is 10A *sin120deg =  8.66A

    in-phase  is 10A* cos 120 + the original 20A.  = (-5A + 20A) = 15A


    use pythag to get the magnitude is sqrt[(15)2 + (8.6666)2] =  sqrt[(225) + (75)] = sqrt(525) = 22.9A


    Edit That is wrong as Geoff points out below - should read ..

    Magnitude  sqrt[(15)2 + (8.6666)2] =  sqrt[(225) + (75)] = sqrt(300) = 17.32A

    Angle is arctan 15/8.6 ~ 30 degrees off the the phase carrying 60A, rotated  in the direction of the phase carrying 50.
    Mike.


     
  • Blimey I got s stress headache now
  • Sorry Mike only 8/10.  225 + 75 = 300.  SQRT(300) = 17.32A.


    Now for your homework - calculate the phase to neutral voltages and the voltage between your neutral terminal and the the star point of the supply transformer ?


    Regards

    Geoff Blackwell
  • Correction - the above should say do the calculation assuming a broken supply neutral


    Regards

    Geoff Blackwell