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I have some difficulty grasping these concepts. Let's say, for example, a power source of 10W operating at 5V is connected to a load of 0.5 ohms. According to Ohm's law, it is expected that a current of 10A should flow through the circuit. However as given above, the power is 10W so a current of 2A is expected, using the voltage-power relationship.

My question is: What is the expected current in this particular case and why?

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    \$\begingroup\$ You can't specify both power, voltage and resistance. You can choose two freely, but those two will determine the third. \$\endgroup\$
    – Wouter van Ooijen
    Commented Jun 17, 2018 at 17:19
  • \$\begingroup\$ @Wouter van Ooijen, how comes, I have seen things like power adapters specify both output power and voltage, so what is the significance of this specification? \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 17:50
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    \$\begingroup\$ What is specified in that case is the maximum power the adapter can provide. \$\endgroup\$
    – Wouter van Ooijen
    Commented Jun 17, 2018 at 18:01
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    \$\begingroup\$ If it maintained 5V, then 10A would flow and it would deliver 50W. Since it's rated for only 10W, you are using it outside its specification, and there is no guarantee what it will do. A fuse might blow, it might shut down, it might overheat, or even explode. It might even deliver the 10A. But nothing is guaranteed (that is, from the 5V 10W specs .alone) \$\endgroup\$
    – Wouter van Ooijen
    Commented Jun 17, 2018 at 18:52
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    \$\begingroup\$ @Sam: The edit is improvement but why the inline code formatting for values? You wouldn't see it done like that (change of font and background) in a published article. \$\endgroup\$
    – Transistor
    Commented Jun 17, 2018 at 20:47

4 Answers 4

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10W is the maximum power that the supply can provide. The actual power (and current) will depend on the load connected to the supply.

In your example, the smallest resistor that can be safely connected to the supply is 2.5 ohms, which will result in a current of 2A and power of 10W. If a resistor smaller than that is used, it will attempt to draw more than 2A, with a power greater than 10W. What happens next will depend on the power supply, but the supply's output voltage will fall below the rated 5V, and the supply may overheat or the protection fuse may trip.

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  • \$\begingroup\$ So is it impossible to calculate current in this case or how low the voltage will fall to? (Assuming we don't have any kind of fuse or protection whatsoever) \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 18:29
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    \$\begingroup\$ @AguThadeus Without specific and detailed knowledge about the circuitry used inside the power supply, it is impossible to work out exactly how that power supply will respond to loads which require more current than the supply is designed to handle. A \$5\:\text{V}\$ power supply rated at \$10\:\text{W}\$ has a "compliance current" of \$2\:\text{A}\$. Any load requiring more current than this exceeds the specifications for the supply and the results are "not specified." (Usually not good, either.) \$\endgroup\$
    – jonk
    Commented Jun 17, 2018 at 19:40
  • \$\begingroup\$ If it's a low quality power supply, it may go bang. \$\endgroup\$
    – user20574
    Commented Jun 17, 2018 at 23:26
  • \$\begingroup\$ Assuming no fuse or protection circuitry: For a typical small AC/DC switching power supply like is often used with electronic appliances, output voltage decreases as output current increases, even before the maximum rated power is reached. The exact relationship will depend on the specifics of that power supply. There's nothing magic that happens when the maximum rated power is reached. As the output current increases still further, beyond the intended limits, the supply will begin to overheat. Eventually it will melt or catch fire, or you will trip a circuit breaker. \$\endgroup\$
    – bmow
    Commented Jun 17, 2018 at 23:52
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Assume power can source infinite current at 5V for the moment.

With 0.5 ohms of resistance, now definitely 10 A will flow through the resistor.

It follows ohms law.

Once power source has limited abilities such as only 10 W, it means that it is able to deliver up to 10 W happily but not more.

If you connect a 5 ohm resistance, 10W power supply will supply 1 A as current. So, the power supplied is 5W though the capacity of power supply is 10 W.

As soon as the power expected by the load goes beyond the power supply capability of source, one can't assume ohms law.. Because the output voltage from the source will not be 5V.

Either the output voltage from the power supply may drop or even cutoff the lower all together (over load protection or short circuit protection circuits may kick in)

Stay below the power capacity of the supply and the ohms law will be valid.

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  • \$\begingroup\$ Let's assume there is no load protection or anything to counter this, surely something must happen, a voltage may drop but by how much?, is there a way to calculate the drop and how much current will be flowing at that drop level? \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 18:34
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    \$\begingroup\$ @AguThadeus Without specific and detailed knowledge about the circuitry used inside the power supply, it is impossible to work out exactly how that power supply will respond to loads which require more current than the supply is designed to handle. A \$5\:\text{V}\$ power supply rated at \$10\:\text{W}\$ has a "compliance current" of \$2\:\text{A}\$. Any load requiring more current than this exceeds the specifications for the supply and the results are "not specified." (Usually not good, either.) \$\endgroup\$
    – jonk
    Commented Jun 17, 2018 at 19:41
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    \$\begingroup\$ @AguThadeus Each generator type has its limitations and specifications. A dynamo has entirely different dynamics and specifications than a solid state power supply. But it still has its own unique behaviors and specifications. So, unspecified, no one can say anything much at all if you exceed the specs. But if you specify something quite specific, then more can be said about the behavior if you exceed specs. Often, a lot more. But as always, details matter. \$\endgroup\$
    – jonk
    Commented Jun 17, 2018 at 20:50
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    \$\begingroup\$ @jonk Thanks, I think I get what you mean now, without knowing the inner workings of the power source it would be impossible in this case to determine what would happen if something tries to draws too much power than its rated capacity \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 21:31
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    \$\begingroup\$ Right. The manufacturer of the power supply guarantees it will give you 5V up to 2A. But they don't tell you what will happen if you go over 2A. (Sometimes they do, such as with the laboratory power supplies that have a maximum current knob, but not your average wall wart) \$\endgroup\$
    – user20574
    Commented Jun 17, 2018 at 23:28
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Firstly, you should not connect a 50w load to a 10w power supply. Another way of looking at it is, it is a 5v 2A power supply. You are likely to overheat, damage or blow up the power supply and suffer voltage drop. If the power supply is particularly well designed with most components and construction overrated by a factor of 500% then the only effect will be much voltage drop on the output and your load will not operate as expected.

Your basic understanding of the operation of Ohms law seems to be correct but what you have overlooked is powers supply ratings. A 10w power supply is only rated to supply 10w of power, which in your example you have correctly calculated for 5v to be 2A. No matter how big the load you connect, the power supply is only rated for 10w.

To connect a 50w load safely you need a 50w (or greater) power supply. If your 10w power supply can provide 10A without blowing up you will see the supply voltage drops by 4v meaning an effective supply of 1v 10A (10w) if the power supply is indeed capable. If it cannot deliver that much current it will go bang or the voltage drop will be greater also.

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Power law and Ohm's law are two different things.

First, power law use the relation between voltage and current (P = VI). You can represent it as the dissipated energy in your system. Usually, a power supply is defined by his voltage and his current output. In your case, the supply is a 5 V and the 10 W would go on your load (resistance). With power law, you get a maximum current of 2 A at 5 V for safe use.

Then comes the Ohm's law. If your resistance is at 0,5 ohms and your supply at 5 V, you get 10 A (V = RI). You can assume that 10 A will flow if your resistance if the supply is able to. Knowing that only 2 A can safely circulate, then your resistance would burn.

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  • \$\begingroup\$ Why would a resistor burn?, say resistor can withstand power greater than 10W, what would happen then, would it be possible to calculate the voltage and current in that state? \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 18:37
  • \$\begingroup\$ Well, it is possible that your resistance is rated at 50W for exemple, then your resistance will not burn and will let the current flow. Let say you still have your 5V battery and 0,5 ohms resistance, 50W resistance would be good according to power law \$\endgroup\$
    – Marc.D
    Commented Jun 17, 2018 at 18:48
  • \$\begingroup\$ So what is amount of current that will flow? \$\endgroup\$
    – AguThadeus
    Commented Jun 17, 2018 at 18:53
  • \$\begingroup\$ According to Ohm's law, 10A. \$\endgroup\$
    – Marc.D
    Commented Jun 17, 2018 at 18:55
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    \$\begingroup\$ @Marc.D: You have a confused power and energy in both your answer and the comments. Power is measured in W, not Wh (which is a measure of energy). Similarly the power supply will be limited in A output and not Ah. (Even a tiny power supply would be capable of hugh Ah capacity if left on for enough hours.) You need to edit and clarify to the OP who hasn't spotted the errors. \$\endgroup\$
    – Transistor
    Commented Jun 17, 2018 at 20:25

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