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Theoretical question:

Say you have a power generator that can supply 1W at 100V DC.

If you connect a 10k Ohm resistor, exactly 0.01 A is going to flow across the resistor.

If you connect a 1 Ohm resistor, 100 A should flow across it, however the generator is limited to supply 1W, making it impossible. In this case, how much current is going to flow through the resistor? And what voltage will be applied?

My hypothesis is that, since power (energy) is conserved:

R * I^2 = 1 W

therefore I = 1 A. This implies a voltage of 1 V which means that the voltage applied has dropped from 100V to 1V. By the way, how quick is this drop?

Is this reasoning (and is result) correct?

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    \$\begingroup\$ If you actually had a power supply that was designed to deliver exactly 1 Watt into any passive load like a resistor, then you'd get 1 Amp delivered into a resistance of \$1\:\Omega\$. But power supplies don't actually observe and control delivered power as their primary specification. Power is a compliance value for power supplies. Not something under direct control. \$\endgroup\$
    – jonk
    Commented Nov 21, 2016 at 23:21
  • \$\begingroup\$ @jonk ok, but suppose you are trying to power a 100W load, say at 10V DC, with a power supply that can deliver at most 50W at 10V DC. In that case, (putting the inductance of the motor aside), the problem would be similar, there should be a voltage drop. \$\endgroup\$
    – mickkk
    Commented Nov 21, 2016 at 23:26
  • \$\begingroup\$ Your question can't be answered because it depends on the power supply. There is no universal power supply design that lets you predict what will happen when the supply's output specification is violated. Some supplies will have a current limit, some will simply shut down, and others may fail. \$\endgroup\$
    – Barry
    Commented Nov 21, 2016 at 23:48
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    \$\begingroup\$ With my bench supply, if it hits the current limit it switches to CC mode and the voltage drops. Unless OCP is turned on, in which case the power switches off. \$\endgroup\$
    – Majenko
    Commented Nov 21, 2016 at 23:53

2 Answers 2

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This depends on the power supply design. There are many options:

  • Power supply burns out and never works again
    • This is especially spectacular with li-ion batteries -- it involves fire, explosions and lots of smoke
  • A fuse blows in power supply, the supply will work again when fuse is replaced
  • Power supply overheats and shutdowns; once it cools down enough it powers up again (and immediately shuts down again if low resistance is still present)
  • Power supply overheats and starts decreasing voltage, until it no longer overheats. The current stabilizes at some value which depends on room temperature, airflow, etc..
  • The power supply has CC (constant current) mode: the voltage is decreased to maintain exactly maximum current (so it will output 0.01A and 0.01V)
  • The power supply has fold-back current limiting: both voltage and current are decreased; once resistance is normal again output raises to normal levels.
  • The power supply has hick-up current limiting: power supply shuts off for pre-defined interval (few seconds); after timeout expires the output is restored, and if the power supply is still overloaded the cycle begins again
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Solar panels, at least for satellites where power is precious, run at the MAX POWER point. In one method, a small dithering signal gets injected into the control loop and the polarity and size of POWER (I*V) is explored.

There is no reason why, with both voltage and current being measured, an control loop does not produce constant power.

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