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According to the equations, $$P=VI =I^2R\,\text{ and voltage } V=IR$$ it seems clear that when the resistance is lower by fixing the voltage at constant, the current is therefore, higher, generating high power. But what confused me was when the resistance is higher by fixing the current at constant, the voltage is therefore, higher, which in turn lead to a higher power as well. Can anyone pull me out of this confusion?

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The answer is "yes, that's what happens." There's no paradox. If you hold the current constant, and increase the resistance, the power increases. This is because of V=IR and P=VI (Ohms law and the power through a resistor). If you put these together, you can see that $P=I^2R$. If you hold the current constant, and increase the resistance, power goes up. That's just how the equations work.

What makes this confusing is that it's not intuitive how to hold current constant. We typically don't think that way. Usually we think in terms of voltages. So one way to think of this is our higher resistance forces the power supply to provide a higher voltage in order to push through the same current. Intuitively, it should make sense that a higher voltage supply can produce higher powers (though you would need the $P=I^2R$ equation to prove it).

You can use metaphors as well. Any metaphor where you can put a load on something works decently well. Take your own body. You can run at a fairly nice pace. The resistance on your body while running is quite low, so it doesn't take much effort. Now add resistance:

Weight sled

If you run at the same pace (the equivalent of keeping the current the same), you're going to have to push much harder (the equivalent of raising the voltage). And, if you notice, you get hot really fast (power is being dissipated). However, that's only because you kept the current the same. It would also be possible to slack off, not running as hard, in which case you could dissipate less power than before. But your original problem declares that you're keeping the current constant, so you're going to have to work harder and have more power!

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  • $\begingroup$ So by regarding on both of my cases, does a resistor with high resistance has high power or a resistor with low resistance has high power (In the real life cases,such as in an electrical appliance.) ? $\endgroup$ – Yenhan Oct 8 '17 at 4:26
  • $\begingroup$ @Yenhan The answer to that question is dependent on the power source. Resistors themselves do not have high power or low power. They must be hooked up to a circuit, and then they can dissipate a high amount of power or a low amount of power, depending on the source. If the source is a current source, high resistances lead to high power dissipation (P=I^2R). If the source is a voltage source (such as a wall socket or a battery), high resistance leads to low power dissipation (P=V^2/R) $\endgroup$ – Cort Ammon Oct 8 '17 at 6:04
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Notice that by fixing a constant voltage $V$, we have that the current $I$ is inversly proportional to the resistence by Ohm's Law: $$I\propto\frac{1}{R}$$ So, your first assumption is right, when we set the resistence to be lower the current is higher. When you set a constant current $I$, we have by Ohm's Law:$$V\propto\frac{1}{R}$$ Setting a lower resistance we produce a higher voltage.

So you're right we can have a higher power by setting a constant voltage and low resistance or by setting a constant current and low resistance.

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  • $\begingroup$ But that(High resistance,low current) is only true if I set the voltage at constant, but in the second case I have set the current to be at constant , not the voltage, so if the resistance is higher, the voltage must be higher to reach an equilibrium between them. $\endgroup$ – Yenhan Oct 8 '17 at 3:47
  • $\begingroup$ Oh. I thought in both cases constant voltage.Sorry. $\endgroup$ – IEDC PHY Oct 8 '17 at 3:51
  • $\begingroup$ In that case, what you're saying is true. You can have a higher power by setting a constant voltage and a low resistance, or by setting a constant current and low resistence. $\endgroup$ – IEDC PHY Oct 8 '17 at 3:53
  • $\begingroup$ Alright then, it's ok :D $\endgroup$ – Yenhan Oct 8 '17 at 3:53
  • $\begingroup$ And that's why I was confused. I know something must be wrong in my point of viewing the circuit and the components, but I just couldn't find that out. $\endgroup$ – Yenhan Oct 8 '17 at 3:59

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