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I'm trying to understand the difference between amperes and watts. I have a string of LEDs that say require 5W of power and 5amps. What's a good way to memorize the difference and significance of these two values?

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  • \$\begingroup\$ also, my friend just suggested a great analogy. The amperes from a power supply are similar to watts from an engine. So if your circuit calls for more amperes than the power supply can provide, its like a scooter engine pushing a mack truck. eventually it will burn out the smaller "engine". \$\endgroup\$ – 4m1r Nov 7 '16 at 14:47
  • \$\begingroup\$ Rough analogy: Ampère is how many buckets of electrons/water/... you carry per hour, Volt is how far you carry them uphill. And the power you need to get the job done in an hour is Watt=Volt x Amp = how many buckets you can carry how far uphill in an hour. Lifting 10 buckets each hour 1m up is still 10 buckets per hour (Amps), but not much voltage/altitude, and hence not much work per hour (Watt). Lifting the same 10 buckets each hour 10m up is 10 buckets per hour too, but 10m is much more than 1m, i.e. more voltage required, and hence more work per hour needed. \$\endgroup\$ – JimmyB Feb 14 '17 at 9:24
  • \$\begingroup\$ Surely watts from an engine are like .. electrical watts? Whereas amperes are more like torque? \$\endgroup\$ – pjc50 Feb 14 '17 at 9:28
  • \$\begingroup\$ The torque is like the voltage (the force to drive the amps) while speed/RPM is like current. \$\endgroup\$ – JimmyB Feb 14 '17 at 13:38
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Well, the definition of an ampere is quite figurative: It is the current through two parallel infinitely long and thin wires, that are 1 metre apart from each other, so that the force between them is exactly \$2 \cdot 10^{-7}\$ Newton per metre (thanks to @rioraxe for checking the numbers).

Admittedly, this is not easy to transfer to reality. In fact, it is very hard to give another example. It is easier, when you think in power, rather than current. At 5 V, you get 5 Watts. At 230 V you get 230 W at 1 A.

Concerning your case with LEDs at 5V, you need to take light efficiency into account. LEDs have around 80 to 120 Lumens per Watt. So if you have 25 watts in LEDs, you can think of around 2500 Lumens, which is around as much as two Lightbulbs with 100 Watts have.

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    \$\begingroup\$ I never knew this is the definition, but the number seems off. Wikipedia gives the same definition except it works out to exactly 1 Newton per 5 x 10^6 meters (or 2 x 10^-7 Newtons per meter). \$\endgroup\$ – rioraxe Nov 6 '16 at 21:47
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To a first approximation, amps is the number of electrons passing a given point in a second (divided by about 6 x 10^18, but it's proportional - twice as many electrons, twice as many amps). I say to a first approximation, since for p-type semiconductors the current is produced by moving vacancies in the electron cloud, and no single electron can be singled out for counting.

Watts is how much work you do in a unit time interval in pushing those electrons, and for electronic systems it's volts times amps.

Think of electricity in a wire as water in a pipe. Amps is the flow rate, volts is the pressure drop from one end of the pipe to the other, and watts is the power needed to move the water - or the power produced by moving the water as in a hydroelectric generator.

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    \$\begingroup\$ you left out "per second" after "work". Power is not energy. \$\endgroup\$ – Sredni Vashtar Nov 7 '16 at 8:39
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5 Watt is 5 Ampere at 1 Volt. Or 1 Ampere at 5V. Or 2.5 Ampere at 2 Volts. Or any other combination which multiplies to 5.

$$P = U \cdot I$$

$$(W = V \cdot A)$$

To make a water-analogy, electrical current (measured in Ampere) is the amount of water you get from your tap in a second. But if you want to calculate how much power (measured in Watt) that stream has, you also need its speed (measured in Volts).

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  • \$\begingroup\$ FWIW, I tend to take voltage as the pressure of the water in the pipe, which, of course, turns into speed once released from the tap. Yet, the pressure/voltage is also there when the valve is closed, while speed is 0 then. \$\endgroup\$ – JimmyB Feb 14 '17 at 9:10
  • \$\begingroup\$ Using pressure for the water-anology gives you another complication, because people assume pressure lowers once the valve is opened. This is only true because real-life piping has a non-neglible "resistance" but at the same time nobody knows how to handle that in their thoughts. \$\endgroup\$ – Janka Feb 14 '17 at 13:20
  • \$\begingroup\$ I can't think of a hydraulic analog where current is flow and "speed" is volts. You might certainly have an analog where one or the other is true, but not both. Please correct me if I'm wrong. \$\endgroup\$ – Scott Seidman Feb 27 '17 at 20:34
  • \$\begingroup\$ The problem you might have is you are thinking with a fixed outlet size. Then, the speed of the stream and the amount of water per second is proportional. A fixed outlet size is a fixed resistance. Think it the other way: double the outlet size. Adjust the stream so the speed is the same as before. Now you have the double "water current". (Same voltage, half the resistance→doubled current) \$\endgroup\$ – Janka Feb 27 '17 at 23:13
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To answer your question about a memorization device, when I was starting out learning about electrical engineering I thought of Amp sounding like "amplitude" because it is the amount of electricity flowing through the conductor in question. Volt can be remembered as being like "vault" since it is like the pressure pushing the current, it can remind you that this relates to how far it can jump. Pneumonic devices work for me, even if they are kinda dumb.

Just as a side note in case I might confuse anyone, these units are named after physicists André-Marie Ampère and Alessandro Volta respectively. They have nothing to do with amplitude or vaulting.

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No, watts from an engine are not like amps. Watts from an engine are like watts from a power supply. Both are a rate; the flow-rate of energy.

Gut-level verbal/visual viewpoint:

"Amperes" are the flow rate of charge: one coulomb flowing per second is one ampere. And, in copper, one coulomb of electrons is about the size of a grain of salt. When you close the circuit, all the coulombs start slowly moving along as one, like a circular drive-belt inside the wires. Electric current is like the speed of the belt. But not exactly the speed, since the flow rate is faster through thinner wire and slower through thicker. But still, if we double the current in a circuit, the coulombs move twice as fast in every section, and the amperes are doubled.

Watts? That's like rubbing your thumb on the moving belt. GETS HOT! Watts is energy flow, is the rate of performing work (and is the output wattage of heaters and light sources.) The harder you push your thumb, the greater is the back-pressure on the "drive belt" made of charge. Voltage is when your thumb produces a tension-difference in the belt on either side of your thumb (that's why the ancient name for voltage was "tension," as in H.T. power supplies, "high tension.") To tap some wattage out of the slow-moving drive-belt, we need the belt to be moving, and also need a tension-difference. Double both of them and you get 4X watts: joules/sec equals coulombs/sec x volts.

Also: the belt, the coulombs, moves slow. Amperes are a slow flow. Yet the energy moves almost instantly. When you rub your finger on a moving belt, you instantly tap energy out of the entire belt, slowing down the whole thing as a unit.

Finally: the power supply speeds up the "belt," increases the current, while all the "rubbing thumbs," the loads, slow it down again. The forces balance out when the belt is moving at a certain rate, where amperes equals the voltage-force of the power supply divided by the resistance of all the loads. Reduce the resistance of the loads, and the speed of the charge within the wires will rise, until it hits a particular faster value (higher amps.)

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