Questions about voltage and amperage

Factual Questions
1

I have a couple of questions about electrical engineering. First, I’d like to know how voltage affects power. Everyone knows that electrical power equals voltage times amperage. I just don’t understand why though. I was taught that amperage is the amount of electricity flowing through a circuit and voltage is the pushing force behind it. It’s easy to understand how amperage would affect power, but I just don’t see how the pushing force behind the amperage could also affect power, other than indirectly by causing more amperage to flow. Are the flowing electrons more energetic or something? I hope someone can explain this in a way I can understand.

Also, heat caused by electricity (such as in a lightbulb) is equivalent solely to amperage, isn’t it? If so, why don’t electrical devices that rely on heat to work, such as lightbulbs and heaters, use a step-down transformer so that they can use that amperage at a much lower voltage thus using less power and costing less? Or am I mistaken and heat is actually equivalent to power?

2

Consider voltage to be equivalent to the height of water behind a dam and the current to be the amount of water/second that goes through the turbines at the bottom of the dam. The amount of work done in one second by the water falling through that height is the height X the weight of water/sec. The units of power are ft-lb/sec which is the same units as height X weight/sec.

The height of water behind the dam is a measure of the energy stored in the water relative to the bottom of the dam. It is just as if the water had been lifted against the gravitational field from the bottom up to the top of the dam and when the water falls back down that energy can be recovered. The voltage between two points is a measure of the potential energy of the electrical charges that have been moved against the electric field from the point of low potential (voltage) to the point of higher potential. When the charges are allowed to “fall” with the field through a conversion device the stored energy is recovered. The power expended depends upon how rapidly the charges fall, i.e. the number of charges/second that leave the high potential point, and that is the current.

Heat is energy and power is energy per unit of time. So the heat generated by the bulb/ second is in the units of power.

The amount of light/second out of the bulb is also power and if you want to get out a certain amount of light/second you have to supply that much power. Whether it is at a high voltage and low current or a low voltage at high current is immaterial. If you use a step down transformer to reduce the voltage to, say, 12V on a bulb made to operate at 120 V. then the current will be very low because of the relatively high bulb resistance and the filament will not be heated by much. The heat generated in a conductor is proportional to the square of the current (actually the current density but that’s a fine point for our purposes) so that when you reduce the current in the bulb because of low voltage you greatly reduce the temperature of the filament.

If you have more questions ask them and either I or someone else will sooner or later find a way to clear things up.

3

If the resistance remains the same, more voltage does cause more amps to flow. More amps thru the same load means more watts (power). I think you just overlooked the resistance portion.

Same thing. The filament in a bulb has a certain resistance. If you drop the voltage, then not enough amps will flow thru the filament to make it glow. You could make a filament that would glow at a lower voltage, it would have a lower resistance. That’s the difference between a bulb in a table lamp and a tail light.

4

Easiest question first. Yes, heat is equivalent to power, or, rather, power multiplied by time.

For the rest of the question, the crucial element is resistance. Resistance is the ratio of voltage to current in any conductor:

R = V / I.

This also gives us current in terms of resistance and voltage:

I = V / R.

And we can also calculate voltage in terms of resistance and current:

V = I * R.

Something like a heating element or the filament of a light bulb has a resistance that is fairly constant - it may reduce slightly when it gets hot, but not by an enormous amount. A typical car lightbulb has a resistance of about 3 ohms.

So, what happens when we apply 12 volts to it?

I = V / R = 12 / 3 = 4 amps.
Power = I * V = 4 * 12 = 48 watts.

Now, what if we want to increase the power? We can either reduce the resistance to (say) 2 ohms:

I = V / R = 12 / 2 = 6 amps.
Power = I * V = 6 * 12 = 72 watts.

The current has increased, so the power is greater. Or, we can increase the voltage - the pushing power - to (say) 15 volts:

I = V / R = 15 / 3 = 5 amps.
Power = I * V = 5 * 15 = 60 watts.

So, yes, you’re right, increasing the voltage increases the power by causing more current to flow.

No - it depends on both current and resistance. Let’s say we apply the same current - 2 amps - to a 10 ohm and a 5 ohm resistance.

For the 10 ohm resistance:

V = I * R = 2 * 10 = 20 volts.
Power = I * V = 2 * 20 = 40 watts.

For the 5 ohm resistance:

V = I * R = 2 * 5 = 10 volts.
Power = I * V = 2 * 10 = 20 watts.

I hope this clarifies things.

5

In a way voltage and current are very much inter-related, so it is kind of difficult to understand. If you consider something like a light bulb, then the voltage and current are related by ohm’s law V=IR. R is the resistance of the bulb, which is fixed, so the way you get more I is to have more V.

Heat, however, is related to power, not amperage. If you have two different coils of wire with two different resistance values, you can get exactly the same amount of heat out of them by varying the voltage driving one of them. For the same heat value, the current will be different, because the resistances are different.

A good example to think about is baseboard electric heat, used in homes. If you run this off of a 110 volt circuit, and your circuit has a 15 amp limit, the most power you can get out of your heater is 1650 watts. However, if you use a 220 volt circuit, and you have the same 15 amp limit (due to the size of the wire that feeds the heaters) then you can now get 3300 watts out of your heater. In this case the current is identical in each case (15 amps) but the power in the 220 volt heater, and hence the amount of heat that it generates for your home, is twice as much.

If you want to double the power, you have two choices. You can double the voltage, or you can double the current. The heater example above doubles the voltage. A 40 watt bulb draws double the power of a 20 watt bulb, but has the same voltage. This example doubles the current, and does so by reducing the resistance of the filament. In some cases it’s not practical to double the voltage, like the light bulb, where the supply voltage is fixed. In other cases it’s not practical to double the current, like in the home heater where the wire size dictates the maximum current that can be used.

6

Well in the case of falling water, a higher dam would result in water which has a greater speed when it gets to the bottom. Consequently, the same amount of water could cause the turbine to spin faster, because of the increased speed of the water. That still doesn’t explain what the speed of the water equates to in relation to electricity though. Would each electron have more energy just like how each unit of water would have more kinetic energy? I can understand the metaphor, but I’m not sure how it relates.

If I understand you correctly, you’re saying that more voltage causes more power because it causes more amperage, but this isn’t necessarily the case. You can double the voltage and resistance and end up with the same current, but consume more power since P=V*A. How is it that you can have the same amount of electricity flowing through something but get more power out of it?

No I didn’t. Like I said in the above paragraph, you can double voltage and resistance, resulting in the same amperage, yet double the power simply because the voltage was doubled. The voltage seems to have a direct effect on power, not just an indirect effect, and that’s what I don’t understand.

You state that the heat is proportional to the square of the current (or current density actually, but hell if I know what the difference is). So why can’t you lower the voltage and the resistance such that the current would remain the same and get the same heat? Perhaps in a heater system, you could use a physically wider conductor, thus lowering the resistance allowing the same amperage to be obtained from a lower voltage. Or would that lower the current density? (I’m thinking perhaps the current density is the current divided by the surface area of a cross section of the conductor, but I’m not sure.)

Also, you seem to contradict yourself (and several others in this thread) because you say it’s proportional to power, and then you say it’s proportional to current density. These are two totally different phenomena, right? I can’t imagine it being proportional to both.

7

On the dam analogy, doubling the resistance is doubling the number of turbines. The same amount of water flows through each turbine as before, but as you now have two turbines rather than one, you get twice the power.

8

Oh, on the electron level - the kinetic energy of each individual electron depends on the voltage, the current is a measure of the number of electrons flowing through the wire.

So, if we double the voltage but keep the current the same, the same number of electrons are flowing through the wire, but each of them has twice the kinetic energy that they did at the lower voltage, so we get twice the power.

9

Direct investigation and experience is the best way to fight ignorance. To illustrate the difference, look at this wire, with current flowing through it at high amperage but low voltage . . . grab it right here, where I’ve stripped off the insulation . . . Now, this other wire . . .

10

They are related. Power is voltage times current. Current density is current per square unit of conductor (actually circular area units are generally used, but let’s not confuse the issue any more than it needs to be). Therefore, power and current density are proportional to each other–that is, as the power dissipated through a given conductor increases, so does the current density in the conductor. proportionality is not the same as equality.

Driving to a destination, your trip time is proportional to the distance you have to drive. It is also proportional to the speed you travel at. Distance and speed are not the same thing but they are related, since speed is distance divided by time. Does that clear this up?

11

Clean up in post #9, please! Clean up in post #9.

12

That’s exactly the kind of answer I was looking for. I was only thinking of voltage as something that pushes amperage. Now that I know the amperage it produces is also more energetic, its direct effect on power makes more sense.

Okay, well let me ask you this: what is heat production in an electrical device equal to? Depending on your answer, I may finally understand how this strange form of energy works.

13

Heat is a form of energy. Energy is power multiplied by time. If you want to express this so you can see how voltage and current fit in, you can expand the equation and say that heat energy is equal to voltage multiplied by current multiplied by time.

If you want to know how hot this will make the conductor, it gets considerably more complicated. This will depend not only on the material the conductor is composed of and its geometry, but also on its thermal coefficient of resistance–how its resistance chages with temperature.

I don’t want to make your head asplode, so I’ll pause here and see if you’re still with me. Get me so far?

14

Okay, I think I get it now. The rate at which heat is emitted is equivalent to the electrical power (at least for a given conductor), not just the current. Well now I understand why step-down transformers won’t help for a heater, but that bring up another question. I’ve been told (and it was mentioned in this thread) that if you raise the voltage, you can use smaller wires because of the lower amperage. But if the heat is equivalent to power, and the power stays the same, how will a higher voltage help prevent the wires from burning up?

15

Yes, each electron would deliver more energy to the load as it “falls” through the greater voltage difference. For example, a voltage increase of 1 volt represents 1 joule of energy exerted on one coulomb of electricity (6.242*10[sup]18[/sup] electrons) so if that number of electrons “falls” through a voltage of 2 volts it will deliver twice as much energy to the load as was the case for 1 volt.

See above. If a certain amount of electricity moves between two points with a voltage difference of 1 it changes its energy level by a certain amount. If that same amount of electricity moves between two points having a voltage difference of 2 its energy level changes by twice as much as the first case. In your question you asked why you couldn’t merely use a step-down transformer to lower the voltage on a light bulb and have the same light out of the bulb because you are using the same current.

If the bulb current is the same and the bulb voltage is cut in half with the current staying the same then we have V/2 as bulb voltage and I as the bulb current where V and I are the bulb voltage and resistance before the transformer was added.

We can work this out mathematically.

P = VI originally and P = VI/2 after the transformer is added.

See above. And we can also work this out on a current and resistance basis rather than a voltage and current basis.

Again P = VI and from Ohm’s Law V = IR Substituting I*R for V in the power equation gives P = I[sup]2[/sup]*R.

Again you say we keep the same current but cut the resistance in half so our new current and resistance are i and r/2

Giving P = i[sup]2[/sup]*r/2

Or half the power again.

No current and current density aren’t different phenomena. The current density is merely the current divided by the area of the conductor and it is important in transformer design, for example. For conductors out in the open a current density of 4000 amp/sq. in. might be perfectly fine, but in a transformer where the wires are stacked up the current density might well be limited to 500 amp/sq. in. in order to avoid overheating. Just as an illustration, #12 copper wire has a cross section of 0.005129 sq. in. and is rated at 20 amp. for ordinary two wire insulated cable. That is a current density of 3900 amp/sq. in.

16

No, I meant power and current density. Okay, they are related, but they aren’t equal. You can achieve 3900 amps/sq in with different power levels. With the #12 wire you mentioned, you could achieve it at 2400 watts with 120 volts. Simply raise the voltage to 240 volts while increasing the resistance on the device at the end of the wires to keep the current the same and you can get 4800 watts. But if the heat production is equivalent to power as has been said, how can a single wire max out at both 2400 watts and 4800 watts? If heat production is equivalent to power, why would a certain wire be rated by amperage? It still seems contradictory to me. You can work the numbers so that you produce 100,000 watts at 10 amps/sq in, or so that you produce 10 watts at 100,000 amps/sq in. Which will melt the wire?

17

A revelation came over me in the shower. (Perhaps that’s where I do my best thinking.) Heaters have no device at the end of the wires on which you can adjust the resistance; the wires are the only resistance. For a given composition of wire, there are only two ways to adjust the resistance of the wire: change the width or change the length. If you cut the wire so that it’s only half as long, it’ll have half the resistance and you can achieve the same amperage with half the voltage and the wire will be the same temperature, but since there’s only half as much wire (and consequently half as much surface area on the outside of that wire), it’ll only release half as much heat. Thus you’re using half as much power but still only getting half as much out of it. If you widen the wire so that the cross area doubles, you can also get the same current from half the voltage, but because you’re sending the same current through a wire with twice the cross area, the current density is only half and the temperature will only be half. The one thing I can’t make sense about this situation though is that the surface area of the outside of the wire will increase (not quite double, but actually increase by the square root of two) so that while you’re using half the power, you should still be emitting 71% (the square root of 2 over 2) as much heat. There has to be a flaw in my thinking. If someone can just find it for me, I think I’ll have this nailed.

18

The heat is generated in the wire and it doesn’t have to be disipated. If you overheat the wire, it’ll burn up.

19

Don’t get too wrapped up in current density. Unless you are designing electrical machinery you will seldom encounter it in practice. If have 10 amp in a wire of a certain size and reduce the wire diameter by half the current density is doubled and the heat goes up by a facter of four. However you can also get the same result by considering the power dissipated in the wire because the resistance of the wire will have gone up by a factor of four and therefore so will the power.

The answer to your basic question is that voltage and current both contribute to the power in an electrical circuit. Voltage is the energy per unit of charge and current is the rate of transport of charge and their product is the rate of energy flow which is power.

20

The tricky part here is that you are talking about power in two different places. In an electrical distribution system, you are mostly concerned with the power that ends up in the load. But, there is also power that is dissipated as heat in the wires. The key is that you want to reduce the power in the wires, and get as much power as possible to the load.

Assume for example that you have a generator with a voltage of V, a bunch of wire with resistance R1, and a load with resistance R2. The wire is in series with the load, so the current that flows is going to be I = V/R = V/(R1+R2). Since the resistance of the wire is very very small compared to the resistance of the load, you can almost ignore it, and the current that flows is mostly going to depend on the value of R2. For example, if V=120V, R2 = 100 ohms and R1 = 0.1 ohms, you can see that R1 is going to have a very small effect in the overall calculation for current. If you ignore R1 completely, I = 120/100 = 1.2 amps. If you don’t ignore R1, then I = 120/100.1 = 1.198 amps (pretty darn close to 1.2).

You have about 144 watts dissipated in the load R2. But, you also have 0.144 watts dissipated in the wires, which is a pretty small value, but this is just to prove a point.

What if we were to stick a transformer before R1 and after R1 so that the voltage on the line was increased by a factor of 10? Then the current would be decreased by a factor of 10 on that section of line. Now, instead of 0.144 watts lost in the lines you only have 0.00144 watts as heat in the line. The load still sees about 1.2 amps (it’s actually a little closer to 1.2 than it was before, if you want to have some fun and calculate it out), so from the load end very little changed. You’ve still got about 144 watts at the load R2. But on the transmission line, you reduced the power lost by a factor of 100. For values like this it doesn’t matter much, which is why you don’t have little transformers all over your house. But for a big power system where the numbers are all a lot larger, it makes a huge difference. Basically, stepping up the voltage by a factor of 10 makes the lines run 100 times cooler.

This does come with a down side. Higher voltages need more insulation around them and more spacing between conductors.

The thing you are missing here is that the amount of energy you are dissipating is dependent on the power. How hot the wire actually gets though depends on a whole bunch of things, only one of which is how much power is going through the wire. Part of it is the surface area of the wire. Another part of it is the temperature of the surrounding materials. You’ve got X amount of wattage generating heat inside the wire, and convection, conduction, and radiation all taking heat out of the wire.

Thermodynamics always gave me a good case of the willies. I try and avoid it whenever possible.