Theory Amps and Watts

[GoldDigger]

So when a light is on how does the meter know how much was unused if the meter doesnt read outgoing on the hot and the rest coming back on the neutral. Probably a dumb question but one I dont know.

The wattmeter senses both voltage and current.
When the light is off no current flows.
The meter cannot calculate watts if all it senses is current and the voltage at one point.
It has to know the voltage across the load (or supply) for that.

To state it another way, there is not energy going out one wire and coming back through the other. There is just the energy delivered as a total by the two, three, or four wire circuit.

 

[kwired]

Are you saying you're not billed based on watts consumption? What formula do you use to find ohms, fraction upside down?

Power is watts, it is sort of an instantaneous value at any given moment, energy which is watt-hours factors time into it. If you use a certain amount of power for a given time period you then can have a value of work that was done. If it takes you half the time to do the same task then your power has doubled - yet energy used has remained the same (not factoring in changes that may occur in efficiency anyhow).

If you draw 1000 watts for one hour you have used one kilowatt hour. If you draw 2000 watts for half an hour you have used one kilowatt hour. If you draw 60,000 watts for one minute you have used one kilowatt hour. All same amount of energy used - just over different time duration in each instance. Of course the same load device whether it be a heater, motor, light source, etc. likely can not handle such an increase in use and needs to be designed differently

To find ohms:

E=IxR

R= E/I

I= E/R

Power formula if interested:

P=IxE

E=P/I

I=P/E

 

[zcanyonboltz]

Correct.

Let's say you have three 100 watt loads.

Factoring in NOTHING ELSE, what is your electric bill going to be?

Well you couldn't calculate the bill because it depends on how many watts are used and for how much time. What I'm thinking is that the bill is calculated from how many WATTS are used and for how long the WATTS the are used, right?

 

[zcanyonboltz]

Power is watts, it is sort of an instantaneous value at any given moment, energy which is watt-hours factors time into it. If you use a certain amount of power for a given time period you then can have a value of work that was done. If it takes you half the time to do the same task then your power has doubled - yet energy used has remained the same (not factoring in changes that may occur in efficiency anyhow).

If you draw 1000 watts for one hour you have used one kilowatt hour. If you draw 2000 watts for half an hour you have used one kilowatt hour. If you draw 60,000 watts for one minute you have used one kilowatt hour. All same amount of energy used - just over different time duration in each instance. Of course the same load device whether it be a heater, motor, light source, etc. likely can not handle such an increase in use and needs to be designed differently

To find ohms:

E=IxR

R= E/I

I= E/R

Power formula if interested:

P=IxE

E=P/I

I=P/E

Thanks for the info. This goes with the way I thought bills are calculated, how many watts you use, am I missing something here? I am familiar with the power wheel, I asked Besoeker what formula he used because I didn't see how he got his numbers in post #12 with an upside down fraction based off the formulas to find ohms. I used E/I same as V/I in this wheel below.

 

[gar]

150405-2344 EDT

zcanyonboltz:

I have not read all the posts so I may be repeating what has been said.

Watts (power) is a measure of the rate of doing work. This is like the speed of a car. This measure is an instantaneous value unles you qualify it by saying average power.

If we have a load that draws a constant power, then we can determine the energy received in the load over some time period by simply multiplying the time duration of said period by the constant power value to obtain the Watt-hours of energy.

If the power is constantly varying during said time period, then we need to do a lot of short time segment measurements, and add all these up to get the total energy used.

Suppose the power is approximately constant over each 1 second increments, and we want the energy used over a 1 hour period, then we measure average power over 1 second and multiply that by 1/3600 to get the increment of energy for that 1 second increment. This is repeated for all 3600 contiguous seconds, and all these 3600 small values are added up to get the 1 hour energy received. Essentially this is what the power company kWh meter does.

.

 

[gar]

150405-2424 EDT

Time out limits are a major problem on this form. Thus, I won't recompose what I had tried to add.

Go to my wesite http://beta-a2.com/energy.html and view graphs 9.8.2.2 and 9.8.2.4 . You will have to figure them out for yourself.

.

 

[kwired]

150405-2424 EDT

Time out limits are a major problem on this form. Thus, I won't recompose what I had tried to add.

Go to my wesite http://beta-a2.com/energy.html and view graphs 9.8.2.2 and 9.8.2.4 . You will have to figure them out for yourself.

.

A way around that is to compose in word processor or even just simple text document on your computer then copy and past text into the composition page of this site.

Sorry I guess that don't help with editing an existing post - you do have a time limit. But I don't think it is right to be able to edit a post for too long after it has been submitted - it just adds to confusion when people refer back to something that isn't there anymore.

 

[kwired]

Thanks for the info. This goes with the way I thought bills are calculated, how many watts you use, am I missing something here?

You are missing the time factor, without it you only have power which is an instantaneous value, throw time factor in there and you have energy.

As said if power is constant it is easier to calculate energy. But if power varies over time then actual energy used is a little more complicated.

Straight resistance loads are simpler just figure the power times run time to get energy, motor loads are more complicated if the driven load varies, as the motor only draws what it needs to try to maintain its speed for the most part. An increase in load will cause the motor to slow down if it didn't draw more power to compensate and try to maintain steady output speed. That increase in power along with the time it happens results in a higher energy level used during that higher load period. Now go through a cycle of lesser load and energy starts to balance out over an even longer time period - which the basic watt-hour meter is just an accumulation of how much energy is used, more sophisticated meters can record the peaks.

 

[ggunn]

Well you couldn't calculate the bill because it depends on how many watts are used and for how much time. What I'm thinking is that the bill is calculated from how many WATTS are used and for how long the WATTS the are used, right?

True enough.

It's like your water bill. You don't get charged by the rate at which you are taking water out of the system (electrical equivalent = Watts), but for the total number of gallons you use in a month (electrical equivalent = Watt-hours).

A Watt-hour is a very tiny quantity, so electrical energy is usually expressed in kilowatt-hours (kWh). 1kWh = 1000 Watt-hours. A kWh costs about 10 cents most places in the US.

 

[Carultch]

So I've searched this online for a while and I keep getting taken to physics websites and forums and long drawn out answers that don't make a lot of sense to a non physicist. My question is how is it that when voltage increases amperage decreases but watts stay the same? To illustrate, lets say you have 10,000 watts being drawn at 120 volts, the amperage is 83.3 amps. Now take that same 10,000 watts at 240 volts and you have 41.6 amps. People often think just because equipment operates at 240 volts the power consumption is less but when I tell them this isn't the case you're billed in wattage consumption and this stays the same at either 120v or 240v they are in awe. Any ways to explain this? Thanks

That is only true when it is a design problem of selecting the optimal characteristics for transmitting a GIVEN AMOUNT OF POWER.

When you build a device at 10 kW to work at 120V, it requires 83.3 Amps.
When you build a device at 10 kW to work at 240V, it requires 41.7 Amps.

If you take a 10 kW device built to work at 120V, and plug it in to a 240V circuit, that is a potentially dangerous thing to do (no punn intended). Assuming it behaves as a linear resistive load, it now becomes a 40 kW device, subject to twice the current and twic the voltage. Its inner-workings are likely not built to work with the extra power, and it will fail.

 

[Strathead]

Thanks for the info. This goes with the way I thought bills are calculated, how many watts you use, am I missing something here? I am familiar with the power wheel, I asked Besoeker what formula he used because I didn't see how he got his numbers in post #12 with an upside down fraction based off the formulas to find ohms. I used E/I same as V/I in this wheel below.

It is funny, because what is simple to one person can be difficult to another. I don't know what your math sites stated, but it is simple for me to understand the concept that wattage and amperage proportional. When one goes up, the other goes up the same percentage when ALL other factors stay the same, period. Voltage and amperage or inversely proportional. When one goes up, the other goes down the same percentage when ALL other factors stay the same, period. Two very important factors here, "all other factors stay the same", and this is the formula for DC circuits, which I didn't note anyone else mentioning. All of the formulas on the wheel are for DC power. That is fine for AC as well to get to general assumptions.

So, to start you back on the road to confusion, The wattage that a unit uses is a measure of the power that it takes to produce the results. The power that must be delivered to the unit to produce that wattage (or work) is probably not the same as the power at the unit. One major contributor to difference between the power inputted to a unit and the power required to be outputted from the generation is the power factor, which is a product of the phase angle between voltage and amperage. Any phase angle other than 1 causes a loss of usable power at the equipment and must be factored in to any Alternating current application. Confused yet?

 

[Dennis Alwon]

I love those charts but I have seen error on them and some don't continue the radical sign so it is confusing. Here is one- Check out the middle formula and the one to the right under V= It can be both the middle should say Sq. root of (P x R) Not (sq. rt of I)x R

Similar problem under current in the middle. The sq. root should extend down to the denominator and the I should be a P

 

[gar]

150406-1903 EDT

Strathead:

Power is not work. Power is the rate of doing work. The integral of power over time is work. Mathematical integration is the summation of small increments over some range.

If I have a 33,000 # block and raise it 1 foot in 1 minute I require a 1 HP motor. That same 1 HP motor can raise a 1 # block 33,000 ft in 1 minute (ignoring the change of gravitational force). In each case I did 33,000 ft-pounds of work. Both cases used a power of 1 HP or 746 W. This required 0.746/60 = 0.0124 kWh of energy. It would take a typical human about 3 to 4 minutes to do this same work.

I can do that same work, 33,000 ft-pounds, in 1 hour with the use of a motor with a power rating of 1/60 HP. But the same amount of energy will be required (work).

.

 

[zcanyonboltz]

Thanks for all the info that's some very interesting material. I think I'll check out a library book on electrical theory to learn more about this stuff.:happyyes:

 

[junkhound]

Any ways to explain this?

try to NOT use any electrical terms.

Say you have a 16# bowling ball and an 8# shotput. You drop both on your foot from 12" high, the 16# ball hurts worse than the 8# ball. Now drop the 8# ball from 24" high, it hurts as bad (or even worse) than the 16# ball at 12".

watts = hurts <G>

 

[GoldDigger]

Any ways to explain this?

try to NOT use any electrical terms.

Say you have a 16# bowling ball and an 8# shotput. You drop both on your foot from 12" high, the 16# ball hurts worse than the 8# ball. Now drop the 8# ball from 24" high, it hurts as bad (or even worse) than the 16# ball at 12".

watts = hurts <G>

A very bad analogy.

The weight of the bowling ball can correspond to the watts. But what causes a large part of the hurt is the energy which is the product of the weight and the height of the drop.
But the hurt also depends on how fast the ball stops when it hits you. Back toward watts again.
And a 500# ball will really hurt just sitting on you without being dropped.

 

[busman]

A very bad analogy.

The weight of the bowling ball can correspond to the watts. But what causes a large part of the hurt is the energy which is the product of the weight and the height of the drop.
But the hurt also depends on how fast the ball stops when it hits you. Back toward watts again.
And a 500# ball will really hurt just sitting on you without being dropped.

In my opinion, it's a fairly good analogy for energy. Potential energy is equal to mass times gravity times height (mgh). If we assume gravity to be constant, then the energy transmitted to your foot is the same for an 8# ball from 24" or a 16# ball from 12". In fact, voltage is sometimes called potential because it is based on a field (electric versus gravity) over a distance. In this analogy, the mass is the current.

Mark

 

[kwired]

It may be a good analogy as far as potential energy, but it is a bad analogy as far as how much it will hurt because if you drop either one on your foot it will still hurt enough that when you drop the second one it is hard to make an accurate comparison, and if you wait for everything to heal before the second attempt - much information from first try will be forgotten.:happyyes: