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Why in some cases a device draws more power than the source is capable of supplying thus damaging it, and in other cases the device will work under-powered?

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    \$\begingroup\$ it appears that you are talking about two different things ... or you are indirectly talking about a power supply .... if you are talking about a power supply, then ask a direct question about a power supply \$\endgroup\$
    – jsotola
    Commented Mar 29, 2020 at 21:35

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If the voltage source has low output impedance, its output voltage will not drop a as much as it continues to supply higher levels of current, so the load will continue to try to draw a higher current. This can cause the voltage source to overheat and damage itself.

But if the voltage source has a high output impedance, it's voltage will droop as it tries to supply more current which in itself will limit the amount of current the load will try and draw (since the load sees less voltage, it will try to draw less current).

An analog is like a person who gets tired when they try to push something too hard. The harder they push the more tired they get so they can't push as hard anymore. But a person who doesn't get tired might just keep on pushing until they break their bones or damage their muscles.

And of course, some sources are just protected with current limiting to begin with.

Whether the load will work with a lower voltage due to voltage droop from the source supplying too much current is independent of the question you actually asked. It's entirely up to the characteristics of the load.

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You have to match two things in a power supply:

  1. the voltage. The voltage rating of the source and the load should be matched, for example, 5 V or 12 V;

  2. the current. The current rating of the supply should be more than the load uses, For example a 5 V, 5 A power source will be able to supply a 5 V, 1.2 A load.

In this example the power supply is a \$5\ V\cdot 5\ A = 25\ W\$ power supply and the load has a \$5\ V\cdot 1.2\ A = 6\ W\$ rating.

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