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On this KhanAcademy article, under the Power in a resistor heading, it states three equations, all of which can be derived from $V=iR$ and $P=iV.$

They are:

  1. $P=Vi$
  2. $P=i^2R$
  3. $P=\frac{V^2}{R}$

Then if you click on [How much has the power dissipated by the resistor been changed? ] at the bottom of the same section, it reveals this puzzling tract:

Power in a resistor is proportional to $v^2,$ so lowering voltage by a factor of $2$ cuts the power by a factor of $4.$

Power in a resistor is also porportional* to $i^2,$ so lowering the current by a factor of two also cuts the power by a factor of $4.$

Overall, Aaron and Beth have reduced the power by a factor of $4\times4=16.$

*they meant proportional

However, if you these changes through equation $1$ in our list above, you find that power has only decreased by a factor of $4.$

Is KhanAcademy right? And if so, why are we taking both equation $2$ and $3$ into account at the same time? Why not take into account all three equations and make power decreased by a factor of $2^3\times2^3=64?$ (One $2^3$ for $i$ and one for $V$.) Is this just another time in science where we discover something and randomly mash equations together to try to model something, and if it works we just don't question it, or am I missing something really big?

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  • $\begingroup$ I think that article is a little unclear, but I think it means Aaron reduces the voltage across the resistor by a factor of 2 and in addition to this Beth cuts the current by a further factor of 2. However Since $V=IR$ in a resistor each of theses changes cuts both the current and voltage by 2, so each is reduced by a factor of 4 by the end. $\endgroup$ – By Symmetry Jun 15 '16 at 19:51
  • $\begingroup$ In equation 1, what happens to $i$ if you cut $V$ in half? $\endgroup$ – garyp Jun 15 '16 at 19:51
  • $\begingroup$ The question is ambiguous because there are three things that can change ($I$, $V$, and $R$) but only one relation between them, $V = IR$. So if you change one of them, you need more information to figure out what actually has happened. $\endgroup$ – knzhou Jun 15 '16 at 20:01
  • $\begingroup$ In this case it should have said "Aaron reduces the voltage by a factor of 2 while keeping the resistance constant (which implies that the current is reduced by a factor of 2), while Beth reduces the current by a factor of 2 while also keeping the resistance constant, etc.." $\endgroup$ – knzhou Jun 15 '16 at 20:02
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    $\begingroup$ This kind of crummy writing is typical for online physics sources, though. I would recommend switching to MIT OCW, EdX, or Coursera; everything else is pretty incoherent. $\endgroup$ – knzhou Jun 15 '16 at 20:03
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I think there is a misunderstanding.
What the article tells you is that if you change either the cuurent or the voltage by a factor of 2 then the power changes by a factor of 4.
And if you change both of them by a factor of 2 then the power changes by a factor of 16.
First equation is $P=VI$, if you change $I$ from $I$ to $2I$ the voltage $V$ also becomes $2V$ (because $V\propto I$) and vice versa. So, $P$ increases by 4.
If you do it again $P$ will increase by 16 and so on.
All the three equations are same. They are just substitutions.
Increasing $I$ by 2 in equation (2) also increases the hidden $V$ by 2 in equation (2) and same is for equation (3) where $I$ is hidden.

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This is my understanding:
You already have $V = IR$, and take note that $R$ is constant because you are not adding any more resistor.
When you lower the voltage by a factor of 2, you could lower the $I$ and keep the $R$ constant, or you can lower $R$ and keep $I$ constant. The latter is not possible, so the only way for us to decrease the voltage is to also decrease the current.
If we use eqn 1, it relates $P$ to both $I$ and $V$ but we only know that V is decreased, so we also have to take into account the decrease in current when the voltage is decreased:
since $V = IR$; $\frac{1}{2} V = (\frac{1}{2} I)R$; which means that current is also decreased by 1/2.
So using eqn 1:$P = IV$, so $P_1 = \frac{1}{2} I\frac{1}{2} V = \frac{1}{4} IV = \frac{1}{4}P$ thus the power is decreased by a factor of 4
Using eqn 2:$P = I^2R$ so $P_1 = (\frac{1}{2}I)^2R = \frac{1}{4}I^2R = \frac{1}{4}P$ thus the power is also decreased by a factor of 4
Using eqn 3:$P = V^2/R$ so $P_1 = (\frac{1}{2}V)^2/R$ = $\frac{1}{4}V^2/R = \frac{1}{4}P$ thus the power is also decreased by a factor of 4
So by reducing V by 1/2, we are reducing $P$ by a factor of 4, using either 3 equations.

Then, in step two, we reduce the current by 1/2. Again, we know that by reducing the current, we also reduce the voltage, while keeping $R$ constant: $V = IR$ so $\frac{1}{2}V = (\frac{1}{2}I)R$.
since $V = IR$; $\frac{1}{2} V = (\frac{1}{2} I)R$; which means that current is also decreased by 1/2.
So using eqn 1:$P_1 = I_1V_1$, so $P_2 = \frac{1}{2} I_1\frac{1}{2} V_1 = \frac{1}{4} I_1V_1 = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is decreased by a factor of 16
Using eqn 2:$P_1 = I_1^2R$ so $P_2 = (\frac{1}{2}I_1)^2R = \frac{1}{4}I_1^2R = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is also decreased by a factor of 16
Using eqn 3:$P_1 = V_1^2/R$ so $P_2 = (\frac{1}{2}V_1)^2/R$ = $\frac{1}{4}V_1^2/R = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is also decreased by a factor of 16
So by reducing $I$ in step two by 1/2, we are reducing the power by a factor of 4 using either 3 equations, and the initial power is decreased by a factor of $4\times4 = 16$.

In the article, they used eqn 3 for step 1 because by using it, you don't need to take account of the decease in current involved, but only the voltage. Also for step 2, they used eqn 2 so that you don't need to take into account the reduction of voltage due to the reduction of current.

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