calculating power

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hi all i am looking to install an outdoor pir security light replacing an old fitting on outside wall obviously this new light will need increased wattage and just wanna make sure im not over-loading the circuit.
fuse is 6 amp mcb i already have 3x100 watt lights, but in kitchen i have 20 spots which are on transformers 12v 50w bulbs can anybody help to calculate my current ampage in this circuit please???

thanks very much
 
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You've been doing 2330 for 2 years and you don't know the relationship between volts, amps and watts? :eek:
 
well i was just making sure as wasnt sure wat effect the transformers had on the power
 
Total watts = (3x100) + (20x50) = 1300

Current = 1300/230 = 5.7A

That's not even allowing for any losses in the transformers, so in reality that circuit is completely maxed out already. Outside light will need to go on a different circuit or perhaps you might consider reducing your ridiculous kW kitchen load by using CFLs or LEDs instead of the halogens. Or more suitable luminaires :)

Liam
 
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probably even a little less amps as i have yet to see true 230 volts supply.

usually around 237-239 still
not 230

check with your meter......
 
If we are going to split hairs then the wattage is at the rated voltage of the lamp. As the voltage is decreased the current will also decrease and vice versa.
 
You've been doing 2330 for 2 years and you don't know the relationship between volts, amps and watts? :eek:

Believe me BAS this is not an uncommon occurence with 'prentices currently.

Very few of them seem to think they even need to actually carry out an IR test!!! They think it will always be >299 :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes:
 
I'm just as concerned when they don't have a clue about an earth fault loop impedance test!
I find nowadays apprentices are trained in how to fill in paperwork and given little knowledge about the actual job.
 
probably even a little less amps as i have yet to see true 230 volts supply.

Nope - more amps. If the lamps are rated 50W @ 230V, and the supply voltage is in fact higher (probably is, as you point out) the lamps will draw more current and run at a higher power.
 
have to disagree. I have just completed my 1st year of 2330 and can work out ohms law just about. ;)
 
You've been doing 2330 for 2 years and you don't know the relationship between volts, amps and watts? :eek:

Believe me BAS this is not an uncommon occurence with 'prentices currently.

Very few of them seem to think they even need to actually carry out an IR test!!! They think it will always be >299 :rolleyes: :rolleyes: :rolleyes: :rolleyes: :rolleyes:
Never mind all that - what about the OP's question?

It is what was 'O' Level Physics. I don't know if it's in the GCSE physics curriculum, but the point is that its complexity is at the early teen level.

And he's been doing 2330 for 2 years? Is he at the same college as kevin1234?
 
have to disagree. I have just completed my 1st year of 2330 and can work out ohms law just about. ;)

Ohm's law - let me see, I think I was about 11 when I worked that one out. Two-way switching was a little harder - I think that took till 13. This stuff is so basic I am amazed how anyone can have any difficulty with it.
 
It is hard to credit, but what really amazes me is that people who claim to be, or think they could become, electricians can have any difficulty with it.
 
liampope:

1300 watt divide 230 = 5.7A

1300 watt divide 240 = 5.41A

the more voltage with the same current is LESS amps as there are more volts to divide the amps by!
 

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