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I understand from the diagram below that the neutral wire provides a return path to complete the circuit so that the current can flow. But once the current goes back to the distribution transformer, what happens to it after that? What I'm trying to say, is where does the current go after it reaches the secondary windings of the distribution transformer? Does the same current flow over and over again through the same circuit, in a continuous cycle like fashion?

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Also, why is the 240V split as 120V+120V? If I want the full 240V for some appliances, how do I make the connection? The diagram shows 240v across two live wires. Where is the return path for a 240V connection?

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    \$\begingroup\$ That is the magic of three phase power: If you have balanced loading, you don't need a return path. \$\endgroup\$ – Oldfart Apr 18 at 10:28
  • \$\begingroup\$ Oh, but why does the 120V live wire require a return path? Is it unbalanced? \$\endgroup\$ – noorav Apr 18 at 10:30
  • \$\begingroup\$ To explain the three phase supply magic you need a phase diagram with three vectors. Basically: The three vectors added up to give zero. A single (none-zero) vector can never produce zero. \$\endgroup\$ – Oldfart Apr 18 at 10:32
  • \$\begingroup\$ Can you clarify which country you are in please? Mains supplies vary between countries significantly. \$\endgroup\$ – Warren Hill Apr 18 at 10:34
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    \$\begingroup\$ @Oldfart This isn't three phase though; it's split phase with the two hot lines 180° out of phase. \$\endgroup\$ – marcelm Apr 18 at 10:42
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Just think of current in a wire as water in a pipe - it just gets pumped in circles by the electromotive force (EMF) measured in Volts

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    \$\begingroup\$ It is like water in a central heating system being circulated by a pump. \$\endgroup\$ – HandyHowie Apr 18 at 10:34
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The primary of this system is 3-phase and may have been generated either star or delta but is a delta connection as it enters the distribution transformer since there is no neutral connection.

On the secondary we have a split phase system. The voltage across each secondary is 120V and since these are \$ 180^o \$ out of phase the line to line voltage is 230V.

Now for simplicity we will assume there is no load on the secondary between either phase and the central star point, or earthed local neutral. If we connect a load between the two phases then this current must flow equally in each secondary.

It will come out to X2 and back in to the other secondary but will not pass through the central, neutral, wire. By transformer action this load passes back to the primary between phases A and B.

If we now add a second load between one of the secondary phases and local neutral one of the secondary windings has more current in it but this is also passed back to phases A and B by transformer action.


To put some numbers to this lets assume both the 240V load and the 120V load are resistive and each draws 1A. We have 1A in one secondary and 2A in the other.

This is equivalent from the primaries point of view to a single winding with 3A in it.

The primary to secondary turns ratio is 20:1 so the current in the primary is \$ \dfrac{3 \text{ A}}{20} = 150 \text{ mA} \$.

For the 240V load alone it would be 100 mA.

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If you only have a load from X1 to X2 (say) then current flows from X1 to X2 (and back again on the other half of the waveform). So the neutral carries all the load current. If you have exactly equal 120V loads from X1 to X2 and X2 to X3 (say two 100W 120V incandescent bulbs) then the current flows from X1 to X3 and no current flows through the neutral. At any given instant the sum of all the currents out of a node (or into a node) is always zero, so the neutral carries the difference in the currents from X1 and X3.

If you have a 240V load (say a 1500W 240V oven element) from X1 to X3 then again no current flows through the neutral and it is not part of that circuit.

Now imagine there are two incandescent bulbs, a 25W one from X1 to X2 and a 100W one from X2 to X3. If the neutral develops a fault and becomes disconnected they will effectively be in series across the 240V and the 25W bulb will become very bright and burn out quickly, while the 100W will barely glow.


The DC equivalent of the secondary would be two 120V batteries in series (adding), with the common point connected to the neutral.

schematic

simulate this circuit – Schematic created using CircuitLab

AC circuits are a bit trickier. They may add or subtract like batteries if they are in phase or 180° out of phase (as is the case in your secondary circuit), but if they are (say) 120° out of phase (as in a 3-phase circuit), the sum voltage is not 2 times, but \$\sqrt{3}\$ times.

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