Lets say I have a 500 watt, 12 volt DC power supply. This means I should be able to draw roughly 41 amps (12V*41A =~ 500).

This is far more current than my mains power can provide. My circuit breakers are like 15 amps.

Well, also the supply side is at a different voltage. It supplies power at 120 volts AC. So, 500W / 120V = 4.1A, it should draw about 4.1 amps to supply the required power. Well within my mains power spec.

Here's where I get confused. Since the ampere is a measure of "charge per second" does this mean there is more charge coming out of the PSU than going in?

  • \$\begingroup\$ Charge in Coulombs is Amps per second! \$\endgroup\$ – Leon Heller Sep 23 '15 at 19:42
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    \$\begingroup\$ @Leon Heller, A is C/s \$\endgroup\$ – Chu Sep 23 '15 at 20:12
  • \$\begingroup\$ 1 Amp flowing through a conductor means that 1 Coulomb of charge is flowing through the conductor each second. This 1 Coulomb of charge equals about 6.241*10^18 electrons \$\endgroup\$ – Bimpelrekkie Sep 23 '15 at 20:27

Current flows in loops (always) and your power supply actually has two loops that matter for this.

Simplifying slightly there is a primary side loop, Live -> Psu -> Neutral (completed by the power companies equipment), and a secondary side loop made up of the 12V wiring, the 12V return and the load.

In both loops the same amount of charge flows into and out of the PSU on its respective loop, so on the DC side it is 41 coulombs/second (at full load), and on the AC side it is an average of around 4 coulombs/second (Actually closer to 5 or 6 probably for various reasons, also on the ac side this is slightly simplified).

In both loops the same amount of charge exits the supply on one wire as enters it on the other, energy is converted by moving the charge thru a potential difference and it is energy that is transferred between the primary and secondary circuits, not charge.

Note that because the same amount of charge exits on one wire of the loop as enters on the other (to a very good approximation), there is little net charge built up in the power supply, it this condition did not hold the charge on the supply would build until something (quickly, and probably explosively) failed.

Regards, Dan.

  • \$\begingroup\$ Ah, so there's a difference between charge and energy. A battery drains because charge is used to create energy as it moves from a "high" potential to a "lower" potential. In this situation, the mains power is being used to create energy to create a potential difference across the secondary loop which causes charge to move in that system which is what creates the energy then consumed by the device. Is this close? \$\endgroup\$ – MichaelB Sep 23 '15 at 19:50
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    \$\begingroup\$ "A battery drains because charged is used.." No this is not how it works. The charge does not contain any energy. Charge is just some electrons moving from one place to another. The movement of the elctrons costs or gives the energy. The energy in a battery is stored chemically. A chemical reaction causes the electrons to flow. The charge (amount of electrons) is always kept constant !!! \$\endgroup\$ – Bimpelrekkie Sep 23 '15 at 19:52

You assume that the 41 amps also flow at the mains side. This assumption is WRONG. A 12V 41 Amps supply would be a switching supply. With such a supply 120 V 4.1 A goes in, 12 V 41 A comes out. Such a supply might be sitting in your PC.

Current is charge flow per time indeed. With emphasis on the FLOW. The charge isn't coming from the mains side. It's just the energy to move the charge that comes form the mains side. Heck, no charge comes to your house from the Power plant ! It is the movement of the charge that transports the energy !

In a switching power supply, the electric energy is converted into a changing magnetic field (magnetic energy) and then converted back to electric energy (at a different voltage and a different current).

  • \$\begingroup\$ A minor typo but I think you mean 120v not 12v? \$\endgroup\$ – BenG Sep 23 '15 at 19:53
  • \$\begingroup\$ What do you mean that "no charge comes to your house from the power plant, it is the movement of charge that transports energy" Doesn't that mean the charge "moves" to your house (or at least the substation/transformer/whatever), crosses a device and then moves back on the return/ground side? \$\endgroup\$ – MichaelB Sep 23 '15 at 20:03
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    \$\begingroup\$ Actually the charge is just moving back and forth through the wires in your house and the transformer outside. Compare it to a system of closed pipes filled with water. The water (charge) is trapped inside the system. The energy is transferred by moving the water. The charge/current actually moves back and forth 60 times a second since your mains will be AC (as opposed to DC like a battery where the current will keep flowing in one direction only). \$\endgroup\$ – Bimpelrekkie Sep 23 '15 at 20:10
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    \$\begingroup\$ @MichaelBeattie the charges are already in the conductor, at any time. Without a potential they just won't move anywhere. Also think about alternating current, the charge moves back and forth 50 or 60 times a second, it's not fast enough to get from the power plant to your home and back (it's moving in the order of cm/hour). \$\endgroup\$ – Arsenal Sep 23 '15 at 20:10
  • \$\begingroup\$ Actually the charge cannot even get to the powerplant as the powerplant's circuits are isolated from yours by a transformer. But that doesn't matter, the energy is in the movement of the charge. The transformer converts the movement to magnetic energy and back. \$\endgroup\$ – Bimpelrekkie Sep 23 '15 at 20:13

Easy explanation comes from energy conservation law. With 100% efficiency the output power equals input power, 500W in 500W out. 120*4.1 = 12*41

  • \$\begingroup\$ You haven't said anything more than I did. If amps are a measure of electrons, then there are more on one side than the other. Where do the extra ones come from? \$\endgroup\$ – MichaelB Sep 23 '15 at 19:41
  • \$\begingroup\$ @MichaelBeattie, There are trillions of electrons just sitting around in every cm of wire, waiting for something to do. \$\endgroup\$ – The Photon Sep 23 '15 at 20:11
  • \$\begingroup\$ @MichaelBeattie but you can only measure the amps if there is a loop ! This loop is needed to move back the electrons to the other side. \$\endgroup\$ – Bimpelrekkie Sep 23 '15 at 20:15

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