Amps through Christmas tree lights


by biferi
Tags: amps, christmas, lights, tree
biferi
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#1
Feb5-13, 01:20 AM
P: 193
I always wanted to know how this worked?

I know AMPs is how many Electrons move past one point in a SEC.

I know that Voltage is the Force that drives the AMPs.

I know that a House Outlet will supply 120 Volts at 15 Amps.

Now take Christmass Tree Lights and to keep this easy lets just talk about the old Lights the ones that would just be Lights in Series.

I know there were say 100 Lights in Series on the strand but still these Lights are so so little.

Why do they not Blow Up when 15 Amps goes through them?
Thanks
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Low-Q
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Feb5-13, 07:37 AM
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Quote Quote by biferi View Post
I always wanted to know how this worked?

I know AMPs is how many Electrons move past one point in a SEC.

I know that Voltage is the Force that drives the AMPs.

I know that a House Outlet will supply 120 Volts at 15 Amps.

Now take Christmass Tree Lights and to keep this easy lets just talk about the old Lights the ones that would just be Lights in Series.

I know there were say 100 Lights in Series on the strand but still these Lights are so so little.

Why do they not Blow Up when 15 Amps goes through them?
Thanks
You would have a very bright tree when 15 amps goes through the bulbs. You'll have about 1.8 kW.
However, even if your fuse is rated to 15 amps it does not mean that everything you are powering draws 15 amps.
When nothing is connected to the 120V outlet, there is no amps at all. The amps depends on the load. My wild guess is that your tree pulls about 0.5 amps (approx 60W total). If you have 100 bulbs on that tree, over each single bulb it rests about 1.2V - which means that each bulb get about 0.6W each.

Vidar
NascentOxygen
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Feb5-13, 08:11 AM
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Quote Quote by biferi View Post
I know there were say 100 Lights in Series on the strand but still these Lights are so so little.
When you connect a series string of 100 identical bulbs to the 120V outlet, it means that each individual bulb 'feels' just 1.2V across its own terminals. So each gets only a low power. The power to each is about the same as if you brought out a single bulb from the string and connected an AA cell* to that single bulb. It would be like a single-battery flashlight. Not very powerful.


* rechargeable NiMH cells have a terminal voltage of nominally 1.2V

russ_watters
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#4
Feb5-13, 05:22 PM
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Amps through Christmas tree lights


Christmas tree lights don't have anywhere near 15A running through them. 15A is just what the circuit is capable of providing if the connected load demands it.

With a quick google, I found a 42' strand of 100 lights and it says you can connect up to 3 together. So the strands pull about 5A apiece.
biferi
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#5
Feb5-13, 11:29 PM
P: 193
So if I have 100 Lights in series the Amps across each Bulb stays the same but the Voltage will ADD right?

So each Bulb will Divid the 15 Amps over 100 Bulbs if I get this right.
sophiecentaur
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Feb6-13, 04:24 AM
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Quote Quote by biferi View Post
So if I have 100 Lights in series the Amps across each Bulb stays the same but the Voltage will ADD right?

So each Bulb will Divid the 15 Amps over 100 Bulbs if I get this right.
Each word in Science has a specific meaning and you have to use them correctly in your answer, if you want to understand what's happening. If someone talked about your favourite sport or music and they used the wrong words, it would just not make sense either and you couldn't have a sensible discussion about it, could you? (You would roll your eyes and give up, probably.)
The string of series connected string of bulbs Share the Voltage between them. They do not "share the current".
The supply Volts are the thing to start with. You then need to know the Resistance of the Load (in this case, it's 100 times the resistance of one bulb, because they are in series). The Current that passes will be the same for each bulb and will be (Total Volts)/(Total Resistance)

I'm afraid you need to be precise or your results can end up as nonsense.
tiny-tim
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Feb6-13, 04:38 AM
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hi biferi!
Quote Quote by biferi View Post
I know that a House Outlet will supply 120 Volts at 15 Amps.
Quote Quote by biferi View Post
So each Bulb will Divid the 15 Amps over 100 Bulbs
sorry, but you're fundamentally misunderstanding what the 15 amps means

15 amps is only the "strength" of the fuse in the circuit

it's the maximum the circuit is allowed to take (if the current goes above 15 amps, the fuse melts, safely in its own little container, which is a lot safer than the wiring melting, and setting fire to the whole house! )

as russ says
Quote Quote by russ_watters View Post
15A is just what the circuit is capable of providing if the connected load demands it.
it has nothing to do with the actual current in the circuit
biferi
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Feb6-13, 01:50 PM
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Ok let me start over.

If I have a 5 Volt Bulb and a 20 Volt power supply I know the my power supply will give to mutch power to my Bulb and the Bulb will burn out.

So I need to find out the Resistane I will need this is E Divided by I = R so this tells me I need a 4 Ohm Resistor.

Just tell me if I have this right?
tiny-tim
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Feb6-13, 02:11 PM
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Quote Quote by biferi View Post
If I have a 5 Volt Bulb and a 20 Volt power supply I know the my power supply will give to mutch power to my Bulb and the Bulb will burn out.

So I need to find out the Resistane I will need this is E Divided by I = R so this tells me I need a 4 Ohm Resistor.
i don't think you can get a 5V bulb

(or did you mean a 5 amp bulb? i don't think you can get that either)

the fixed quality of a bulb is its resistance

unfortunately, in my experience, the resistance isn't marked on the bulb!

instead the bulb is marked eg 120V 60W, meaning that if you use it where the mains voltage is 120V, then the power will be 60W

now suppose you bring that bulb over here, to london, where the mains voltage is 240V, and you still want to get 60W out of it how much resistance should you place in series with it?

well, the resistance of the bulb is V2/W ohms, ie 1202/60 = 240 ohms,

and you want to halve the voltage, ie have half the voltage drop across the resistance, and the other half across the bulb
so obviously, you need 240 ohms
(i've done this for DC i think AC works out differently)
Windadct
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Feb6-13, 02:44 PM
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Hello Biferi - your resistance idea works in theory, but you need to know something else about the bulb - usually Wattage (power) but some smaller DC bulbs may list the current.

You then need to calculate the bulb's rated current and then select a resistor.

Also note - if you measure a light bulb's resistance with a meter it will not give you the resistance you need for this calculation, the bulbs resistance will be too low, as the bulb heats up ( and subsequently glows) the resistance increases - this is called a Positive Temperature Coefficient -
sophiecentaur
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Feb6-13, 04:08 PM
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Quote Quote by biferi View Post
Ok let me start over.

If I have a 5 Volt Bulb and a 20 Volt power supply I know the my power supply will give to mutch power to my Bulb and the Bulb will burn out.

So I need to find out the Resistane I will need this is E Divided by I = R so this tells me I need a 4 Ohm Resistor.

Just tell me if I have this right?
This post isn't clear. Are you suggesting that a resistor in series with the bulb could allow the bulb to operate correctly on the 20V supply? If you are, then you would want to 'drop' 15V across the resistor, leaving 5V for the bulb. You would need to know either the Wattage or the Resistance of the bulb first so that you would know what resistance to use for this method to work. But the best way would surely be to put four bulbs in series.They would operate perfectly with a 5+5+5+5V supply (20V) as long as all the bulbs have the identical rating (i.e. same resistances)
biferi
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Feb6-13, 04:47 PM
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I know that you can not get them.

I just ment this is the right formula to find out what kind of Resistor I need
E Divided by I = R am I right?
sophiecentaur
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Feb6-13, 05:08 PM
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Quote Quote by biferi View Post
I know that you can not get them.

I just ment this is the right formula to find out what kind of Resistor I need
E Divided by I = R am I right?
You did the wrong sum - you need to subtract the lamp volts first! But, of course, in general, R=E/I
benjaminbailes
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Feb7-13, 08:39 PM
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Quote Quote by biferi View Post
I always wanted to know how this worked?

I know AMPs is how many Electrons move past one point in a SEC.

I know that Voltage is the Force that drives the AMPs.

I know that a House Outlet will supply 120 Volts at 15 Amps.

Now take Christmass Tree Lights and to keep this easy lets just talk about the old Lights the ones that would just be Lights in Series.

I know there were say 100 Lights in Series on the strand but still these Lights are so so little.

Why do they not Blow Up when 15 Amps goes through them?
Thanks

I saw your follow up question too I just want to explain these basics the way I understand them.

Voltage- You can think of voltage as pressure, like water pressure in a pipe. The water being the electrons. The pump being comparable to a generater and the size of the pipe being comparable to the size of a conductor. Also though not quite correct you can think of voltage as the speed that the electrons are travelling. Maybe a little of both just to help visualize it.

Amps- Amps are as you wrote above: electrons moving through a point in a conductor per second. Just think about amps as the number of electrons to help to visualize everything.

Resistance- Different materials have different resistances to the flow of electrons. A small wire has more resistance to the flow of electrons than a larger wire. Similarly a small water pipe has more resistance to the flow of water than a larger pipe.

Watts-Watts are the combined power of amps and volts. A*V=W

Say you have a 120 volt ac power supply and a 60 Watt lightbulb. The tungsten filament in the lightbulb will resist the flow of electrons and in doing so create a sort of friction that gives of heat and light. That circuit would draw 0.5 amps. Using r=v/I (r= resistance in ohms, v=volts, and I= current in amps) you would get a resitance of 240 ohms from that lightbulb.

120/0.5=240

In this circuit the electrons flow through the wire with little resistance until they get to the filament. The small size of the filament wire slows the flow of the electrons.

I could go on and explain more and in detail but if you have some specific questions it would be a lot easier to write.


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