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Let's say I have a single phase power supply at 100 V and 10 amps at my wall outlet. If I were to connect a very low resistance element (say 0.1 ohms), using Ohm's law I could say the resistance draws V/R = 100/0.1 = 1000 amps. This is what some people around me say happens. However, I think the current is limited to 10 amps and cannot go above that otherwise it violates the law of conservation of energy as follows.

The electrical power supply is 100V × 10 amps = 1000 W. If the resistance were to draw a current of 1000 amps, the heat dissipated would be I^2×R = 1000^2×0.1 = 100000 W = 100 kW, which is not quite possible. So the actual current drawn should be 10 amps, and the heat dissipated by the resistance is 10^2×0.1 = 10 W. Therefore, the resistance is actually dissipating only a fraction of the power available.

Any explanations or clarifications will be appreciated.

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  • \$\begingroup\$ Have you a typo in "(say 0.1 amps)"? Did you mean 0.1 ohms? \$\endgroup\$ – Transistor Sep 23 at 17:54
  • \$\begingroup\$ Yes that was a typo, I have corrected it now. Thanks for pointing it out. \$\endgroup\$ – Sine Nomine Sep 23 at 17:59
  • \$\begingroup\$ You are just going to blow the fuse or breaker on any decent power supply. Fire and damage on anything badly designed \$\endgroup\$ – Hilmar Sep 23 at 18:40
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This is a good opportunity for a bit of experimentation!

I pick a dual outlet in my home, stick my multimeter probes in one of the sockets, and plug a 2200W electric kettle in the other.

Kettle off: 232.3 V AC RMS

Kettle on: 227.4 V AC RMS

The kettle's heating element is a resistor, about 24 ohms.

The mains' impedance and the heating element form a voltage divider, so I can calculate the "output impedance" of this main outlet. Voltage dropped by 4.9V for a current of about 9.6 Amps, so this electrical outlet has an impedance of about 0.5 ohms. That's an approximation, but it's good enough.

In this case, power used by the kettle is RI^2 = 2211W (with R=24 ohms). Using the same formula with R=0.5 ohms gives an upper bound of 46W for power lost in the electrical grid. Actual losses will be lower, as the impedance I measured at my electrical socket is not purely resistive, it has an inductive component, but a simple multimeter measurement won't distinguish between resistive and inductive impedance.

I'll pretend the source impedance is 0.5 ohms resistive from now on, and forget about the inductance.

If a very low impedance was connected to the socket, like the 0.1 ohms in your question, it would still be in series with the 0.5 ohms source impedance, so at most it would draw 230/0.6 = 383 Amps.

Since this is a voltage divider with 0.5 ohms source impedance on top, and 0.1 ohms on the bottom, there would be 230*0.1/(0.1+0.5) = 38.3V across the 0.1 ohm resistor (dissipating 14.7 kW) and the rest of the voltage up to 230V would be accounted for by losses across the electrical grid.

(230V-38.3V) * 383A = 73.5 kW would be lost in the electrical grid.

So the answer to your question is: you forgot to account for the voltage drop caused by the current in the wires, thus you got the wrong current.

However, the 0.1 ohm resistor will still dissipate a HUGE amount of power, and the wires will, too.

What determines the legal current limit for electrical outlets is the insulation's maximum temperature rating. Consider a 1.5mm2 wire (about 15-16 AWG). In France these are rated for 16A when used with an accurate circuit breaker, 10A when used with fuses. Such a wire can carry much more current if it is air-cooled, say by a fan. But inside a wall, surrounded by thermal insulation, heat has nowhere to go, and PVC insulation melts at a rather low temperature. So the ratings are rather conservative... but wires do have some thermal mass, so a huge overcurrent lasting a couple tens of milliseconds won't melt things, at least not too much. It might spot-weld switches in a permanently closed state, and other side-effects, though.

What determines the actual current limit is the source impedance, wire resistance, output impedance of the transformer in the street, etc. When using batteries, also consider their internal impedance.

For minimal losses you want this source impedance to be low, so the actual current limit will be much higher than the legal, safe rating. In this case, 380 amps is about 24x the trip current spec for a 16A breaker, and according to its datasheet, at this current, it will cut power in about 8 milliseconds. That's pretty quick, but still enough to vaporize the tip of a screwdriver blade. It also makes a nice bang. Wear safety glasses when working on mains voltage.

So, unless you use a bench power supply or other power supply that is designed to safely limit current when needed without wasting power when it does not, the need for efficiency (low resistance wires) dictates that actual current limit is always much higher than the safe current "limit". The first is enforced by physics (ie, whatever explodes or melts first opens the circuit) and the latter is enforced by circuit breakers, fuses, and other safety devices.

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Let's say I have a single phase power supply at 100 V and 10 amps.

We don't know from your description what will happen if you exceed 10 A. If it is fused or fitted with a circuit breaker then it will blow or trip at some current and time. (See the datasheets for details.) If it has a current limiting regulator then the current will limit at 10 A by decreasing the output voltage.

If I were to connect a very low resistance element (say 0.1 amps) [I think you mean ohms], using Ohm's law I could say the resistance draws V/R = 100/0.1 = 1000 amps.

Your calculation is correct.

This is what some people around me say happens. However, I think the current is limited to 10 amps and cannot go above that otherwise it violates the law of conservation of energy as follows ...

If the power is coming straight from the national grid then the current is limited only by the resistance of all the wiring from the various generators to the load. Given that it's a 10 A circuit it is unlikely to be wired with a heavy gauge wire so there will be some series resistance - say 1 Ω. That in itself would limit the maximum current to 100 A.

If the power is coming from, say, a 100 V battery bank then it too will have some internal resistance but this can be quite low and very large currents can flow and discharge frightening and dangerous amounts of energy.


Comments:

I was under the impression that the power supplied from a grid comes with a voltage and current rating. So I was saying if the electricity supplied to my outlet is rated at 100 V and 10 amps.

The current limit is determined by design and local regulations. The following are some of the factors:

  • The maximum voltage drop you can tolerate. (The voltage drop will increase with current.)
  • The maximum temperature you can tolerate in your wiring.
  • The rating of switches and contacts in the circuit.

All of those determine the rating. The fuse or circuit breaker are added to ensure that the design specification is not exceeded but they take some time to trip. Relatively huge currents can flow in the meantime.

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  • \$\begingroup\$ Hi Transistor, I was under the impression that the power supplied from a grid comes with a voltage and current rating. So I was saying if the electricity supplied to my outlet is rated at 100 V and 10 amps. \$\endgroup\$ – Sine Nomine Sep 23 at 18:05
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Sep 23 at 18:13
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Is the maximum current a resistance can draw limited by the maximum current available (current rating) in the power supply?

The current rating of most power supplies isn't a promise not to deliver more than that much current. It's a requirement for the user not to try to draw more current.

Depending on the design of the power supply, when you draw more current than the rating the behavior could be

  • Work normally, delivering more than rated current (this one would only happen for a small over-current load)

  • Blow a fuse or trip a breaker

  • Overheat and reduce the operating life

  • Fail in some other way, permanently damaging the source

  • Overheat and catch fire, risking life and property around it

  • Sharply reduce the output voltage and current. To get this behavior, buy a source with a foldback circuit.

  • Reduce the output voltage to maintain the output current at the specified limit. This is what your question implies is the normal behavior, but it's actually very uncommon. If you need this behavior you'll need to find a source that specifically offers it as a feature, which might be called current limiting.

If the resistance were to draw a current of 1000 amps, the heat dissipated would be I^2×R = 1000^2×0.1 = 100000 W = 100 kW, which is not quite possible.

Why not. Just make your resistor very big, and you should be able to dissipate 100 kW easy. You could used forced air, water, or oil cooling if necessary. But there's not theoretical reason a resistor can't be made to dissipate 100 kW.

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  • \$\begingroup\$ Thanks for the answer photon. The reason I said the heat dissipation of 100 kW would be impossible is that the power being supplied is 100 V × 10 A = 1000 W = 1 kW. But if current higher than 10 A can be drawn, as you suggested, then that would work. Regardless, for conservation of energy to apply, there should be an upper limit on the current that an element can draw, I feel. \$\endgroup\$ – Sine Nomine Sep 23 at 18:21

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