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What are the advantages of each? What situation would I want higher amperage than voltage, and vice versa?

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    \$\begingroup\$ When situation would you want high water pressure and when would you want high flow? What are the advantages of each? When would you want higher flow than pressure, and vice versa? \$\endgroup\$ – Transistor Feb 28 at 7:12
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    \$\begingroup\$ "Higher amperage than voltage" isn't meaningful. Volts and amperes are different kinds of units. They can't be compared, no more than you can compare meters and kilograms. \$\endgroup\$ – duskwuff Feb 28 at 7:50
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    \$\begingroup\$ I'm sorry you've been downvoted. This seems like a reasonable question. \$\endgroup\$ – Sean Feb 28 at 9:00
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    \$\begingroup\$ @Isaac It isn't that bad question, just add: "To provide effective power transmission in large/small distances..." or: "When I want to begin my welding process..." or something like that. \$\endgroup\$ – smajli Feb 28 at 11:02
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    \$\begingroup\$ This may be a broad question, but I think it's an absolutely reasonable beginner question that can be answered well, as @Huisman has done. +1 and voting to reopen. \$\endgroup\$ – JYelton Feb 28 at 18:17
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You want a high voltage in e.g. electrical power transmission. The transmission losses are \$I^2 R\$. So, for the same power (e.g. 10kW) it is better to have 10kV and 1A than 100V and 100A.

For welding you want to have high current and low voltage, because the current determines the energy in the arc.

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In digital electronics, where CMOS logic is used, you control everything just with voltage and ideally you don't want any current to flow. As you may know, to turn on or off the MOSFET transistor, you need to apply positive (or negative) voltage on its gate. In BJT transistors you need current because they are turned on with current applied.

To transmit energy on the long distances we use very high voltage. To understand why, consider the following

enter image description here

I is the current, U is voltage, P is power and t is time. We can calculate the heat Q according to the Joule–Lenz law. You can see, that to decrease heating you can either decrease the resistance R of the wire or increase the voltage. It is very expensive to decrease the resistance since you have to change the material of the wire or to increase its cross-section area.

However, you can just increase the voltage, because increasing voltage 10 times will decrease heating 100 times.

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    \$\begingroup\$ Be careful with MOSFET's when using fast signals though. Not considering the reloading gate capacitance time/currents may cost you hours of debugging. \$\endgroup\$ – smajli Feb 28 at 10:58

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