# Effective Power and Load Behavior

• posted

Suppose you have voltage = 110 volts and current 2 Ampere

From P = V I

Power = 220 Watts

Suppose you have voltage = 12 volts and current 18.3 Ampere

Power = 220 Watts

What is the behavior of the load in each case? They both have the same 220 watts power yet the second one with a

18.3 Ampere current should be affected more.. yet if the voltage in the first is 110 volts.. the power is similar...

What is the behavior of loads with more voltage or more current. Which can it take more. What part is affected more, etc.

Pls. elaborate (also the physics part of it). Many thanks.

Qude

• posted

no answer possible from what you are asking as your terms are not defined. what do you mean 'affected more'?? what type of 'behavior' are you thinking of?

• posted

For example, what creates brighter light bulbs. A 110V/2Ampere bulb or a 12 Volts/18.3Ampere bulb, both produce 220 watts of power. what's the occasion when you must choose the 110V/2A or 12V/18.3A in designing loads.

qude

• posted

you pick the voltage based on other considerations... usually by what is available for whatever you are designing. i.e. if you are designing for a house or office you use 110v, if you are designing for a car you use 12v. or from what the load you have to power needs. i.e. if the load is designed for 110v you give it 110v, if its designed for 12v you give it 12v. from that you calculate current needed and how big of a wire you need for it.

• posted

defined.

thinking

You design it based upon what voltage the user will have available. If you manufacture an amplifier for use in homes in North America, and design it so it will work off of +-31Vdc, you're gonna have trouble selling it and anyone who buys it will have trouble using it. If you make it so it works off of

120Vac or thereabouts that's a great idea. If you make an amplifier for use in cars in North America, then you would probably design it to work off of 12Vdc or thereabouts.

If there is some situation whereby there is no voltage constraint like that, then typically a moderate voltage and moderate current work well. High voltages require lots of insulation. High currents require heavy conductors and lose a lot of power to parasitic resistance.

j
• posted

I am so bad at maths I'll let you do P = I^2 R and see the resistance must be much lower for the 12 volt load to consume the same power.

That also means a conductor like nichrome or tungsten will be fatter and shorter than the 110 volt load. This can be a real mechanical advantage for hot envelope halogen lamps because a long thin coil of wire can flop around, shedding, fatiguing or actually striking the envelope.

This is why you don't see this type of lamp below ~100 watts in the

110 volt variety. Above ~300 watts the filiment can be short an fat so doesn't require too much mechanical support.

That isn't quite what we call physics and I am guessing... lamps.

Sue...

• posted

I'm just interested in the theoretical side of it.

Suppose my source can equally do a 220v/2A and 12V/18A. If I get a light bulb for each. What would be brighter.

Or better yet. I'm interested in the physics side of what's going on inside the wires. In the 12V/18A. More electrons flow in the wire.. however the voltage is only 12V. In the case of the 220v/2A. Less electrons flow in the wire. So what is the function of the voltage or potential difference. In the 220v/2A case. There are fewer electrons but bigger potential difference. What is the effect of it on electron flow compare to the larger 18A but lower voltage or potential difference. Since electron flow has same speed in all ampere rating. What is the function of the voltage in the behavior of the current?

qude

• posted

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conductors

Well, the bulbs will be different to suit the voltage and current to be applied. These construction and materials differences would determine brightness and other light characteristics more than the voltage and current per se. In either case the voltage and current deliver energy at the same rate to the filament.

Hmmm. I think the electron flow does not have the same speed in all ampere ratings. I would think that

I = v * A * p

I current (C/s, or A) v drift velocity of electrons (m/s) A cross sectional area of conductor (m^2) p density of free electrons in the material that the wire is made of (C/m^3, or # free electrons/m^3 * 1.6E-19)

A and p would basically be fixed for a certain conductor. So drift velocity should vary with current. Though I may be proven wrong. Thinking about Ohm's law, this all seems to work out nicely, in my head.

Are you familiar with the difference between (a) the speed of light in a conductor (on the order of 2*10^8m/s), (b) the instantaneous speed of an electron as it pinballs its way down a conductor (on the order of 10^6m/s), and (c) the "drift velocity" or overall average speed with which an electron makes its way down a conductor (I'll say 10^-4m/s to give a feel of the magnitude but I think it would vary hugely)?

j
• posted

you

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Don't get wound up in electron flow. The individual electrons just drift along and, in case of AC, wobble around a bit.

Note that power depends on both current and voltage, not just current alone. The theoretical answer is that both the 12V/18A and the 220V/2A lamps the power and the brightness is the same assuming the same filament temperature (-which depends also on non-electrical factors). What appears to be more electrons passing a given point at 18A than at 2A really does not determine this.

The voltage is the driving force behind the current-Your 12V lamp has a resistance of 12/18 ohms while the 220V lamp has a resistance of 220/2 ohms. If you applied 220V to your 12V lamp- it would be extremely bright momentarily just before it blew up as the current would be 330A.

For any application there is an optimal voltage range. For automobiles 12V was it but more recently moving to 48V has been considered because of higher electrical loads. For homes, 12V is impractical and 120/240V is better. For power distribution, this is too low, etc. -- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

• posted

------------------------ There isn't one same load. They are two different loads, their resistances differ wildly.

R = V/I

-Steve

```--
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Electronics Site!! 1000's of Files and Dirs!!  With Schematics Galore!!```
• posted

------------------------ They would be two bulbs of different resistance, both the same brightness.

------------------------ Two push their currents through two totally different resistances.

-------------------- Oh no, that's totally wrong!! Electrons don't all "go at the speed of light" or some such baloney. They travel at the drift velocity for those circumstances. You need a physics course.

-Steve

```--
-Steve Walz  rstevew@armory.com   ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!!  With Schematics Galore!!```
• posted

E=IR or I=E/R

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• posted

The effective resistance of each load is different for the two cases you describe...

1) 110/2 = 55 Ohms 2) 12/18.3 = 0.66 Ohms

If these were light bulbs the wire filament needed to make the 12V lamp (0.66 Ohms) would be much thicker than the 110V lamp (55 Ohms). Therefore the 12V lamp would be more robust and resistant to the thermal shock that occurs when it's switched on/off. Indeed low voltage Halogen downlight bulbs do last longer than mains voltage lamps.

• posted

Thanks for the answers but I want understand the physics side.

When 18A is said to flow into a wire. What is the difference of it in terms of electron flow compared to lets say 2A.

What does the voltage do to the current. It drives them you will say. But how come you can drive a 18A current with only 12 volts whereas in a

110 volts, the current is only 2A.

It may have to do with the load. So in loads with lower resistance. More electrons can pass thru it, right. In the case of the 18A/12 volts supply, more electrons pass thru the load. In the case of the 2A/110 volts. Only few electrons pass thru it.

Now here's the problem.

That means using higher voltage didn't mean pushing greater amount of electrons. So what is the function of voltage and current in terms of electron movement?? Let's focus on the physics side of it.

qude

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In response to what qude posted in news: snipped-for-privacy@f14g2000cwb.googlegroups.com:

It is 9 times greater. But current doesn't flow 'into' a wire, it flows THROUGH it.

A guy called Ohm explained this ages ago. I = V/R. Current is proportional to voltage, and inversely proportional to resistance.

Yes.

Voltage difference - otherwise known as EMF (electromotive force) is the force that drives the electrons, resistance opposes that movemnt. So more emf, or less resistance, both mean more current flow.

-- Joe Soap. JUNK is stuff that you keep for 20 years, then throw away a week before you need it.

• posted

Clicking around these pages should get it on your own terms:

Sue...

• posted

After thinking it over. I got the main idea. So the load is related to it all. But you said lower resistant load requires thicker material.. What's the physical basis? Is it because when the material is thicker, there are more spaces for the electrons to pass through that's why resistance is less. While in high resistance load, it needs thinner material so the electrons would have to struggle to pass thru it and hence the entire electron line is adjusted to be lesser (lower current)??

If right. Well. How do you select material where thicker means more electrons can pass thru and less resistance yet have same lighting power to the thinner wire with more resistance? Can you use same material to build thicker or thinner parts or does it depends on material properties for each (higher or lower resistance) application?

I want to be able to visualize it all and not just memorizing formulas and equations. Thanks.

qude

• posted

That is pretty much what is happening. Instead of "more spaces", people usually think in term of more area (cross-section), but you have the right idea.

The material selected is the one that can withstand the highest temperature and conduct electricity. Of all known materials, tungsten is the one that fits this description. As far as I know, all light bulb filaments made these days are made from tungsten, regardless of power or voltage or other requirements.

Then the wire's length and thickness must be chosen. This can be done by knowing the necessary resistance and operating temperature of the wire. The resistance can be calculated from the available line voltage and the power.

Operating temperature is usually chosen to be around 2200 to

2800 C for tungsten. Hotter temperatures result in unreasonably short burn-out lifetime. Lower temperatures result in less efficient operation (less light produced per Watt of electricity).

Even though many length & thickness combinations will give the necessary wire resistance, only one combination of length and thickness will give BOTH the desired resistance AND operating temperature for the wire. A wire of that length and thickness will only "work" for a given supply voltage. At lower voltages, the wire will run less efficiently. At higher voltages, the light will be more efficient BUT will burn out faster.

For resistance, visualize that a longer path length OR smaller cross-section area will increase the resistance:

Resistance is proportional to ( Length / diameter^2 )

For temperature, a smaller outside surface area will increase the temperature, since the power leaves the wire mainly by radiating from this surface.

Temperature depends (approximately) on ( Diameter / Length^2 )

Mark

• posted

It has to do with more electron collisions in resistors to produce the increased temperature or heat (and hence light). But something eludes me. If you make the resistance of the tungsten load higher, meaning thinner... there is more electron collisions to the lattice hence more friction and light. Now in lower resistance, the area is larger so less frictions. Isn't one wasting electrons in this latter since more electrons pass thru the lower resistance and didn't strike much lattice but some just passing thru the space in between??

qude

• posted

You can think of a lower resistance meaning that there are fewer collisons between electrons and the atoms in the lattice, BUT: If the wire is connected to a constant VOLTAGE source, then each collison will be more energetic (the electron gains more energy in between collisions than it would in a higher resistance material). This higher energy per collision actually results in more power dissapated, so more heat and/or light is generated.

If the wire is connected to a constant CURRENT source, then the lower resistance results in less voltage pushing on the electrons. The net result is less power (and heat, and/or light).

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