LOL! I just can't get you guys to forget my original analogy.

Please forget about the electric shock factor. Forget the body, the skin, the Independence of the body etc. Just stick to the 2 circuits with the same power in watts but different voltages and different resistors.

I think @atrinao92 is getting close when he said: " The point of watts is a measure of power and over time, energy. So in your example if the load is an actual resistor (ohms) then both circuits will produce the exact same amount of heat per second and will "burn" you the exact same amount if you touch the resistor. "

If we can stick to the 2 circuits, one conducting 2 amps with using a 5 volt power source and the other conducting 1 amp using a 10 volt power source. The circuit with a 10 volt power supply has to "push" current through a 10 ohm resistance, which has 4 times more resistance than the 2.5 ohm resistance of the 2 amp circuit which has a 5 volt power supply.

So the 1 amp circuit still uses 10 watts of power because but has to push current 4 times harder then the 2 amp circuit to get through the load so it needs 10 volts of power supply to push 1 amp of current though a greater resistance by 4 times.

So the power usage is the same on both circuits. One needing a greater voltage in the 1 amp circuit to pass less current through a resistor with 4 times more impedance than that of the other 2 amp circuit with half the voltage at 5 volts and a quarter of the resistance at 2.5 ohms. So the power in usage in both circuits is the same. So what gives here? The amount of current that is throughput.

So back to the car analogy. When we say "power to weight ratio" we mean a heaver car with the same power of a lighter car will go slower but still consume the same energy aka power. Its the weight (aka the load aka the ohms) that determines the speed and the speed is equivalent to the amperage.

I think that's it.

 

More current means more power. Much more power.
Power increases as the square of the current, i.e. current x current.

Do the math in your two examples.

2A @ 5V vs 1A @ 10V

Is 2A four times more powerful than 1A?
No. The two consume the same power of 10W.
Why?
Because the two situations are different.
One has a 2.5Ω load while the other has a 10Ω load.

If the loads were the same, going from 1A to 2A would increase the power by a factor of 4.

But the load resistance is different.
Power = current x current x resistance
2A @ 5V = 2A x 2A x 2.5Ω = 10W
1A @ 10V = 1A x 1A x 10Ω = 10W

The power is the same.

 

Hopefully it has been adequately explained, that electrical power is the product of voltage and current, and that those two are certainly dependent, but not at all the same.
This is why understanding basic circuit theory is important. All of the facts matter, and they are all needed to achieve an adequate insight and understanding. These threads include a whole lot of examples of ideas that develop from only being aware of part of the information. The explanations of why we can not have perpetual motion have lead to disappointment for quite a few folks who were lacking part of the picture.
That is one big reason to keep on learning.

 

Can you please forget the electric shock analogy. Its just an analogy, and not a very good one because its distracting from the real question.


My new example talks about watts as a measure of power.


The real question: How can 2 circuits with the same power have different amperage's yet be considered to have the same power?

It's basically analogous to how a small projectile fired at a high rate of speed might have the exact same overall energy content as a much larger body moving very slowly and yet the former will most likely cause more damage upon impact due to the concentration of forces.

 

I recall studying the concept of mass * speed (cantidad de movimiento in Spanish) Experts studying collision of ships use the concept quite in detail.

 

In general, you get to chose either the voltage or the current and the circuit will dictate the other.

In general, yes.

Something that doesn't - a 1000 W arc lamp.
80V open circuit initially with a 40 kV pulse to start.
Then current limited and intensity controlled by controlling power.
Operating is around 22V 40A

 

More current means more power. Much more power.
Power increases as the square of the current, i.e. current x current.

Do the math in your two examples.

2A @ 5V vs 1A @ 10V

Is 2A four times more powerful than 1A?
No. The two consume the same power of 10W.
Why?
Because the two situations are different.
One has a 2.5Ω load while the other has a 10Ω load.

If the loads were the same, going from 1A to 2A would increase the power by a factor of 4.

But the load resistance is different.
Power = current x current x resistance
2A @ 5V = 2A x 2A x 2.5Ω = 10W
1A @ 10V = 1A x 1A x 10Ω = 10W

The power is the same.

Mr.Chips brilliant! Thank you, I think this is the explanation I was hoping for but also a big thank you to other responses that have been very helpful too.

I always thought power as P = V x I. I did not think of expressing power as Power = current x current x resistance.

But yes, if V = I x R then P = V x I becomes P =(I x R) x I or also expressed as P = I x I x R. Just algebra.

But alas I still remain a little confused. You said "More current means more power. Much more power." So the 2A circuit has more current yet the same power as the 1A circuit that has less current? What??

Should you not have said something like "power is determined by current and resistance"?

 

In general, yes.

Something that doesn't - a 1000 W arc lamp.
80V open circuit initially with a 40 kV pulse to start.
Then current limited and intensity controlled by controlling power.
Operating is around 22V 40A

An arc is about the most non-linear circuit element that there is. Thus it does not fit this discussion. Simple nath works for linear devices only. Power is the work done pushing the amps through the load, which offers some level of resistance. So it is the voltage that pushes the amps. that is as basic as it gets if we stick to common language.

 

There are 3 variables, V, I and R.
Only two are independent. The third variable is determined by the other two variables.

I = V / R
V = I x R
R = V / I

Hence, in effect there are only 2 variables.
Pick any two and you can determine the power.

P = I x V
P = I x I x R
P = V x V / R

Hence all of the three statements are correct.

Power is determined by current and voltage.
Power is determined by current and resistance.
Power is determined by voltage and resistance.

 

... Power is the work done pushing the amps through the load, which offers some level of resistance. So it is the voltage that pushes the amps. that is as basic as it gets if we stick to common language.

I really like this definition too.

I guess originally I was trying to find out if amps alone really should be the determinant of power, but the equations make this clear.

As for the electric shock factor (dare I bring back the bad analogy...nooooooo) while one circuit may carry more amps than the other, the energy (aka power) is the same in the 2 examples. So it isn't really power or voltage that dictates "shock danger", its the amps which are controlled by the voltage and the resistance in the circuit. The power is really just a statement of how much energy (not current) is in the circuit.

I think we have pretty much come full circle and please excuse me for reinventing the wheel, but I now have a clearer understanding of power vs current . Yes the equations say it all really but circuits come in many forms and its easy to make assumptions...

 

There are 3 variables, V, I and R.
Only two are independent. The third variable is determined by the other two variables.

I = V / R
V = I x R
R = V / I

Hence, in effect there are only 2 variables.
Pick any two and you can determine the power.

P = I x V
P = I x I x R
P = V x V / R

Hence any one of the three statements are correct.

Power is determined by current and voltage.
Power is determined by current and resistance.
Power is determined by voltage and resistance.

Spot on Mr Chips. Thank you. Very helpful.

 

Spot on Mr Chips. Thank you. Very helpful.

I posted that information a dozen posts earlier...

 

I posted that information a dozen posts earlier...

Yes you did. Sometimes the newcomer needs to hear the same thing multiple times before they get it.

You always find what you are looking for at the last place you looked.

 

Given that a major aim of this forum is education, I am pleased that my explanation served to be educational. Some concepts are not so easy to explain well enough.

 

Hello,

Read chapter 3.2 of the attached handbook.

Bertus

 

Be careful, electricity can hurt you...

 

The Jolt of the Volt is proportional to both the Large of the Charge and the Arc of the Spark. Or something like that.

 

In general, yes.

Something that doesn't - a 1000 W arc lamp.
80V open circuit initially with a 40 kV pulse to start.
Then current limited and intensity controlled by controlling power.
Operating is around 22V 40A

What does it mean to be controlled by the controlling power? I'm not at all familiar with the details of how arc lamps behave, but if you are saying that an arc lamp limits the current to, say, 40 A and that this doesn't change much as the applied voltage changes, then that still follows what I said. Consider a diode. The voltage, when conducting, changes very little as the current changes. But if I power it with something that forces a particular voltage across it, the diode will dictate the current -- it just might be extremely high (until the device is destroyed). I image it would be similar with an arc lamp. If I power it with a device that forces a particular current through it, then the lamp will dictate the voltage that is needed to make that happen. It may well be that operating at the point will either extinguish the arc or destroy the lamp. But I don't see how either the controller (or the device) can dictate both the voltage and the current -- it is the interaction between the two that establishes the operating point.

 

Mr.Chips brilliant! Thank you, I think this is the explanation I was hoping for but also a big thank you to other responses that have been very helpful too.

I always thought power as P = V x I. I did not think of expressing power as Power = current x current x resistance.

But yes, if V = I x R then P = V x I becomes P =(I x R) x I or also expressed as P = I x I x R. Just algebra.

But alas I still remain a little confused. You said "More current means more power. Much more power." So the 2A circuit has more current yet the same power as the 1A circuit that has less current? What??

Should you not have said something like "power is determined by current and resistance"?

The same thing could be said about voltage. More voltage more power, much more power. The power goes up as the square of the voltage.

P = V x I, but V = I x R and so I = V / R, hence P = V x V / R, Just algebra.

You are still stuck on the notion that two circuits having the same current have to have the same power. They don't! Current, alone, does NOT determine the power. You also need the voltage difference through which that current flows. No matter how you calculate it, you need the product of those two things.

 

..You are still stuck on the notion that two circuits having the same current have to have the same power. They don't! Current, alone, does NOT determine the power. ...

Nope, I think that you may have misunderstood where I got stuck. But I could have misunderstood you too. Can you show 2 circuits with the same amps but different wattage in P = V x I??

I WAS stuck on the notion that two circuits having DIFFERENT currents can have the SAME power in P = V x I:

Circuit 1: 10 watts = 10 volts x 1 amp
Circuit 2: 10 watts = 5 volts x 2 amps

So because the 2 amp circuit carries more "shock" current it made me wonder how this can have the same power shown in the 1 amp circuit? Ohms law clearly shows that amps are just part of the equation and power is not determined by amps alone, voltage and resistance are needed too. I knew this but the "why" still wasn't clear to me.

I think the closest answer was about the energy and/or burn factor in the resistors in both circuits being the same.

My question was really about what "power" actually means in abstract terms and how can a circuit with more amps than another circuit have the same power?

One of the best responses explained that power represents the amount of energy in the circuit so both circuits use the same power, both resistors would burn you the same because they consume the same energy. And energy is power, not amps, which is the essence of my question/confusion.

Circuit 1 has half the amps of circuit 2 yet twice the electro motive force (emf aka voltage). Its resistor (aka load) would burn you the same as in circuit 2 funnily enough. The resistance/load in circuit 2 is 25% of that of the resistance in circuit 1. Both resistors would get equally hot as they consume the same power/energy.

But alas I concede that I have not expressed the question well enough from the beginning but hope that this response makes it clearer?

In my simple mind I thought that more amps should mean more power, but the equations prove that to be incorrect. I was asking how/why?

I take your good point about voltage and algebra around ohms law, another good insight. Thank you.