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I know that watts are equal to amps times volts, and that watts are a rate of energy consumtion. I have looked at some circuits, but I am unsure of how the total energy use is being distributed. My guess—correct me if I’m wrong—is that the energy used by a component is the voltage across the component times the amps through it. Since the volts add up to the power source voltage, this maintains the total amount of watts.

No amps through something, no energy being moved. No volts across something, no pressure difference is necessary to traverse it.

Does this mean that a resistor is not just slowing the flow, but turning the energy into heat? An inductor that is “charged up” and has little resistance is not using any energy because there is no voltage across it, even though there are many amps? The vast energy loss in a short is in the existing resistance, the battery/power source?

I always imagined (but knew better) that less resistance should be easier. Perhaps it is; the wire is hardly drawing anything. But is the resistance in the source?

One last thing, maybe a bit off topic: is it easier to turn a generator connected to an open circuit than a closed one simply because no energy can be transferred? A water pump would get harder in that scenario. (In terms of work, less energy would be used if I am always applying the same pressure to the crank. It wouldn’t move, so no work.)

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    \$\begingroup\$ Yes, resistors turn electrical power into heat. This is called Joule heating. And yes, the majority of the energy lost in a short is normally in the internal resistance of the power supply, though with a sufficiently low-resistance source you can burn a lot more power in the wire--because we still aren't quite to the point of having superconducting wire in everyday applications! \$\endgroup\$ – Hearth Dec 3 '18 at 1:35
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I know that watts are equal to amps times volts,

Instead, let's say that Power is equal to the product of current and voltage. It's beneficial to keep the terms seperate from their units.

and that watts are a rate of energy consumtion.

Instead we'll say that Power, measured in watts describes a rate of energy consumption. By the way, this is known as Watt's Law, P=IE.

I have looked at some circuits, but I am unsure of how the total energy use is being distributed. My guess—correct me if I’m wrong—is that the energy used by a component is the voltage across the component times the amps through it. Since the volts add up to the power source voltage, this maintains the total amount of watts.

Yes, this is correct. Note that there are different methods of measuring and describing voltage, current and power, so if you want to actually figure things out, you need to differentiate between, for example, Vrms(root mean square voltage, a form of average) and Vavg(A more literal average voltage), and if you want to work on circuits containing capacitance or inductance, you have to differentiate between P(Real power, measured in Watts, a measure of actual power consumption), S(Apparent power, measured in var(Volt-Amps reactive), the power you will read if you actually measure the circuit) and Q(Reactive power, measured in VA(Volt Amps), which describes the portion of the apparent power that is not actually being used (something capable of storing power is feeding it back into the circuit)

No amps through something, no energy being moved. No volts across something, no pressure difference is necessary to traverse it.

Or, put another way, given that P=EI, due to the multiplication property of zero, if E or I equals 0, so too must P.

Does this mean that a resistor is not just slowing the flow, but turning the energy into heat?

Yes. Note that same as the different types of power mentioned above, for resistance, R is measured in ohms, but if you have capacitance or inductance in the circuit, you have to differentiate between different "types" of resistance. So we add the terms Z for impedance, measured in ohms, and X for inductive or capacitive reactance, also measured in ohms. "Resistance" is used to describe the portion of the impedance that dissipates(uses) power, and "Reactance is used to describe the portion of the impedance that is caused by power being stored and returned to the circuit.

An inductor that is “charged up” and has little resistance is not using any energy because there is no voltage across it, even though there are many amps?

Ermmm... This part seems a bit sketchy. An inductor resists changes in the current flowing in the circuit. If you try to increase current, it will store energy in it's magnetic field, decreasing the rate at which the current increases. If you try to decrease current, it will release the energy stored in the magnetic field to maintain the current flow. Inductors do have (ideally) low resistance, but can have high impedance. The actual resistance and the actual current flow will determine how much power an inductor wastes, whereas the voltage and impedance of the inductor will determine how much current flows in the first place. Note that constant DC power will not experience impedance from an inductor, only it's resistance.

Superconductors are the only conductor available to humankind that does not have resistance. In electrical theory, components are often treated as being "ideal", meaning they have no resistance at all, but in real life, they have as low a resistance as is practical. In order to have power dissipation, either voltage or current must be nonzero, so if there was as you describe, a large current flowing but no voltage drop, you would be describing a superconductor. You're probably wondering about a large current and a tiny voltage drop or a tiny resistance, perhaps too tiny to measure, and yes, if a device has a high current and a tiny voltage drop, that would indicate low resistance, low impedance and low(comparatively) power dissipation.

The vast energy loss in a short is in the existing resistance, the battery/power source?

I'm not sure what you're asking here. A "short circuit" is by definition a circuit with low impedance. Low impedance means a high current, which, without protection mechanisms to break the circuit or limit current, will produce tremendous heat and safety hazards.

When a large current is permitted to flow, it will dissipate power in everything it flows through, proportional to the resistance of each part.

I always imagined (but knew better) that less resistance should be easier. Perhaps it is; the wire is hardly drawing anything. But is the resistance in the source?

Hmm I think you need to clarify what you're asking here, "less resistance should be easier"? Easier for who, the user, the installer, the owner? You could also be referring to the fact that less resistance means electricity will flow more "easily", but if that's the case, "but knew better" should not be there.

Every conductor other than superconductors has resistance. So your electrical source, the wires that connect it to the load and the load itself, has resistance, and they all contribute their resistance to the circuit. So if you have a battery, 2 wires, and an LED forming a circuit, your total resistance would be the internal resistance of the battery, plus the resistance of the two wires, plus the effective resistance of the LED.

One last thing, maybe a bit off topic: is it easier to turn a generator connected to an open circuit than a closed one simply because no energy can be transferred?

Yes. The less electrical load is placed on the generator, the less countertorque it will place on the energy source, so if you're driving it with a gas motor, less gas will be used to maintain the rotor speed.

A water pump would get harder in that scenario. (In terms of work, less energy would be used if I am always applying the same pressure to the crank. It wouldn’t move, so no work.)

The only thing about this I would note is that an absence of work output in a system does not indicate an absence of energy use.

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  • \$\begingroup\$ Forget the third yellow bar from the bottom. I was sort of rambling. I’m no good at AC circuits or impedance because I have not gone through any education with it, but I see what you were saying with the inductor. I just meant that because if there is a noticable voltage difference and amps, it is storing energy, and when there wasn’t a noticable voltage difference but still amps, it has essentialy stopped draing energy. \$\endgroup\$ – Scott V. Dec 3 '18 at 15:26

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