3

Having worked with an electrician in the past, I have seen how house electrical circuits are laid. Sockets are laid out in a ring back to the fuse board and the live, earth, and neutral of each end of the ring are wired in to the same corresponding point on the fuse board.

This ring is used so that if a there is as split in the cable, you can still use each of the sockets within the ring. This begged the question to me, in which direction is the current going around the ring? Now, I know domestic electricity supply is alternating current. Does this mean the current alternates and goes both ways, or is it alternating in "strength" from a maximum to a minimum and then back to a maximum in a particular direction alone. https://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Ring_circuit.svg/800px-Ring_circuit.svg.png

4 Answers4

2

Starting from the consumer unit / fuse box you have wires taking two different paths but ending up at the same place - a socket with a device attached.
So in effect you have the two live wires in parallel and the two neutral wires in parallel.
So if the current in one live wire is flowing from the consumer unit to device the same will be happening in the other wire.
When the current in one wire is reversed so the current will reverse in the other wire.

The magnitude of the currents flowing will follow the normal rules for parallel conductors.

Farcher
  • 104,498
0

The direction of current flow in the wires is unimportant. What is important is the direction current flow through the device connected to a socket. And even that isn't really important because all you really care about is the energy transfer, and that's always into the device (unless you have a generator; then you better be careful when you connect it!).

You will have a "hot" wire a the circuit. It will have an alternating electric potential going sinusoidally from $\sqrt{2}V_{rms}$ to $0$ to $-\sqrt{2}V_{rms}$ and back. We will consider the neutral wire to be stable. This means that there is a conventional current through the device always travelling from the higher potential wire to the lower potential wire. Sometime the higher is the "hot" wire, and sometimes it is the neutral wire. The product of $i(t)v(t)$ is going to be positive, so power is consumed by the device.

The ring simply provides a way to change the potential across the device.

Bill N
  • 15,522
0

In general, with alternating current, AC, systems the current does go backwards and fowards all the time, but that does not mean that it goes one way around the ring main and then the other. In fact it should never go around the ring, because as you pointed out in another answer the potential is going positive and then negative on all parts of the ring at the same time.

So if nothing is connected to the sockets on the ring main almost, but not quite, zero current will flow into the ring and out of the ring the the voltage goes positive and then negative in the ac cycle. You are correct in your comment - the 'potential' or voltage of the wires will go up and down.

Now if you connect an electric heater and turn it on then current will flow into the heater and then out of the heater for each part of the ac cycle and current will flow through the ring main - you can think of it like a river with a large island in the middle and narrow channels each side of the island - water can flow around the island both ways and in the same direction on both sides of the island - how much flows on each side will depend on the length of the channels and other factors..... except becaue the current is ac in our model of the island in the river the water has to flow downstream - and then upstream... this could happen in tidal river where the direction of current can change due to the tide coming in or going out.

Now when we think about ac current it is aways going one way and then the other and if we took an average of the current we would always get zero, which is not very helpful. So to measure ac current we normally look at the `rms' value... the root mean square - we square the current so that both positive and negative currents give a positive value, take the average and then take the square root. So if an ac device needs 5 Amps of current then that probably refers to the rms value.

So when we talk about ac current it is always changing direction, but if we use rms values we can actually get a useful quatitative measure for the ammount of current going backwards and forwards in each cycle.

... sorry now rather long answer.. hope it is useful

tom
  • 7,034
0

RMS values gives an explanation for available, constant, power for both positive and negative sinusoidal voltage. Current flow in a ring main is still a difficult concept and in my mind has to boil down to the preference ( restriction) of electron flow (agitation). RMS negates the idea or cancellation of positive and negative cycles but does not answer the question of initial current flow which must be a function of attached resistive load defining flow precedence. Basically, if a load is attached then current flow takes the path of least resistance (no matter how small this may be). But the second a load is attached then the other path is presumably preferable and so the swing between paths may occur (hence RMS answers this as paths must/may be continually switching and supplying posive and negative cycles of the voltage AC Sine wave). The idea of a single pipe (piece of copper) being able to facilitate flow in both directions is difficult to imagine unless one assumes that that the return path is via the neutral (a separate conductor to the live). Multiple devices attached to the live will only take so much power due to their resistive loads, hence more electron agitation occurs as more loads are attached. That's the physics I guess, how does "The mains" know to provide more power? if power basically equals V x I then as V is a constant (usually say 240v) then as more devices are attached to a ring main more amps are required and this can only be achieved by a mechanism of regulating voltage ( keeping 240v constant from mains) and by increasing the amperage (quite a task over a national network).

Hope these thoughts are more help than hindrance?