# bulbs connected to a three phase system - what is the current in neutral conductor?

I wonder about the three phase system. 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 then two bulbs (60W each) between L3 and N .. What is current in the neutral conductor N?

my answer : if we had just one bulb between L3 and N then we get N=0 Now if the two bulbs between L3 and N , if the two bulbs are connected in series then the current through N = 0,26 A * 3 = 0,78 A Is my solution correct?

thanks

• It is not same question .. they told me that I have to post a new post if I want to ask another question .. – Joe Aug 18 '16 at 15:37
• The best way to understand what is happening is to draw three vectors at 120 degree offsets. One for each phase with N in the middle. Vector length is the current in that phase. Sum of vectors is the neutral current. – Majenko Aug 18 '16 at 15:44
• Not right now, I am out. – Majenko Aug 18 '16 at 15:54
• electronics-tutorials.ws/accircuits/phasors.html – Majenko Aug 18 '16 at 16:02
• @Joe: No, your answer is incorrect. This is a follow on from your other question. There's a strong smell of homework so we need to see your effort. (We don't do people's homework for them but we will give them some tips on how to solve it themselves.) Show your phasor diagram as prompted in the other question. – Transistor Aug 18 '16 at 17:43

If you put 2 bulbs in series, the current is more than 50% since the R vs temp changes over a 1:10 range from cold to hot due to the PTC characteristic of tungsten.

What ever the difference in current between L3 and the average of L1 & L2 is the net neutral current as these all have pf=1 phasor currents.

if bulbs between L3 and N are connected in parallel than the current in N is I=0.26A ( same as of single bulb ). If bulbs between L3 and N are connected in series than it is effectively a 30W bulb ( I=0.13A ),and the current in N is the difference of L3 current ( 0.13A ) and vector sum of L1 and L2 current ( 0.26A ). The final answer is 0.13A

• Welcome to EE.SE. As this appears to be a homework question we generally avoid giving a complete answer. Instead prompt the OP on how to solve it. You could improve the impression given by your answer by editing to fix incorrect capitalisation, punctuation and spacing (there should be none after a '(' or before a ')'. – Transistor Aug 21 '16 at 17:12