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plug socket
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i have a double plug socket in the wall.what i would like to know is how much can i plug into them,surely there must be a limit.i seem to think that 100w is the most but am confused can someone tell me an easy way to calculate how much cos at the moment i have one of the plug sockets with a four way extension lead and the other a single plug but i need to have another plug in.would it be safe to have an extension lead in both.each of them enabling 4 plugs in each.look for to an educated reply as per normal
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If your sockets are part of a standard UK ring main system then the maximum load per ring main is 7,200 watts, the ring being protected by either a 30A fuse or 32A circiut breaker at the consumer unit.. This can be distributed over all of the sockets or it can all be taken by one socket. If it is not part of a ring main then check the fuse/circuit breaker rating at the fuse box / consumer unit.
Assuming it's part of a normal ring-main circuit, you can draw 13 amps, that is, 3000 watts (3kW) from each socket, either with a single appliance, or the total drawn by appliances using an extension lead.
Bear in mind, though, that this only applies IF the extension lead is a 13A (3kW) one - most are less - typically 10, 6, or 3 amps, which would carry 2400, 1400 and 700 watts respectively.
W = V x A A = W / V V = W / A
Bear in mind, though, that this only applies IF the extension lead is a 13A (3kW) one - most are less - typically 10, 6, or 3 amps, which would carry 2400, 1400 and 700 watts respectively.
W = V x A A = W / V V = W / A
Well 4candles is right each socket outlet can only take a maximum of 13 amps as the plug top that's plugged into a BS1363 socket has a maximum rated fuse of 13 amps. A 4 way extension is a good idea as I get asked loads of times, "if I plug my tv, video, table lamp, sky or dvd into all of the 4 sockets can I overload the extension lead" well the answer is no as the plug top on the ext lead has a fuse no bigger than 13 amps so you can't overload the lead or the socket that the lead is plugged into.
A simple way to work out the loadings of appliances is look at the wattage: i.e a kettle is 2200watts so 2200watts divided by the voltage 230 = 9.56 amps.
Watts divided by voltage = amps
Easy isn't it. BTW ive been a sparky for over 20 years....
A simple way to work out the loadings of appliances is look at the wattage: i.e a kettle is 2200watts so 2200watts divided by the voltage 230 = 9.56 amps.
Watts divided by voltage = amps
Easy isn't it. BTW ive been a sparky for over 20 years....
It is not correct that the Watts divided by the Volts will give the current (Amps) used by an appliance. While the simple case is true for devices like heating elements it is not that simple for things like electric motors and transformers. These appliances draw more current than would be expected for their Watts.
The "Power Factor" must also be taken into account. It represents the "true power" divided by the "apparent power". Apparent power is measured in VA which is the Volts times the Amps.
Amps are what matters when it comes to the maximum allowable load.
The "Power Factor" must also be taken into account. It represents the "true power" divided by the "apparent power". Apparent power is measured in VA which is the Volts times the Amps.
Amps are what matters when it comes to the maximum allowable load.
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