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This question has been bugging me for a while. I know that theoretically its like - infinity to + infinity and it depends on the load that is connected to the mains supply plus the load at which the fuse/circuit breaker can handle max.

Lets take an example.Assume that I bypass the power meter and everything and directly connect to the pole utility. Will my current ability increase as I connect a bigger cable to draw more and more current? What will be the max current is possible to be drawn?Is it limited by the max current capability of the utility transformer?

Also what could be the max current at the source of generation of electricity(lets take an example of the generator at nuclear power plant, its not infinite anyway as it could handle only a particular amount of current )?Thanks in advance everyone. Cheers :)

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closed as unclear what you're asking by Andy aka, Chris Stratton, uint128_t, Dmitry Grigoryev, laptop2d Apr 26 '17 at 3:36

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ can you please structure your question a little more? A paragraph or two (a paragraph can be made by inserting empty lines between a logical section) would really help the readability! \$\endgroup\$ – Marcus Müller Apr 24 '17 at 6:15
  • \$\begingroup\$ @Marcus sorry for the cluttering. \$\endgroup\$ – Rahul Salin Apr 24 '17 at 6:21
  • \$\begingroup\$ also, your question is (unintentionally) incredibly broad. The answer is that you would have to know the immediate current draw of every "tap" on your local sub-grid; then you'd need to compare that to the total generating capacity of all connected, active (or activatable) generation plants. \$\endgroup\$ – Robherc KV5ROB Apr 24 '17 at 6:22
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    \$\begingroup\$ Generators are required to have safeties in place to disconnect them from the grid & shut-down in case of any overload. \$\endgroup\$ – Robherc KV5ROB Apr 24 '17 at 6:30
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    \$\begingroup\$ I'm tempted to flag this question as "unclear". Do you want to know the maximum rating currently used on earth, what currently limits the grid or what's necessary to construct a grid (generator, power lines, transformers) that delivers twice, thrice, ... the power? So is your question on "applied" current laws (P = I x V) or how the grid currently limits the maximum available power. This might be an X/Y problem - asking for limits while wanting to know how the grid from plant to socket works. \$\endgroup\$ – try-catch-finally Apr 24 '17 at 9:00
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The answer is not infinity, even 'theoretically'.

There are at least 3 defined currents for any point on the supply, and that point can be anywhere from the terminals of a nuclear power station, down through transformers and transmission lines, right to the socket on your wall.

1) Rated current

This is the maximum current it's designed to deliver, 24/7, without overheating. All upstream components are designed to carry this indefinitely.

2) Minimum fault current

This is the minimum current that should flow if you short circuit the output. It is designed to be big enough to open fuses 'quickly', before thermal damage occurs to upstream wiring. It's often used to define the minimum resistance of safety earth bonding, so that in the event of a line to chassis fault, enough current will flow to open the live breaker. It's no good have a (let's say) 10 ohm ground connection, that just sits there and cooks in the event of a short to ground, while failing to open the breaker.

3) Maximum fault current

This is the maximum current that should flow if you short circuit the output. It is designed to be small enough to not damage up stream wiring while the breakers open, and to specify how much current the breakers should be able to stop successfully when choosing the breakers for that link. Damage could occur thermally, or on big feeders through Lorentz forces tearing them out of mountings. This is achieved by having a high enough impedance in the upstream wiring and through transformers, both their resistance and their leakage inductance. Distribution transformers are installed to a minimum impedance specification. It often has to be increased locally to protect areas of the grid.

Obviously there is an economic tradeoff here. A high impedance supply will generate lower fault currents (easier to protect), but will lose more power in line resistance, and have more voltage drop with varying load. The cost of the supply rises as its quality rises, which is kinda what you'd expect.

It is necessary to have a defined maximum fault current, because short circuits will happen, and replacing installed cable and transformers is far more expensive than replacing breakers. Lightning strikes can initiate arcs, and I once had a very entertaining grandstand view of a digger putting its backhoe through an 11kV feeder on an adjacent building site. Our lights dimmed for about one second before going out, a puff of smoke rose from the hole, as he tripped the breaker to the entire business park.

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An example: http://www.power-technology.com/projects/drax/

[The generators] generate 19,000A at 23,500V. A transformer increases the voltage to 400,000V before sending it via cables to the National Grid sub-station for distribution.

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