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I do not understand the answer of the following question :

A three-phase power in the lab have symmetrical three-phase voltages 400 / 230V and terminals marked L1, L2, L3 and N. A 60W bulb is connected between phase L1 and N. Another 60W lamp is connected in the same way, but the phase L2 and N.

Calculate the current in the neutral conductor N.

The answer is 0,26 A.

Why it is 0,26 A , should not it be 0,26 A + 0,26 A (from the both lamps)?

Thanks

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    \$\begingroup\$ Calculate the current flowing in L3 too \$\endgroup\$
    – PlasmaHH
    Commented Aug 18, 2016 at 12:52
  • \$\begingroup\$ what happens if we have one bulb between L1 and N then one bulb between L2 and N then two bulbs between L3 and N .. What is current in the neutral conductor N? thanks \$\endgroup\$
    – Joe
    Commented Aug 18, 2016 at 13:33
  • \$\begingroup\$ Remember these are sine waves, and you don't add the pk to pk values to find out how much current is there because the phase comes into play. \$\endgroup\$
    – Voltage Spike ♦
    Commented Aug 18, 2016 at 16:28

2 Answers 2

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To confirm, from \$ P = VI \$ that the current would be \$ I = \frac {P}{V} = \frac {60}{230} = 0.26~A \$. No problems there.

A very simple way to consider this problem is that if we connected a 60 W lamp from each phase to neutral then the neutral current would sum to zero.

Now consider what happens if we remove one of the three lamps: the neutral current must change by that amount, 0.26 A. That's the simple way to calculate for this problem.

The more general way would be to add the vectors. (And this is your problem: you forgot that they are not in phase.)

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1 (a) The phase A and B current vectors. (b) A and B vectors summed to give the resultant current.

Clearly from the vector diagram, since A and B were at 120° then in (b) they must be at 60°. Since they're the same size the triangle is equilateral. Therefore the sum must be 0.26 A.

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  • \$\begingroup\$ what it will be if we had a bulb is connected to L3 and N ... What is the current in the neutral conductor N ? and why ? My answer is N=0 because we get a triangle of three vectors and there is no sum ,, sum is zero .. \$\endgroup\$
    – Joe
    Commented Aug 18, 2016 at 13:19
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Commented Aug 18, 2016 at 13:20
  • \$\begingroup\$ what happens if we have one bulb between L1 and N then one bulb between L2 and N then two bulbs between L3 and N .. What is current in the neutral conductor N? thanks \$\endgroup\$
    – Joe
    Commented Aug 18, 2016 at 13:24
  • \$\begingroup\$ That's a different question. Draw the vectors as shown in Figure 1. Add them. That will give you a graphical representation. Now repeat by doing the vector addition mathematically. \$\endgroup\$
    – Transistor
    Commented Aug 18, 2016 at 13:27
  • \$\begingroup\$ I understand how to draw the vectors for L1 and L2 but how to add the vector of L3? How you do ? thanks \$\endgroup\$
    – Joe
    Commented Aug 18, 2016 at 13:30
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Because the currents are not in phase. So they partially cancel.

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  • \$\begingroup\$ what happens if we have one bulb between L1 and N then one bulb between L2 and N then two bulbs between L3 and N .. What is current in the neutral conductor N? thanks \$\endgroup\$
    – Joe
    Commented Aug 18, 2016 at 13:34

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