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I realize this is probably a super common question, but I have to put it in my own words.

I try to understand ohm's law with the water analogy. Two tanks with water, one with a higher level than the other, and a pipe connecting the two of them. Water wants to flow. There is a valve which represents a resistor.

The thing that begins my confusion is when I start thinking about heat dissipation in an electrical circuit. Where does this heat actually come from?

It can't come from the pressure, the voltage, because if it did, the valve should be extremely hot if there is simply enough water in the higher of the two tanks, thus exerting a lot of pressure on the valve.

I've read that the heat comes from the actual flow of electricity, the current. At first this seems intuitive. But then I progress on to consider what power is. This is where the confusion sets in. Because if I double the pressure and double the resistance, the current stays the same. I would think this would mean that the heat dissipation would stay the same.

But the power doubles. So what does that actually mean then?

Has my tank drained into the other tank at a different rate because of the higher power, even though the actual flow of current has stayed constant?

What is power?

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  • \$\begingroup\$ Sometimes it helps to step away from the analogy and look at the concrete facts. Power is energy over time. Voltage is energy over charge, and current is charge over time. Multiplying both gives power. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 7 '13 at 4:38
  • \$\begingroup\$ A watt-hour is energy over time, right? How does that make a watt energy over time? \$\endgroup\$ – Kelsie Dec 7 '13 at 4:39
  • \$\begingroup\$ Or am i misusing terms? \$\endgroup\$ – Kelsie Dec 7 '13 at 4:39
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    \$\begingroup\$ A watt-hour is energy. 3600 joules, to be exact. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 7 '13 at 4:39
  • \$\begingroup\$ I'm not sure I understand the term "energy over charge". \$\endgroup\$ – Kelsie Dec 7 '13 at 4:41
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You can think of what happens to electron flow in the wire. Although not true, try to think of the electron as a mechanical particle. Whenever it tries to move whithin the wire, it hits something and that collision generates heat. So you can think of a energy being transfered from kinetic energy of the electron to heat (so electron speed falls down in that moment). So the electrons do not have constant speed all the time altough we may say they have an average speed and this average speed depends on the resistance of the wire which is exaclty this obstacles that the electron hits.

The wire would not became hot if it does not have any resistance. So no power would be consumed by the wire.

When you double voltage and also double the resistance, you may think that the electric field within the wire is higher so the electron may reach a high speed faster than with lower voltage. But also the resistance is higher so it may hit its obstacles in a stronger way. So the average speed may became the same (current is the same) altough now you are dissipating more heat because collisions are stronger.

This is a very gross way to think but it may help you imagine why things are the way they are by some analogies.

Also, you may think power as Joules per second (Watt). So it relates some energy unit per time. In ohms law example, it applies to heat dissipation. In other words, how much energy is being wasted in heat throughout the wire. If you think of mechanical systems, power may represent how much energy is necessary to move something (you can calculate it's minimum kinetic energy to achieve a desired speed and so calculate how much energy you should transfer to this object to reach that speed). So, as power is direclty related to energy, you may think that energy is always being transfered from one manner to another. Power may indicate how fast this things happen.

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  • \$\begingroup\$ You are welcome. I'm glad it was useful. \$\endgroup\$ – Felipe_Ribas Dec 7 '13 at 5:01
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    \$\begingroup\$ "The wire would not became hot if it does not have any resistance. So no power would be consumed by the wire." - superconductivity. \$\endgroup\$ – sherrellbc Jun 26 '14 at 14:49
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For a mechanical analogy, think of electrical resistance as mechanical friction, voltage as force, and current as speed.

Assume that there is an object, subject to friction, moving with a constant speed (this is analogous to an resistive circuit with a constant current).

There must be an applied force (analogous to a voltage source) and an opposing frictional force (analogous to the voltage across the resistor).

Now, as you've surely observed, friction converts kinetic energy to thermal energy (think of how brakes get hot when stopping your car quickly from a high speed).

The associated power is the rate of this energy conversion; it's how much kinetic energy is converted to thermal energy per second.

It's easy to show that the power associated with a frictional force \$F_f\$ and a speed v is given by:

$$P_f = F_f \cdot v$$

This should be intuitive to you. If you move your hands slowly together, you will not feel much if any heat. If you move your hands rapidly together, you can warm them up quickly.

The frictional force is given by:

$$F_f = \mu \cdot v$$

where \$\mu\$ is the coefficient of friction. This is analogous to Ohm's Law

$$V = R \cdot I$$

Finally, let's address your question:

Because if I double the pressure and double the resistance, the current stays the same. I would think this would mean that the heat dissipation would stay the same.

But the power doubles. So what does that actually mean then?

In our mechanical analogy, what happens if we double the friction (which is analogous to doubling the resistance) and assume the object's speed remains the same (which is analogous to the current remaining the the same)?

The frictional force doubles and thus, the power due to the frictional force doubles.

Mechanically, this is intuitive. If you're in a car at a constant speed and the rolling friction suddenly doubles, you'll need to double the engine power output (press the gas pedal harder) to maintain your speed.

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  • \$\begingroup\$ Well explained Alfred and excellent analogy!!! \$\endgroup\$ – AKR Jan 17 '14 at 10:55
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Power is first and foremost a rate of change in energy. If energy was money then power lost would be your monthly expenses, and the power gained would be your monthly income. If they are both equal then there is no net energy change each month.

But what is energy really? Energy is the thing you need to make work, like lifting something heavy (against a gravitational field), or pulling apart two magnets (against a magnetic field), or displacing charged particles (against an electric field). It is this last example that applies in basic electricity.

You can usually define some kind of particle that is sensitive and can get pushed around by being within a field, and a field is just a way of visualizing and quantifying within the degrees of freedom of this particle (like spatial coordinates), how strongly and in what direction it gets pushed.

So physically moving this particle across the field requires energy. If you define an arbitrary point A within the field and calculate the energy to get a particle to another point B, you could say that point B has a potential equal to this energy. Since A was arbitrary, it only makes sense to talk about potential differences.

In the context of an electric field, the sensitivity of particles (like electrons) to this field is called charge, and the units are called Coulombs. So the potential has units of energy/charge, or [Joules]/[Coulomb], which is the same as Voltage.

So if you have a potential difference between to points A and B in a circuit (a voltage), and there is a certain amount of charge going from A to B at a certain rate (a current), then there is a rate of energy being used up (power). It doesn't really matter how they went from point A to point B (via a wire, resistors, diodes, transistors, air, a pencil, etc), all that matters is the voltage and the current, and power is their product:

$$ Power = Voltage\cdot Current $$

You can check the units:

$$ \frac{[Joules]}{[Second]} = \frac{[Joules]}{[Coulomb]}\cdot \frac{[Coulombs]}{[Second]} $$

When you talk about resistance, you are just talking about how a material affects how much current goes through it, given a potential difference across it, but only resistors have such a simple linear relationship, so the \$P=V^2/R\$ relationship that was implied in your question is not true for anything else but idealized resistors, and that power equation is just a result of their property of the current being directly proportional to the voltage across them. The good news is that this can be part of a model of many real devices at particular operating points, so it is a very useful concept, I just want to clear up that it is not a complete model of any real device. In other words \$P=V\cdot I\$ is universal, \$P=V^2/R\$ is not.

Hopefully by now it should be clearer why without current there can't be any power (you aren't displacing any charged particles, so no work is being done), and why power doesn't depend just on the current (moving charges across zero potential doesn't require any 'effort'). It is really about how much charges you move per unit of time, and across how much potential difference.

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By definition, power IS the rate at which energy is transferred or varied.
If you take that as a fundamental essential, all other questions need to make sense with respect to this.
If the question does not "respect" that definition then the question does not make sense.
Trying to understand the answers to nonsensical questions is fraught with danger :-).


It can't come from the pressure, the voltage, because if it did, the valve should be extremely hot if there is simply enough water in the higher of the two tanks, thus exerting a lot of pressure on the valve.

You have pressure but no flow. Energy is not being transferred - there is no power required.

I've read that the heat comes from the actual flow of electricity, the current. At first this seems intuitive. But then I progress on to consider what power is. This is where the confusion sets in. Because if I double the pressure and double the resistance, the current stays the same. I would think this would mean that the heat dissipation would stay the same. But the power doubles. So what does that actually mean then?

Follow the energy.
As I = V/R = 2V/2R the current will not change when both V and R are doubled.
BUT the energy required to push the same current through a pipe of twice the resistance is double. Yes?
ie double both pressure and resistance -> the current is the same but the energy flow rate is doubled so power is doubled.


Note that Power

= VI = V^2/R = I^2R.

These formulae are functionally identical and are interchangeable.
You can go from one to the other just by substituting variables.
If any one of them makes sense to you then the rest can be derived from it just by plugging in variants for the variables based on Ohm's law.

eg
P = V x I But V = IR
So P = IR x I = I^2R

P = I^2R But I = V/R So P = (V/R)^2 = V^2/R

If you are happy with power being 'explained' by any of VI or V^2/R or I^R then the above allows you to show that the others are identical.


P = V x I
Energy rate is proportional to amount of stuff pushed and how hard it is pushed.

P = I^2R
Energy rate is proportional to how hard stuff is being pushed BUT proportional to the square of how much stuff is being pushed because when you double the amount of stuff pushed through a given pipe not only do you get twie as much stuff per time BUT it is twice as hard to push it.

P = V^2/R
Energy rate is proportional to the square of the pushing force BUT inversely proportional to how hard it is to push it.
1/R is easy as less effort = less energy needed.
If you double the force used you double the amount of force used so energy rate rises BUT flow rate also doubles (I=V/R) so you have to push twice as much twice as hard, hence V^2 term.

It all makes sense.
It is all consistent.
It can all be converted between various ways of saying it.
Any time any of these 3 do not appear to be true, attack the "reason" that it does not seem so and you will find that the reasoning has a flaw.
eg in the first example given there was no current flow so no energy transfer so no power.

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