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I have some struggles understanding how amperage works.

What I understand so far is that amps are the amount of electrons which get pushed by the voltage.
But if I have an outlet with 230V~ which is fused to 16A.

When I plug in a load of 1000W, the outlet supplies circa 4,3A.
But why doesn't it supplie 16A?

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  • \$\begingroup\$ I think current rating always confuses people, a current in a power supply rating doesn't means it constantly will provide that amount of current, rather it means the power supply will provide UP TO that current. Then your load will only, if you will "suck" as much current as it needs. (The amout of "current sucking" is a characteristic of the impedance of the load) \$\endgroup\$ – Kvegaoro Sep 29 '15 at 14:09
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The outlet does not supply 16 A because it does not behave as a current source but as a voltage source. Imagine the socket would always try to supply 16 A, then you switch off your toaster, what would happen ? The current needs to flow so the socket would increase the voltage to several thousands of volts until the current just sparks over ! Hmm, that does not sound very practical to me.

So instead the socket just maintains the voltage. If you have a load of 1000 W, at 230 V that would be 4.3 A. Ohm's law then states that it has a resistance of V/A = 230 / 4.3 = 53 ohms. When you connect this 53 ohms load to the 230 V mains only 4.3 A will flow and your load will consume 1000 W.

Now what would happen if you would connect a load of 230V / 23A = 10 ohms. Then 23 Amps would flow and the 16A fuse would blow to protect the wires !

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  • \$\begingroup\$ "Current supplies as much as it can." is getting closer to the answer I need. But my problem with this concept, I might missthink something(?), is that voltage * amperage is energy. with enough energie you can destroy stuff, which eventually leads to blown out electronics. Is really any load which needs 4,3A (just for example) robust enogh to hold off 11,7A(16-4,3)? For example the charger of my smartphone. Shouldn't it get really hot, holding of the spare amps? \$\endgroup\$ – Sempie Sep 29 '15 at 10:20
  • \$\begingroup\$ @Sempie: There aren't any "spare amps" to hold; only 4.3A gets in. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 29 '15 at 10:22
  • \$\begingroup\$ Voltage * Current is power. Voltage * Current * Time is energy. \$\endgroup\$ – JRE Sep 29 '15 at 10:37
  • \$\begingroup\$ The thing it's "holding off" is not amps but volts, really. \$\endgroup\$ – pjc50 Sep 29 '15 at 10:41
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    \$\begingroup\$ Also note that you are using 16 amps as the amount of current the outlet is trying to supply only because it is fused for 16 amps. What would happen if the fuse was replaced by a 20 amp fuse? Do you think the outlet would then try to supply 20 amps to every load? No, the outlet will only supply what the load needs in accordance with the impedance it represents. \$\endgroup\$ – Barry Sep 29 '15 at 13:32
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What I understand so far is that amps are the amount of electrons which get pushed by the voltage.

Incorrect. Current is what a resistance allows to flow when a voltage is placed across it. If the resistance and voltage are fixed, then the current will be as well.

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  • \$\begingroup\$ This answer is as clear as schoolbooks "here you got the formula, insert numbers and you get other numbers." How can a load blow out then, if no more current can flow through, than resistance allows for that voltage? Of course I know that these 3 variables are related in a fix way, but the question is WHY is it like that, how does it work? \$\endgroup\$ – Sempie Sep 29 '15 at 10:11
  • \$\begingroup\$ "Blowing out" is an excess of heat. Heat is power over time. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 29 '15 at 10:13
  • \$\begingroup\$ Afaik, electronics blow out if current is too high. correct? How can current be too high, if no more current pass through than resistance allows? the 230V~ aren't getting 500V~ out of magic. \$\endgroup\$ – Sempie Sep 29 '15 at 10:14
  • \$\begingroup\$ No. Too much power for the device's power handling capacity is what causes it to blow out. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 29 '15 at 10:15
  • \$\begingroup\$ this is exactly what I'm saying. Voltage is fix, so power can only result in higher amps. \$\endgroup\$ – Sempie Sep 29 '15 at 10:20

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