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Discussion Starter · #1 ·


in the image above is two ideas
the one below is how i wired 2 normal lamp fittings together ( for spiral 650k energysavers) and they turn on

how ever im concerned that this is the wrong way to wire it and the top one is how i feel it should be done theres no way for me to do the top way as where the smaller wires connect to the lamp theres no room for an extra wire leading to the 2nd lamp.

any advice is good!
 

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Hi NoRp,

They both accomplish the same thing, but the first way there is a possibility that the first bulb in the circuit will burn brighter than the second bulb.

The second way utilizes more connections which could result in connection failures. How about a third choice?

Just like your top diagram, blue wires stay where you show them. Brown wire goes to the far bulb first, then to the closer bulb. Same number of connections, but the bulbs should burn equally bright!
 

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Discussion Starter · #3 ·
thank you for the advice

so from the plug the brown goes to one light the blue to another and then a normal wire joining the 2?
 

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Hi NoRp,

No, that is not what I said. Blue wires stay just as you show in the first drawing. The brown wire goes from power supply to the light on the right first and then from there the brown wire goes the light on the left.
 

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If you are using the right size wire the two connecting methods are exactly the same. Both bulbs will be equally bright. Use which ever method will work in the space you have.
 

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I second hoppy. The first diagram is how all house wiring is done. The bulbs are connected in what is referred to as "parallel", rather than in "series". The order of wiring really only matters if the wire is real long (20-40 feet), even then the problem is easily solved with a thicker wire.

If you notice that one bulb is brighter, swap the bulbs. If the bulb is brighter in the new location, don't sweat it. Some bulbs do that, esp if one is older than the other. If the bulb is brighter in the same fixture, sweat. You need a thicker wire or your light fixture is a fire hazard. Please note that any power cord wire should work just fine.
 

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Discussion Starter · #7 ·
wow guys thanks for all the help.

so using my method ( the 2nd drawing) would it be posible to conect up 3 lights like that?


withing my lighting unit i wanted the lights to go

2,3,3,2

recon that would be ok
 

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I have done a lot of electric wiring on various houses I have remodeled, and I used whatever wiring method would fit well when I put light fixtures in parallel, or outlets in parallel. You can use your circuit for as many lights as you want, as long as you are using the size wire that goes with the circuit breaker rating. The problem with your circuit is that it gets increasingly hard to keep all of those wires in good contact with each other, and a wire nut gets to be very hard to use. Three wires, or 4 wires together isn't too bad, but more can be a nightmare.
 

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Hey guys,

Electrically they're wired exactly the same. They are both in parallel. So hook it up either way, you won't notice a difference.

I've attached a pic of what it would look like in series.
 

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Obviously you shouldn't connect lights in series, but if you did you wouldn't cause a fire. You would just have very dim light, as the current through each bulb would be half of normal. And, the power used by each bulb would be one quarter of normal, so the bulbs would be very dim.
 

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well, ok, so maybe no fire....

but it would still be bad ;)

cfl's probably wouldn't even light that way anyway.
I'm pretty sure the CFL's would just sit there giggling at the idea that they would even think about getting fired up that way! Seriously, it is possible they would get hot, since this would be comparable to trying to use them with a dimmer, a big no-no.
 

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Obviously you shouldn't connect lights in series, but if you did you wouldn't cause a fire. You would just have very dim light, as the current through each bulb would be half of normal. And, the power used by each bulb would be one quarter of normal, so the bulbs would be very dim.
You're right that the light bulbs would be dimmer in series but your math is all wrong.

Current is the same at all points in a series circuit. Voltage on the other hand will be shared among the lights thus causing the lights to be dimmer.

Ok lets say there's 120Volts and you have 2 60 Watt bulbs.

Current = Power/Voltage
I = 120Watts/120Volts
I= 1Amp

This circuit will have 1 amp flowing through it.

So take one of the 60 Watt bulbs in this circuit and you want to find the voltage for that bulb.

Voltage = Power/Current
E = 60Watts/1Amp
E = 60 Volts

As you can see, each bulbs only gets 60Volts per light bulb. (Dimmer Lights)

Power rating will be the same.
A light bulb rated at 60 Watts will continue to be 60 Watts.

Power = Voltage * Current
P = 60Volts * 1Amp
P = 60 Watts

Ok, now for parallel.

Voltage is the same in a parallel circuit so each light bulb with the same wattage rating will have the same intensity.

Current = Power/Voltage
I = 60Watts/120VAC
I= .5 Amp

So each light will get half an Amp and receive 120V. Current for this circuit is still at 1 Amp if you have 2 60 watt bulbs in parallel. The same as in series.

Power = Voltage/Current
P = 120V/.5Amp
P = 60 Watts

Power is the same per light bulb as in series.

The difference is that voltage is halved with a series circuit.

Add too many lights without increasing the gauge of wire and yes you'll have a fire. Just make sure you have the proper gauge for the current you'll be using.

You'll see an increase in current with added lights.

Current = Power/Voltage
1 60 Watt bulb
I = 60Watts/120Volts
I = .5 Amps

2 60 Watt bulbs
I = 120Watts/120Volts
I = 1 Amp

3 60 Watt bulbs
I = 180Watts/120Volts
I = 1.5 Amps

Sorry Hoppy, Couldn't help myself. :mrgreen:
 

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Snydaleid:
There's a slight problem in your math. A 60W bulb only draws 60W when connected to a 120V source. If the applied voltage changes, the power draw changes accordingly. Power is a derived unit of voltage and current P =VI. The Voltage and Current are related by the electrical impedance of the bulb. V = IR. You must determine the impedance of a bulb to determine the power of a bulb connected in series.

My math:
Single 60W bulb:
V = 120
P = 60W

Current = Power/Voltage
I = 60Watts/120Volts 0.5Amp

So the impedance of a single bulb is:
R = V/I = 120V/0.5A = 240ohm


Two Bulbs in Parallel.
Itotal = V/(1/(1/R1 +1/R2) = 120V /(1/(1/240ohm + 1/240ohm) = 1amp
1amp through circuit.

current per bulb = Ibulb = Itotal/#Bulbs = 1A/2bulbs = 0.5amp/bulb.
Voltage per bulb = Vbulb = Ibulb x R = 0.5 x 240ohm = 120V

total power = Vbulb = V x Itotal = 120V * 1A = 120W.
power per bulb = Vbulb x Ibulb = 120V*0.5amp = 60W

Two Bulbs in series.
Itotal = V/(R1 + R2)= 120V/(240ohm + 240ohm) = .25 amp
0.25Aamp through circuit.

Current per bulb = Ibulb = = 0.25amp
Voltage per bulb = Vbulb = Ibulb x R = 0.25 x 240ohm = 60V

total power = V x Itotal = 120V x .25A = 30W.
power per bulb = V x Ibulb = 60V x 0.25amp = 15W

The math is valid for incandescent lights. I don't know how under-voltaging affects a CF bulb. I also don't know if the effective impedance of a CF bulb is constant over applied voltage.
 
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