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Answers Please. Electrical Technology.

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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.
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    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.


     
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  • 0
    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.


     
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