Fusing theory

I too am curious :-)

However, an appliance only draws the current that it requires doesn't it? I you plug a 2 watt appliance into a supply with a 100 Amp fuse it will only demand 2 watts won't it?

If your 'phone needs charging and requires 10 watts to charge you can use a charger rated at 10 watts or more, it will only demand 10 watts. If you used a 5 watts charger in these circumstance it would get hot and may catch alight.

You do have to get the voltage right for the charger.

Genuinely interested and awaiting expert input!

If it's a 16-amp supply, you better not draw more than 16 amps off it. If the plug is rated at 2.5 amps, your device you're connecting the plug to better not draw more than 2.5 amps. 2.5 is a lot less than 16, so why d'ye think it would be dangerous? Just because the supply is rated at 16 amps, doesn't mean that it's gonna force 16 amps through your plug. The current taken is determined only by the device you've connected the plug to.

The amount of current it draws is defined by the voltage of the supply divided by the resistance of the device (which may vary depending on whether it's hot or not).

Most domestic supplies are fused at either 100A or 80A but you will not see that at any socket or light fitting as each circuit has its own maximum breaker. The amount of current in each circuit is determined by current draw of the load. So for instance a lighting circuit may be fused at 6A but if you only switch one bulb on then the current measured on that circuit will be dependant on the wattage of the bulb it will not be 6A.

A 5A rated cable can be connected to a 32A ring main as long as the load on that cable does not exceed approx. 1150W. A correctly fused plug ensures that

So what happens if a fault occurs in the appliance which draws 12A? The

The point of a fuse is to protect anything downstream of it from a fault current which may be dangerous. The point in this case is that a Euro-type plug is not fused, and that the flex leading from it must be able to safely handle fault currents up to 1.5 times the rating of the next upstream fuse, in this case, the circuit fuse. I believe the plug itself will probably handle this safely, but it seems unlikely that a domestic lamp will be fitted with a flex able to handle 24 Amps for up to four hours.

This is the question which is being asked, and I don't know the answer. It may be that not all European circuits are rated at 16A, they may have differentiated socket capacities as we did before the 13A plug and socket. That is obviously no longer the case in the UK, and the short answer in this particular case is to fit a 13A plug fused appropriately to the cable size, rather than just using an unfused adaptor. Since the cable end will be stripped, the strands can be measured and counted to determine the cable capacity. It must be assumed that any wiring inside the lamp is of equal capacity.

I think the question is why in the UK are we required to fit a fuse when in the rest of the world they rely on the Circuit Breaker in the "consumer unit".

Here in Spain the Air Fryer I have just purchased from Aldi has no fuse. As its 1.4kw in the UK it will have a 13 amp fuse.

WHY?

Dave

Then the fuse in your plug blows.

Perhaps we're more safety conscious? The Yanks, with their cheese-paring 110V, are even worse off, with four times the heating effect (for a given device wattage) in the cabling and plugs.

They can get 240V as most houses are supplied by 2 phases. A bit fiddly, but you can work with it.

I think the theory is that it is very unlikely that a fault in the appliance will result, in your example, in a 12A continuous draw. Usually it will either work properly, or it will develop a short circuit.

In the case of a short circuit, the appliance wiring and plug will be able to carry the short-duration very large fault current without failing, as the heat generated in the short time will not be sufficient to melt anything.

But the UK idea of a fuse in the plug is obviously better.

nib

Can you use European appliances designed for one live and one neutral when there would be two live inputs (presumably out of phase)?

This case was a Europlug, no fuse.

What now?

If it required a fuse, then it would need to be in the lamp. In most countries there won't be the option of a fused plug, so *fault* protection will need to be provided by the circuit breaker at the origin of the whole circuit (typically 16A) [1].

For something like a lamp there is no real need for overload protection since the characteristics of the load limit the maximum load. There is not much you can plug into a ES or BC lamp holder that would draw more than saw 150W unless you want to get very creative. (The lamp itself may also specify a maximum wattage, but this is likely to be a thermal constraint to stop the shade melting etc).

The key is the different types of over current protection. Normally you consider two scenarios: "overload" and "fault".

Overload is when you can expect a scenario that could cause more than the expected design current to be drawn for an extended period of time.

Fault current is when there is a (close to) short circuit event - typically L connected to N or L connected to E caused by an abnormal event (like nail through a hidden cable, chair leg crushing a flex etc).

So all circuits can experience fault currents. But not all are liable to overload. A lamp on its own is very unlikely to need to be protected against overload. A circuit with 100 lamps on it however may.

With a fault current the only limit on the current flow is the resistance of the fault path - this could be fractions of an ohm, and the current could be 1000s of amps. That means thing will get very hot very quickly (in a really bad case it could even explode (aka an arc fault)). So here you need a very fast active protection since you need to limit the amount of energy that is "let through". Fuses can do this if they are the right type. Circuit breakers normally have a separate mechanism for handling fault currents - typically a magnetic solenoid that will fire at a particular threshold and interrupt the circuit. (for a normal "type B" circuit breakers this is 5x nominal current - so to trip a B32 MCB "instantly" you would need >= 160A of fault current.

An overload situation can occur on something like a socket circuit. The user can do things that the designer has no contrl over. To many large loads plugged in and running at once for example. However if you load up a 32A ring circuit with 50A of load, bad stuff is not going to happen immediately. Things will start warming up, but it is not until the cable conductors get to over 70 deg C that you might start degrading the life expectancy of the cable. So the protective device can impose a more leisurely protection scheme - tripping on a thermal response that takes account of the size of the load and the duration. They usually use a bi-metal strip like in a thermostat for this.

If you look at :

I have no idea. But it's how they get the heftier kit to run (HVAC etc).

The general European idea is the socket would be on a 16A MCB radial, so fault currents would be limited to 16A. The UK has a 32A ring main, from which you could draw a 32A fault current - that's bad.

So instead plugs are fused at 13A. When you plug in an appliance, either it has an internal fuse (often non replaceable) or an internal design that inherently limits the current. So really the plugtop fuse protects from a nail through the flex or similar short circuit before the appliance.

Once you have a plug fuse, you can then replace it with a lower rated one, to match a thinner cable. You can't do that woth a Europlug, but the cable may ultimately act as its own fuse.

It's about risks, failure modes and cost/benefit. UK and other countries' designs just land in different places on the spectrum.

Theo

Yes, assuming they'll work on 60Hz. And they don't do things like tie 'neutral' to earth/ground (the appliance neutral will be hot in this scenario) - which would trip an RCD nowadays anyway.

Importing European kettles to the US is a thing people do.

Theo

While the lamp will only take what it needs normally, there is no guarantee that under a fault condition, it may not take considerably more and overload the flex.

That is when you apply the adiabatic equation... [1]

You can do a "fault withstand" calculation to check that the amount of metal available in the conductors will safely handle the fault condition.

For example, let's say you have a table lamp with 2m of 0.5mm^2 flex on it. It is on the end of a long circuit with a 32A MCB at the origin. The plug has a 13A fuse in it. What happens when you cut through the flex next to the lamp with wire cutters when it is powered on?

The first thing you work out the prospective fault current. So you need to add up the external earth loop impedance, the circuit resistance to the socket, and the resistance to the far end of the flex.

Let's say that is 0.35 ohms for the external earth loop, another ohm in the circuit wires, and another 0.15 ohms for the far end of the flex. That is 1.5 ohms. So the fault current at the far end of the flex could be 230 / 1.5 = 153A. Sounds like a lot for a 3A flex!

However on the bright side that is comfortably enough to blow the fuse in the plug, so it will only carry that current for a very short time. Let's assume 0.1 secs (in reality it could be much less - fuses can be very fast for massive fault currents).

We can assume all the cables have PVC insulation - so we can use the "k factor" of 115, in the equation:

s = sqrt( I^2 . t ) / k

s = sqrt( 153^2 x 0.1 ) / 115

s = 0.42mm^2

So you need at least 0.42mm^2 of copper cable area to be able to safely clear the fault by blowing the plug fuse, and we can conclude that the

[1]

UK standards are WAY above EU standards

As has been pointed out the idea of a plug fuse is to protect the cable from catching fire under fault conditions.

The unit itself should have an internal fuse to protect itself.,

Frankly US style and Europlug style electrics are sketchy as f*ck