Ohm's Law: I $=\frac{V}{R}$: Increasing voltage increases current.
Power Law: P $={V}*{I}$: Increasing voltage decreases current.
Am I missing something?
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$\begingroup$ @AlfredCentauri It does not exactly, as my doubts are slightly different. $\endgroup$– El FleaCommented Mar 14, 2020 at 15:14
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$\begingroup$ @ElFlea In what way are your doubts different? $\endgroup$– sammy gerbilCommented Mar 15, 2020 at 20:27
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$\begingroup$ @sammygerbil Why must you always be on the lookout, wanting to close my questions. You already closed two of my questions for being "off-topic". $\endgroup$– El FleaCommented Mar 17, 2020 at 3:43
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$\begingroup$ Nothing personal! I am not the only user voting to close your questions.11 others have voted. 3 other users also voted to close 2 of your 5 closed questions. And 3 more have voted to close 3 of your questions. ... You have had 3 answers, one of which you have accepted. $\endgroup$– sammy gerbilCommented Mar 17, 2020 at 3:58
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$\begingroup$ @sammygerbil If a question got answers, it's got to stay. Others have put effort into answering it. $\endgroup$– El FleaCommented Mar 17, 2020 at 4:00
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3 Answers
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Am I missing something?
Yes, you are missing what is held constant in each case.
Ohm's Law: I=V/R. Increasing voltage increases current.
Should read “Increasing voltage increases current for a fixed resistance.”
Power Law: P =𝑉∗𝐼. Increasing voltage decreases current.
Should read “Increasing voltage decreases current for a fixed power.”
The two statements are not contradictory since they refer to different scenarios. Usually resistance is fixed and power is not.
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As the others say, you are implicitly holding one factor constant and seeing how the others are related but a different constant factor in each case.
A real world example may help. Consider an old style filament light bulb. It is intended to be used in a $220V$ country and consume $110W$. So, it should draw $0.5A$ and needs a resistance of $440 \Omega$.
We can keep the resistance constant and vary the voltage by taking it to a $110V$ country. Ohm's law tells us that it will now draw only $0.25A$ hence the power will be only $27.5W$.
For the constant power scenario, imagine that the manufacturer wants to make a similar $110W$ bulb for $110V$ countries. He needs it to draw $1A$ so he must arrange for its resistance to be $110 \Omega$. Note that this will be a different bulb.
For some fun, take the bulb intended for $110V$ to a $220V$ country. I leave that as an exercise.
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$\begingroup$ God Bless people who give examples! Makes it so easy to understand. $\endgroup$– El FleaCommented Mar 14, 2020 at 15:12
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$\begingroup$ For the exercise, I suppose the bulb, as its resistance is fixed, will burn out? Or maybe it glows way too brightly? :) $\endgroup$– El FleaCommented Mar 14, 2020 at 15:16
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$\begingroup$ If we pretend that it will cope then it will draw $2A$ and hence $440W$. Four times its design load. It will probably glow very brightly for a short while before going pop. Don't try this at home. I have seen it in the Philippines where they use US style plugs and sockets but $220V$. It looks as if you can use US devices but you can't. $\endgroup$– badjohnCommented Mar 14, 2020 at 18:55
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$\begingroup$ Some modern devices, e.g. phone chargers and laptop power supplies, will seem to disobey this. However, they use more complex components and don't obey Ohm' s law. That's why I used an old style filament light bulb for my example. $\endgroup$– badjohnCommented Mar 14, 2020 at 19:21
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You have missed that power does change since voltage is changing, the easiest way to see it is to rewrite P = VI = V(V/R) = (V^2)/R. So , as you see if voltage is increasing the current is increasing and power is increasing as well.
