Which Amperage is correct?

[AntHalls]

I need to work out the amperage concerning a light. It is an IKEA single LED spot clip on light. The wire from the light plugs into a transformer plug. The details on the plug are as follows.

PRI: 100-240V 50/60Hz 0.05A
SEC: 7V 0.43A 3W

I fully understand the equation and calculation converting Watts to Amps, but, which of the two voltages do I need to use, as I have 10 of these lights and need to submit my amperage for all 10.

Do I use 3W divided by 240V X 10 = , or 3W divided by 7V X 10 = ?

 

[Ian1981]

Deleted

 

[Pete999]

I Amps = P watts divide by V voltage 3/7=0.428 Amps

 

[mhar]

1 light draws 0.05A max @ 240V according to what you have written above, so 10 lights will draw .5A max.
Are you trying to calculate the load on the circuit or trying to size 1 transformer to supply all 10 lights?

 

[AntHalls]

Thanks for replying. I can only plug a set number of lights into a provided supply of a 3KW socket. I understand that it is a 13A supplied socket. I have other lights as well. If I calculate the amps using the transformed 7 volts, I will be drawing 4.3amps. If I use the 240 volts the amperage is as you say o.5 amps. I am using the lights singly plugged in one at a time. I have permission to daisy chain 3 extension cables at no more than a total of 7 metres between the 3. The total amps must not exceed 7 amps at the third cable. If I am able to calculate the amps via the 240 volts the amps will obviously be low. If however if I have to calculate via the 7 volts, I am already using over half my allowance. I am hoping someone can tell me which of the two I am allowed to use as I have to accurately give my amperage. It will determine how many lights I can use overall.

 

[davesparks]

The transformer will draw a maximum of 0.05A at 240V at its full load output of 0.42A at 7V.
However this does not tell you what it will actually draw with the particular light you connect to it.

 

[davesparks]

If I calculate the amps using the transformed 7 volts, I will be drawing 4.3amps. If I use the 240 volts the amperage is as you say o.5 amps.

Yes the output current will be 4.3A at the 7V and the input current will be 0.5A at the 240V.

 

[Leesparkykent]

Why do you need the total if your only concerned with the 3rd extension lead?

 

[AntHalls]

Thank you for your replies. If I only use one extension cable, I must not exceed 12 amps if only using 1 cable, using 2 cables reduces it to 10 amps, and using a 3rd cable, my amperage must not exceed 7 amps. I have many lights and would have to use 3 cables with several sockets. By the time I have plugged into the sockets available, I might exceed my allowed amperage usage of 7 amps after filling the 3rd cable. That is why I am asking if I have to calculate my amps with the lower voltage. If so, I might have to re-think the type of lights I would need to use. That would mean buying new lights.

 

[telectrix]

your current calculation is based on the 240V, as that's the current drawn at that voltage. end of.

 

[AntHalls]

Thank you telectrix.

 

[AntHalls]

Thanks for all your input.

 

[davesparks]

Thank you for your replies. If I only use one extension cable, I must not exceed 12 amps if only using 1 cable, using 2 cables reduces it to 10 amps, and using a 3rd cable, my amperage must not exceed 7 amps. I have many lights and would have to use 3 cables with several sockets. By the time I have plugged into the sockets available, I might exceed my allowed amperage usage of 7 amps after filling the 3rd cable. That is why I am asking if I have to calculate my amps with the lower voltage. If so, I might have to re-think the type of lights I would need to use. That would mean buying new lights.

Why does the number of cables reduce the maximum current?

As I said earlier, the input current will be the 240V rating, the output current will be the lower voltage rating.

 

[AntHalls]

I take part in craft fairs and the Health and Safety Officer has specified the drop in amps through the cables as they are added. Daisy Chaining is, for obvious reasons frowned upon. They are limiting the risk of overload. I just wanted to make sure I stayed within the required limit. Thanks for explaining the situation to me. My assumption was that the output current was higher amperage, therefore I had to use that.