I am confused about what exactly wattage is.  

ON THIS PAGE:
http://science.howstuffworks.com/electricity3.htm
IT SAYS:
In an electrical circuit, the number of electrons that are moving is
called the amperage or the current, and it is measured in amps. The
"pressure" pushing the electrons along is called the voltage and is
measured in volts. ...  1 amp physically means that 6.24 x 10^18
electrons move through a wire every second, and the voltage is the
amount of pressure behind those electrons.

ON THIS PAGE:
http://science.howstuffworks.com/electricity5.htm
IT SAYS:
Volts * Amps = Watts 

ON THIS PAGE:
http://science.howstuffworks.com/electricity6.htm
IT SAYS:
Let's say you have a tank of pressurized water connected to a hose
that you are using to water the garden. What happens if you increase
the pressure in the tank? You probably can guess that this makes more
water come out of the hose. The same is true of an electrical system:
Increasing the voltage will make more current flow.


As I understand it, this is saying that if you double voltage (and
resistance is the same) you also double the current.  And, since volts *
amps = watts - doubling both volts and amps, increases the watts by 4
times.  How is this possible if you are only doubling voltage and
resistance dosen't change?

I assume there is a missing element here, like maybe I'm not
accounting for any load.  Or maybe the amount of current drawn has an
inverse affect on the voltage, so that voltage only remains doubled if
the current remains the same, and voltage actually drops some as the
current increases, such that wattage remains only doulbed.

Is there a good example, such as with water hoses, that can illustrate
the difference between current and wattage?

Hi paulshanks!!


Simple mathematics can help you to understand this. Let me start with
the Ohm's law:
"The potential difference (voltage) across an ideal conductor is
proportional to the current (amperage) through it.
The constant of proportionality is called the "resistance", R. 
Ohm's Law is given by:
V = I.R"

Note: if you doubled V and R remains constant, to mantain the equality
I must double also.
 

To find any one of the three variables if the other two are known you
can manipulate the original equation. The three forms of the Ohm's law
are:
1) V = I.R 
2) R = V/I 
3) I = V/R


You also know that the formula to calculate Power is:

P = V.I =
  = V.(V/R) =      by eq. 3)
  = V^2.R

So if V is doubled and R remains constant, then P increases 4 times.  


See the following to understand the difference between current and power:

"In electricity, current is the rate of flow of electrons, usually
through a metal wire or some other electrical conductor."
From "Current (electricity) - Wikipedia":
http://en.wikipedia.org/wiki/Current_(electricity)

Current is the flow of charge and the unit of current is the ampere.
We say that the current in a conductor is one ampere when a charge of
one coulomb passes through a cross section of the conductor each
second.
From this definition I = Q/t (amount of charge over time...) [a]


"The potential difference is defined as the amount of work per charge
needed to move electric charge from the second point to the first, or
equivalently, the amount of work that unit charge flowing from the
first point to the second can perform."
From "Potential difference - Wikipedia":
http://en.wikipedia.org/wiki/Electric_Potential_Difference#Electrical_definitions

From this definition V = W/Q (amount of work per charge...) [b]


"Power is a measure of how much work can be performed in a given
amount of time. Work is generally defined in terms of the lifting of a
weight against the pull of gravity. The heavier the weight and/or the
higher it is lifted, the more work has been done. Power is a measure
of how rapidly a standard amount of work is done...
The power of a mechanical engine is a function of both the engine's
speed and it's torque provided at the output shaft... Neither speed
nor torque alone is a measure of an engine's power.
A 100 horsepower diesel tractor engine will turn relatively slowly,
but provide great amounts of torque. A 100 horsepower motorcycle
engine will turn very fast, but provide relatively little torque. Both
will produce 100 horsepower, but at different speeds and different
torques...
In electric circuits, power is a function of both voltage and current...
In this case, however, power (P) is exactly equal to current (I)
multiplied by voltage (E), rather than merely being proportional to
IE. When using this formula, the unit of measurement for power is the
watt, abbreviated with the letter "W".
It must be understood that neither voltage nor current by themselves
constitute power. Rather, power is the combination of both voltage and
current in a circuit. Remember that voltage is the specific work (or
potential energy) per unit charge, while current is the rate at which
electric charges move through a conductor. Voltage (specific work) is
analogous to the work done in lifting a weight against the pull of
gravity. Current (rate) is analogous to the speed at which that weight
is lifted. Together as a product (multiplication), voltage (work) and
current (rate) constitute power.
Just as in the case of the diesel tractor engine and the motorcycle
engine, a circuit with high voltage and low current may be dissipating
the same amount of power as a circuit with low voltage and high
current. Neither the amount of voltage alone nor the amount of current
alone indicates the amount of power in an electric circuit."
Extracted from "An analogy for Ohm's Law - Chapter 2: OHM'S LAW - Volume I - DC":
http://www.allaboutcircuits.com/vol_1/chpt_2/2.html

Then P = W/t (power is defined as the rate at the work W is done) [c]


Combining [a], [b] and [c]:

P = W/t = (V.Q)/t = V.(Q/t) = V.I 


---------------------------------------------------------

For additional references see the followings pages:

At "Basic Car Audio Electronics" you will find nice pages with
examples, visit the pages 3 through 13 at the directory to the right
of this page:
http://www.bcae1.com/


PHYSCHEM:
At this site visit the following modules:
"Resistors and Ohm's Law":
http://www.physchem.co.za/Current%20Electricity/Resistors.htm

"Heating effects of currents":
http://www.physchem.co.za/Current%20Electricity/Heating.htm


"Ohm's Law" at the12volt.com:
http://www.the12volt.com/ohm/ohmslaw.asp


"Energy Through Our Lives - 3. Watts, Volts, and Amps, Oh My!":
http://www.uwsp.edu/cnr/wcee/keep/Mod1/Flow/watts2.htm

----------------------------------------------------------

Search strategy:
ohm law watt
"electric power" watt current


I hope that this helps you. If you find something unclear or
incomplete, fell free to request for an answer clarification. I will
gladly give you further assistance on this topic if you need it.


Best regards.
livioflores-ga

---

Request for Answer Clarification by paulshanks-ga on 27 Sep 2004 13:02 PDT

I forgot about Ohm's law.  So that helps me with part of the problem.

However, there is still somthing that I am missing. 

Let's say I push a lever that lifts a weight of 10 pounds, at a rate
of 1 foot per second.  If I got a second person to push on the lever
with me, we could either 1.) lift twice as much weight (20 pounds) at
1 foot per second, or 2.) lift 10 pounds twice as fast (2 feet per
second).

Since there are only two people, I assume, in both cases, that the
amount of work being done (wattage) is doubled.

I realize that, in example 1, the additional weight doubles
resistance, so even though the pressure on the lever (voltage)
doubles, the rate of lift (current) remains the same.  However, in
example 2, it seems that both voltage (amount of pressure on the
lever), and current (rate of lift) are doubled, even though the amount
of work has only doubled.  (I seems it would take four people to
quadrouple the amount of work.)

So I assume that either votage or current is doubled, or that they are
both increased by 41% (sqrt of 2) so that when multiplied, the wattage
is doubled, not quadroupled.

Am I still way off?

Thanks,

Paul

---

Clarification of Answer by livioflores-ga on 28 Sep 2004 07:02 PDT

Hi!!


Sometimes analogies help us to understand a topic, but if the analogy
is not well chosen the result is more confusion:

You said:
"...in example 1, the additional weight doubles resistance, so even
though the pressure on the lever (voltage) doubles, the rate of lift
(current) remains the same.  However, in example 2, it seems that both
voltage (amount of pressure on the lever), and current (rate of lift)
are doubled, even though the amount of work has only doubled."

In your analogy change the men by motors, each motor can perform the
work to lift the weight in T seconds as their maximum output. So when
you put two motors instead one you are doubling the power not just the
force applied (if you put to work together two motors of 1 HP you will
be able to get 2 HP of power from the set). So what you call the
"pressure on the lever" is not the analogous to voltage but the
analogous of wattage.

Regarding to "even though the amount of work has only doubled" that
you said at the end of the referred paragraph I must tell you that in
example 2 the work to do has not changed, only the time taken to
perform the job has changed to its half, then the power is doubled. In
the example 1 the work to do was doubled, but it is performed in the
same time, so the power is doubled.

So you must change your above paragraph by this:
"...in example 1, the additional weight doubles the work to do, so
even though the pressure on the lever (WATTAGE) doubles, the rate of
lift (current) remains the same.  However, in example 2, it seems that
both WATTAGE (amount of pressure on the lever), and current (rate of
lift) are doubled."


I suggest you to read the following article:
"The Physics Classroom - Power":
http://www.physicsclassroom.com/Class/energy/U5L1e.html

After reading the recommended text you will see that:
Power = Force * Velocity

Now you can use this analogy: the weight to be lifted plays the role
of the voltage and the Velocity plays the role of the current.
Note that you have not an analogous to the Ohm's law that relates the
weight to be lifted with the Velocity, so the analogy is not complete.


I hope that this helps you. Please feel free to request for further
assistance if you need it.


Regards.
livioflores-ga

As livi said, We can have pressure with no current = zero watts. We
can have 1000 amps with only two volt pressure = 2000 watts or we can
have 2000 amps at 500 volts = one million watts.   Neil

let us take the example of water in a tank and is flowing through the
hose.here the height of tank can be called potential difference. ie
voltage ie pressure in the tank . higher the tank more the
voltage.flow of water is the current.The voltage causes the flow of
current as in the case of water higher the tank water flows with more
pressure and more water flows . Resistance to the flow of water is
caused by the hose . if hose length is more water flown will be less
and if the cross section area of hose is more resistance to water is
less and more water will flow.
    Same way the resistance  in the wire or load decreses the
current.if resistance is decresed current increases and vice versa.

watt ie power = voltage * current        
voltage and current is propotional when resistance is constant.
ie as long as the crossection of hose is same the water  flow
depending on the height of the tank.
if we need more water similar to more current we increase the height
of the tank or voltage; or by reducing the resistance or incresing
crossectional area of hose.
power =voltage * current
current = voltage / resistance
so by decresing the resistance we can increse the current when voltage
is constant. and hence the power will be incresed.

so the 40 w bulb will have more resistance than 100 w bulb. so for
same voltage more current will flow through 100 w producing more
power.