Timeline for Intuitively, Why is Power Proportional to $I^2R$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2013 at 5:04 | comment | added | pho | @Val, yes. Look up the kinematic equations for more info | |
| Oct 24, 2013 at 14:14 | comment | added | N. Virgo | Just for clarity, the only thing that breaks down in this analogy is $V=IR$. The hole in the tub behaves as a weird nonlinear resistor that doesn't obey Ohm's law, but $P=IV$ is really the point here. | |
| Oct 24, 2013 at 13:51 | comment | added | Val | Do you mean that the longer is the acceleration, the higher is the end speed? | |
| Oct 24, 2013 at 13:40 | comment | added | Egon | @Val: Fall time is not independent from height (otherwise jumping from a plane or jumping at ground level would take the same time). Fall time is independent of mass. More height, means that you need more work to carry the water up (so more power). | |
| Oct 24, 2013 at 12:58 | comment | added | N. Virgo | Ah, I see the point. This particular picture does break down if you vary $V$ - I ignored that because the OP asked about constant $V$. It is possible to make the analogy work with varying $V$ as well, but the picture looks different and is slightly less intuitive. The two tubs have to be on the same level surface and connected by a tube at the bottom. Then the pressure is proportional to $\Delta h$, where $\Delta h$ is the difference between the surface levels, and the flow (assuming it's laminar) will then be proportional to $\Delta h$ as well. | |
| Oct 24, 2013 at 12:54 | comment | added | Val | I wonder why higher height => higher fall speed? Isn't fall time independent of the height en.wikipedia.org/wiki/Galileo_Galilei#Falling_bodies? | |
| Oct 24, 2013 at 12:54 | comment | added | N. Virgo | @Val it's not really anything to do with pressure. The analogy in this case is $g\Delta h \leftrightarrow V$, not $\Delta p \leftrightarrow V$. | |
| Oct 24, 2013 at 5:55 | vote | accept | CommunityBot | ||
| Oct 24, 2013 at 5:55 | comment | added | user24082 | Strange... I thought of comparing the current and resistance, but it's so much more elementary. Thanks. | |
| Oct 24, 2013 at 5:36 | history | answered | N. Virgo | CC BY-SA 3.0 |