Suppose I have a PV panel. On it's peak time it produce 7 A with 14 V. So to find out how many watts I get from it I would simply apply this formula:

\$P = I \cdot V\$

So, by applying above formula:

\$ P = 7~A \cdot 14~V = 98~W\$

So, I would get 98 Watts of power, right.

Ok, here are the diagrams how I took these reading of amps and volts: Measurement of Voltages

Measurement of Amps

Now, If I calculated the power coming from my wall socket by this same method, I could easily measure voltage but when measuring current, the short circuit will cause (because resistance of ammeter is very low) and BOOM! (I think so, this will happen). As current from PV was pretty less than the current from outlet so I assumed it would be safe.

So, basically my question is that there are standard 220 V in my wall socket, how much maximum current can I get from it?

Mentioning of 220 V reminded me that some countries have 110 V house supply. What is the reason for it? What's the benefit of having 220 V over 110 V or 110 V over 220 V?

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    \$\begingroup\$ Hopefully, you had a load when you measured the current or your reading is incorrect. \$\endgroup\$ – StainlessSteelRat May 22 '17 at 21:08
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    \$\begingroup\$ The proper answer is something along the lines of "How much are you willing to pay for?". You can get high-power outlets installed, if you need more power. \$\endgroup\$ – Hearth May 22 '17 at 21:09
  • \$\begingroup\$ Incidentally, I don't know of anywhere that uses 140V mains. It's 110V-120V in most of north america, and a few places use 100V. \$\endgroup\$ – Hearth May 22 '17 at 21:10
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    \$\begingroup\$ The fundamental flaw in the question is that shorting the PV panel is NOT the way to measure the maximum power it can provide at any particular insolation. Hence when the op suggests trying to measure the AC outlet in the same way that is equally flawed. The AC outlets are on a circuit (defined by wire size and breaker) that limits the maximum current from the circuit. At any given current rating (wire, breaker) for the circuit increasing the voltage (110, 120, 230, 240 etc) will increase the power that can be drawn from the circuit. \$\endgroup\$ – Jack Creasey May 22 '17 at 21:18
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    \$\begingroup\$ Read this: edge.rit.edu/edge/P12407/public/MPPT%20Testing.pdf If you need it explained, ask. From this you can see you measured short circuit current and open circuit voltage ...multiplying these does not give the maximum power output of the panel. Maximum power transfer is when the internal impedance of the panel is equal to the load impedance (electronicshub.org/maximum-power-transfer-theorem). Then read this: altestore.com/howto/solar-panels-pv-and-voltages-a98 \$\endgroup\$ – Jack Creasey May 22 '17 at 23:30

Don't put the AmMeter's probes in your outlet! Never! You are right about the BOOM :)

1) The local electricity company when connected your home made a design and gave you some reasonable allowed maximum power. It can be 5kW, 6kW, 10kW... depending on house size and local regulations. Find your contract and this information should be stated there. If you have no contract call and ask them. This is your legal limit.

2) In the way of the electicity to your home there are circuit brakers. Usualy they are rated for about 50-70% more than the power by your contract. A circuit braker may be for 32A, 50A, 63A, an so on. This (+some % overrating) is the maximum current you can draw from all outlets in your house at a same moment.

3) The wires that go from the circuit breaker to your outlet are calculated upon a prediction for a maximum current that will be drawn by this particular outlet. If you draw much more current for extended period (hours) the wires may heat up, melt isolation, accidently make short circuit and produce fire. A wire's current rating depends mainly on its cross section, but also on type of cable and how it is been layd in the wall, whether the wall is concrete or wooden and so on.

People study 5 years engeneering to design electrical installations. This cannot be explained in a few lines.

The benefit of 220/230/240V is that at a higher voltage a less current is needed to produce the same power. The losses in the carrying cables are proportional to the square of current flowing, so conduction losses will be 4 times less in a 220V system compared to a 110V system. This gives the possibility to have longer range between the local downstep transformer (for example one which takes 20kV and converts them to 220V) and the end-user.

  • \$\begingroup\$ 5kW? That's a bit mean isn't it? \$\endgroup\$ – Paul Uszak May 22 '17 at 21:24
  • \$\begingroup\$ It's not just outlets - don't short out a PV system, or in practical terms any source, with an ammeter. The only legitimate reason for doing that would be to measure short circuit current, and then only if you know that all of the source and the meter and the cabling and the way the experiment is activated can handle it. \$\endgroup\$ – Chris Stratton May 22 '17 at 22:17
  • \$\begingroup\$ @PaulUszak Why would that be mean? \$\endgroup\$ – marcelm May 22 '17 at 22:18
  • \$\begingroup\$ @marcelm 5kW power allowance for a house? In the USSR? \$\endgroup\$ – Paul Uszak May 22 '17 at 22:56
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    \$\begingroup\$ @PaulUszak The house I live in has a main breaker of 25A, single phase, at 230V, yieldinng 5.8kW. That is a pretty common configuration here (The Netherlands, western Europe - other common ones are 1x35A and 3x25A). It's plenty for many houses. I can't speak for the former soviet union. \$\endgroup\$ – marcelm May 23 '17 at 0:10

There are a lot of strange non-answers here!

In a typical 220V country, like England, the outlets are on 13A breakers. That is 3100 Watts from a single circuit, or a single outlet if nothing else is connected to that circuit.

In the US, nearly all outlets are on 15A breakers (not 20A as someone said in the comments. 20A outlets have a different socket.) at 120V RMS (Not peak-to-peak as someone said in the comments, which is 340V). This is 1500W, which is the value noted on devices like space heaters or power tools. 20A US outlets can be 2000W and there is often one in a kitchen for a microwave. Again, provided noting else is using power from the same circuit.

You can take more power briefly until the circuit breaker heats up or your wiring catches on fire. The source of the power doesn't care.

A typical US household has a 100A or 200A power panel and quite often have an additional panel for a shop for power tools and welders, etc. The power company will deliver 50kW and often more at an increased rate. Business and farms can get a 440 volt service at much higher power.

  • \$\begingroup\$ While 20 amp sockets are a different connector than 15 amp ones, in fact most breakers in recent construction are 20 amp, with multiple 15 amp sockets on the circuit. \$\endgroup\$ – Chris Stratton May 23 '17 at 3:27
  • \$\begingroup\$ In the UK (including England) we are on a nominal 230V (often more like 240V). Nobody sells 13A breakers. The sockets are often on 32A breakers, or sometimes 20A or 16A. That's why there's a fuse in the plug. \$\endgroup\$ – Simon B May 23 '17 at 8:17
  • \$\begingroup\$ @Simon B What current are the plugs fuzed at? What gauge wire is used? I picked up the 13A figure somewhere. \$\endgroup\$ – C. Towne Springer May 24 '17 at 5:48
  • \$\begingroup\$ @C.TowneSpringer The standard values are 3A or 13A. 5A fuses are also common in IT and similar equipment where there's a 6A IEC connector on the other end of the lead. \$\endgroup\$ – Simon B May 24 '17 at 7:28

So, basically my question is that there are standard 220 V in my wall socket, how much maximum current can I get from it?

The current is limited by the resistance of the wires to the outlet. Since these heat up when a lot of current is drawn through them we put restrictions on how much current can be drawn through the wires. I have no idea what the standard is in your area, maybe 6-10A on a 220V. If you got rid of the breaker you could put enough current through the wires to turn them into molten metal. So all you have to do is check your panel.

If you want to find the power for a different device, and you can figure out how to put an ammeter in series between the hot side and the device, then go for it. This is dangerous however. There are also power meters available so you don't have to strip wires, and the meter runs inline between the wall and your device.

Mentioning of 220 V reminded me that some countries have 140 V house supply. What is the reason for it? What's the benefit of having 220 V over 140 V or 140 V over 220 V?

There is no rhyme or reason behind most electrical standards, other than someone thought it would be a good voltage. No one runs 140V, its 100 to 120V and 220 to 240V. I believe these came from stepping down 440V and transformer ratios.


This is not an answer, but an attempt to address an incorrect assumption in the question.

So, I would get 98 Watts of power, right.

No. You can't determine the power you get that way.

You measured the open voltage (at 0 current) and the short-circuit current (at 0 voltage). From these you first need to calculate an equivalent source to determine how much power you can get out (see Thévenin's theorem).

An ideal voltage source of 14V in series with a 2 ohm resistor (the internal resistance of the source) gives the values you measured (14V and 7A). The following schematic shows this Thévenin source with a load connected.


simulate this circuit – Schematic created using CircuitLab

The maximum transfer of power from a source takes place when the load resistance is equal to the internal resistance. At that point, both the source (its internal resistance) and the load dissipate the same amount of power.

In this case, the current at that point is 14V/4ohm = 3.5 A. And the power that gets to the load 7V*3.5A = 24.5 Watts. Next to that, the panel itself (its internal resistance) disspiates another 24.5 Watts.

Please note: This is only valid when the panel is a linear element; I still think it shows how much you could be off.


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