Arca and sparks are very non-linear and so they do not fit. The resistance of an arc depends on the power in the arc and the explanation is tedious. Diodes are non-linear because the conduction depends on the population of charge carriers, and that depends on the current. For easy to understand linear, use a resistor.

 

And don't ever equate energy and power. They are not the same.
Look at the units.

Power = Volt x Amp = Watt
Energy = Volt x Amp x Second = Watt x Second

In other words Energy = Power x Time

 

In order to answer which "Shocks the Most" Because it is a combination of both voltage and current both quantities are involved. The question could be reworded. What quantities of voltage and current produce the most shock ?
Understanding the functional relationship between Voltage and Current involves using both terms together in describing an electrical behavior.


What are some of the differences between voltage and current ?
https://www.diffen.com/difference/Current_vs_Voltage#:~:text=Voltage,-Diffen › Science › Physics&text=Current is the rate at,electric current between two points.


(Voltage and Current)
Teaching approach revisited again :

Sparky1 Upon reflection I think your answer is the most comprehensive. Special thank you!

All answers have been very helpful too.

Thanks everyone, I think I get it now and some of the abstractions are very helpful.

 

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??

Trivially easy.

Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.

Take a 20 V source across a 4 Ω resistor. You have 5 A and 100 W.

Take a 10 mV source across a 2 mΩ resistor. You have 5 A and 0.05 W

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.

So why doesn't it then follow that two circuits can have the same current but different powers?

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.

Power doesn't represent the amount of energy -- it represents the amount of energy per unit time. This is the same as saying that a flow rate of 10 gallons per hours out of a particular tank represents the size of the tank when, in fact, it tells you absolutely nothing about the volume of the tank.

Similarly, knowing the power being dissipated tells you very little about the resulting temperature rise.

I can easily take two resistors that are the same value and apply the same voltage and get the same current in each and have each dissipate the same power and one will create serious burns if you touch it while the other will just be quite warm to the touch. The first will be a small resistor that concentrates all that power in a small volume, resulting in a very high temperature rise. The other will be a very large resistor that spreads that heat out over a large volume resulting in a minor temperature rise.

EDIT: Fixed typos pointed out by @MrChips

 

Trivially easy.

Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.

Take a 20 V source across a 5 Ω resistor. You have 5 A and 100 W.

Take a 10 mA source across a 2 mΩ resistor. You have 5 A and 0.5 W

LOL! Brilliant! I asked for that one and I am kicking myself now. Doh!

So why doesn't it then follow that two circuits can have the same current but different powers?

Power doesn't represent the amount of energy -- it represents the amount of energy per unit time. This is the same as saying that a flow rate of 10 gallons per hours out of a particular tank represents the size of the tank when, in fact, it tells you absolutely nothing about the volume of the tank.

Similarly, knowing the power being dissipated tells you very little about the resulting temperature rise.

I can easily take two resistors that are the same value and apply the same voltage and get the same current in each and have each dissipate the same power and one will create serious burns if you touch it while the other will just be quite warm to the touch. The first will be a small resistor that concentrates all that power in a small volume, resulting in a very high temperature
rise. The other will be a very large resistor that spreads that heat out over a large volume resulting in a minor temperature rise.

Yes WBahn, I am not worthy, I know.

Just when I think I understand it....

Thank you so much. Excellent response, really helpful.

 

The answer is something like 2 or 10 mA ? however to get that to where it's harmful voltage is required, and how much varies VERY WIDELY.

On the other hand high voltage alone will rip through cells no matter what the amperage.

Another topic is RADIATIVE. sure 10mA would be low heat radiation for most components. but the box on your PC is all metal because PC chips radiate ALLOT (the metal enclosure is required for your protection from radiation). radiative energy is not damaging by "only" voltage or "only" amperage. instead it is by frequency and amplitude and shear amount, and there are different kinds tolerable in different amounts classified by frequency. Like voltage, E&M over a certain threshold a high frequency (or amplitude?) will simply tear though cells though there is zero amperage.

for this reason, you really don't want to build high power electronics without protecting yourself from radiation (ie, by covering your project with a case)

 

...

In the case of electrical current, it is NOT the case that more current means more power. The voltage determines how much energy EACH electron delivers, so having 2 A of current in which each coulomb of charge delivers 10 J of energy will deliver considerably less power than having 10 A of current but where each coulomb delivers only 0.1 J of energy.
...

So in this example the 10 A circuit has less voltage and/or resistance than the 2 A circuit? Is the voltage determining the energy each coulomb has? What am I missing here? If 10 A means 10 coulombs per second does that have more flow/energy that the 2 A??

Sorry I do not quite get this, but a I do find the water tank analogy very helpful where the load/resistor might be a valve or a tap that is partly open of fully open...and the volume of water that flows as the rate at which it flows is the amps...

 

So in this example the 10 A circuit has less voltage and/or resistance than the 2 A circuit? Is the voltage determining the energy each coulomb has? What am I missing here? If 10 A means 10 coulombs per second does that have more flow/energy that the 2 A??

Sorry I do not quite get this, but a I do find the water tank analogy very helpful where the load/resistor might be a valve or a tap that is partly open of fully open...and the volume of water that flows as the rate at which it flows is the amps...

Don't go there. Many of us have come to the conclusion that using the water tank analogy will only get you stuck in the mire.

Stick to volts, amps and ohms.

Only two things you need to know at this point:

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

 

Using the car/engine analogy avails the opportunity to highlight a related stumbling block, which may or may not be hidden below the surface here.

A 100kW engine is capable of delivering 100kW of mechanical power. It is almost always not delivering that much power though.

A 100kW electrical source is capable of delivering 100kW of electrical power. It is almost always not delivering that much power though.

A fixed speed engine runs (ex.) 5000RPM, but with no load coupled to its shaft, it delivers 0kW of mechanical power.

A fixed voltage electrical supply runs (ex.) 5000V, but with no load coupled to its output, it delivers 0kW of electrical power.

When you couple a load to your 5,000RPM spinning engine and begin to increase the mechanical resistance, torque goes up while RPM stays constant (more or less), therefore power output goes up.

When you couple a load to your 5,000V electrical supply and begin to lower the resistance, amps goes up while voltage stays constant (more or less), therefore power output goes up.

In either case, mechanical or electrical, it is the load, not the supply, which determines the power output of the source. And that is the point of my post. It's all about the load. A 100kW generator set does NOT deliver 100kW, mechanical or electrical, if all you have plugged into it is a phone charger. I hope this is helpful and something you needed to hear, even if you didn't know you needed to hear it.

 

Another thing to remember is that

So in this example the 10 A circuit has less voltage and/or resistance than the 2 A circuit? Is the voltage determining the energy each coulomb has? What am I missing here? If 10 A means 10 coulombs per second does that have more flow/energy that the 2 A??


Sorry I do not quite get this, but a I do find the water tank analogy very helpful where the load/resistor might be a valve or a tap that is partly open of fully open...and the volume of water that flows as the rate at which it flows is the amps...

When you increase the resistance of a circuit, current diminishes, and thus the power moving through it.

So it really all boils down to resistance. And by that I mean frequency. Because they are essentially equivalent. Increase the frequency of an electrical impulse and the resistance goes up, total power decreases. For a DC signal it's somewhat difficult to comprehend the relationship but it is there. Internally, all resistances are nothing more than a function of frequency.

 

In the case of electrical current, it is NOT the case that more current means more power. The voltage determines how much energy EACH electron delivers, so having 2 A of current in which each coulomb of charge delivers 10 J of energy will deliver considerably less power than having 10 A of current but where each coulomb delivers only 0.1 J of energy.

Okay, I cannot let this one go.

If current is the "number of electrons per second" (coulombs aka amps) does it not follow that the more current you have the greater the emf has to be or there needs to be less resistance - or both.

It seems like WBahn is saying the opposite, that in his example the 10 A circuit has less energy/J than the 2 A circuit. How can this be if 10 A means more electrons per second than 2 A, which should mean more energy is needed to push more electrons per second. Yet is seems WBahn is saying the 2 A circuit has more energy than 10 A, which is where I get confused.

If more water molecules per second flow in a pipe (10 A example) then it must have greater energy than less water molecules flowing (2 A example) in the pipe. I am making molecules of water analogous to coulombs/amps...

 

The change related to FREQUENCY is caused by the change in impedance, which is the vector sum of resistance and reactance. The two terms can be confusing .

So really continuing on to AC circuit theory is what the TS needs to do. And also more study of DC circuit theory. It is not that hard nor that complex, but it is a very important part to understand because it is the foundation of all of the rest of the understanding. The relationship between voltage, current, and resistance need to be understood completely before looking at power and energy.

 

Firstly, let us correct the typos in the example posted by @WBahn.

He wrote:
Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.
Take a 20 V source across a 5 Ω resistor. You have 5 A and 100 W.
Take a 10 mA source across a 2 mΩ resistor. You have 5 A and 0.5 W

Here are the corrections in red:
Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.
Take a 20 V source across a 4 Ω resistor. You have 5 A and 100 W.
Take a 10 mV source across a 2 mΩ resistor. You have 5 A and 0.05 W

 

Now I will give you two different examples.

Take 2A flowing in a 10Ω resistor.
Take 10A flowing in a 0.4Ω resistor.

Does the 10A circuit generate more power than the 2A circuit (intuitively)?

Current requires voltage in order to flow. The flow is opposed by resistance.

I = V / R

Without resistance, the current would be infinitely large. There would be no voltage require to generate current.
Real materials have resistance. It takes voltage to push the current against the resistance.

Hence we have three parameters. V, R and I.
Since these are interrelated by Ohm's Law,

I = V /R

we need to know two of the three parameters.

Knowing current alone does not tell us anything about the resistance or voltage.
Neither does it tell us anything about the power.

In the example I have given, we can calculate the power required to push the current.

P = I x I x R

The 2A circuit requires 2A x 2A x 10Ω = 40W.
The 10A circuit requires 10A x 10A x 0.4Ω = 40W

The two circuits require the same amount of power.


btw, energy and power are not the same thing.
Both circuits require 40W of power.

Energy = power x time

In order to run the circuit for one hour, total energy required is 40Wh.
This energy is lost as heat energy by the resistors. Both resistors generate 40Wh of energy for the duration of one hour.

Now that we know R and I, we can calculate the V required in each case.

V = I x R

Case 1, V = 2A x 10Ω = 20V
Case 2, V = 10A x 0.4Ω = 4V

To verify,

P = I x V

Case 1, P = 2A x 20V = 40W
Case 2, P = 10A x 4V = 40W

The 2A circuit requires 5 times more voltage than the 10A circuit.
The power is still the same. What is different is the resistance.
It takes 5 times more voltage to push 2A through the higher resistance.

The coulombs/second is lower but the power is the same since you are pushing harder against a higher resistance.

 

Now I will give you two different examples.

Take 2A flowing in a 10Ω resistor.
Take 10A flowing in a 0.4Ω resistor.

Does the 10A circuit generate more power than the 2A circuit (intuitively)?

Current requires voltage in order to flow. The flow is opposed by resistance.

I = V / R

Without resistance, the current would be infinitely large. There would be no voltage require to generate current.
Real materials have resistance. It takes voltage to push the current against the resistance.

Hence we have three parameters. V, R and I.
Since these are interrelated by Ohm's Law,

I = V /R

we need to know two of the three parameters.

Knowing current alone does not tell us anything about the resistance or voltage.
Neither does it tell us anything about the power.

In the example I have given, we can calculate the power required to push the current.

P = I x I x R

The 2A circuit requires 2A x 2A x 10Ω = 40W.
The 10A circuit requires 10A x 10A x 0.4Ω = 40W

The two circuits require the same amount of power.


btw, energy and power are not the same thing.
Both circuits require 40W of power.

Energy = power x time

In order to run the circuit for one hour, total energy required is 40Wh.
This energy is lost as heat energy by the resistors. Both resistors generate 40Wh of energy for the duration of one hour.

Now that we know R and I, we can calculate the V required in each case.

V = I x R

Case 1, V = 2A x 10Ω = 20V
Case 2, V = 10A x 0.4Ω = 4V

To verify,

P = I x V

Case 1, P = 2A x 20V = 40W
Case 2, P = 10A x 4V = 40W

The 2A circuit requires 5 times more voltage than the 10A circuit.
The power is still the same. What is different is the resistance.
It takes 5 times more voltage to push 2A through the higher resistance.

The coulombs/second is lower but the power is the same since you are pushing harder against a higher resistance.

Thank you MrChips. Excellent examples that really demonstrate the point without the need for analogies. Very well done.

I take your point that the 10 A circuit has the same power as the 2A and yet the 2A needs more voltage, the key here being the 2 A circuit has a much greater resistance to push through so it uses more voltage for the same power as the 10 A circuit. Point taken and very well expressed. This really has cleared it up for me.

Can you please provide one more example that shows how a 2 A circuit can have/need MORE power than a 10 A circuit? At least this is what I think Wbahn is saying in his example although I may have misunderstood. In this case the 2 A circuit.

This is my attempt at explaining WBahn's example leveraging your fine examples:

V = I x R

Case 1, V = 2A x 20Ω = 40V
Case 2, V = 10A x 0.4Ω = 4V

To verify,

P = I x V

Case 1, P = 2A x 40V = 80W
Case 2, P = 10A x 4V = 40W

If my modification of your examples is correct case 1 shows a circuit with more power but less amps than case 2 which has less power but more amps.

Okay, it would appear that we are going in circles here I know. Its all really just proving ohms law, my "mistake" has been in thinking that more amps means more power, but it doesn't because P = I x I x R and amps are more a less the by product.

Looks like P = I x I x R will be my new mantra for a while . Thanks again.

 

So in this example the 10 A circuit has less voltage and/or resistance than the 2 A circuit? Is the voltage determining the energy each coulomb has? What am I missing here? If 10 A means 10 coulombs per second does that have more flow/energy that the 2 A??

Sorry I do not quite get this, but a I do find the water tank analogy very helpful where the load/resistor might be a valve or a tap that is partly open of fully open...and the volume of water that flows as the rate at which it flows is the amps...

Flow and energy are not the same thing! You need to stop equating them.

The definition of voltage is energy per unit charge (joules per coulomb). So, yes, the voltage drop determines the energy that each coulomb has.

 

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.

Be careful, power and energy are not the same thing. Power is energy divided by time. I can make two circuits that consume the same amount of energy in a resistor, and one gets red hot, while the other barely gets warm. How? By dissipating the energy in one circuit over a second, while doing so in the other circuit in an hour.

Because power is energy divided by time, the first circuit will be operating at 3600 times the power of the second, but both use the same amount of energy.

The second resistor does not get as hot because the heat can be taken away by the air in that longer time.

Bob

 

Okay, I cannot let this one go.

If current is the "number of electrons per second" (coulombs aka amps) does it not follow that the more current you have the greater the emf has to be or there needs to be less resistance - or both.

No, it does not follow. It only follows if you insist on equating flow with energy. You need to stop doing that.

I used to work with superconducting magnets. Once you got the current to the level you wanted you could put them in persistent mode (basically short the power leads on the magnet) and remove the power supply completely. You now had hundreds of amperes flowing in the circuit with no power at all and it would stay that way for months with very little drop in current due to minute parasitic losses. So current flow simply does not equate to power or energy. They are two different things.

It seems like WBahn is saying the opposite, that in his example the 10 A circuit has less energy/J than the 2 A circuit. How can this be if 10 A means more electrons per second than 2 A, which should mean more energy is needed to push more electrons per second. Yet is seems WBahn is saying the 2 A circuit has more energy than 10 A, which is where I get confused.

In the SAME resistance, it takes more energy to push more current. That's what Ohm's Law is all about. But in two DIFFERENT resistances, it can take considerably less energy to push more current through the low resistance than it takes to push a smaller current through the greater resistance.

If more water molecules per second flow in a pipe (10 A example) then it must have greater energy than less water molecules flowing (2 A example) in the pipe. I am making molecules of water analogous to coulombs/amps...

You are fixing the resistance by using the same pipe.

But make one of those pipes 1/16 inch in diameter and the other 16 inches. Or use the same diameter pipe but put a tight fitting sponge material in one of them.

 

Firstly, let us correct the typos in the example posted by @WBahn.

He wrote:
Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.
Take a 20 V source across a 5 Ω resistor. You have 5 A and 100 W.
Take a 10 mA source across a 2 mΩ resistor. You have 5 A and 0.5 W

Here are the corrections in red:
Take a 10 V source across a 2 Ω resistor. You have 5 A and 50 W.
Take a 20 V source across a 4 Ω resistor. You have 5 A and 100 W.
Take a 10 mV source across a 2 mΩ resistor. You have 5 A and 0.05 W

Fixed the typos -- thanks for catching them.

 

Fixed the typos -- thanks for catching them.

Oops. You missed one more.
Second line is 4Ω, not 5Ω.