3 ms·
It's too bad that automakers didn't make this change a long time ago. One thing the article didn't mention is that power losses in a wire are a function of the
by PhantomGremlin 9y ago
It's too bad that automakers didn't make this change a long time ago.
One thing the article didn't mention is that power losses in a wire are a function of the square of the current. So, let's say one needs to deliver 48 watts to a load. In a 48 V system, that's 1 A of current. Using 12 V, that's 4 A of current.
Power loss is 16x greater at 12 V for same diameter of wire, loss = (4 A * 4 A) * resistance of wire.[1] Therefore, wires are made thicker in 12 V systems to avoid this. That's a lot of extra weight in a car.
Before the nitpickers chime in, I know the above is simplified. But it also clearly explains why long distance power transmission lines run at such high voltages, 230 KV or higher.
[1] https://en.wikipedia.org/wiki/Copper_loss https://en.wikipedia.org/wiki/Copper_loss
- bsder 9y agoYou're not wrong, but you should probably tell the whole story if you're going to post on HN. Primary Objection: Um, Ohms law gives V=IR. And P=IV. So, P=I^2R. Except that P ALSO can be calculated as P=V^2/R. So, we got squared losses anyway and R should go the other direction. WTF? The answer is that a copper wire is a distributed system, not a lumped system. So, you derive the power loss from the generated heat across a differential slice of the line and then integrate over the whole length of the line--which gives you Power=RI^2 in the limit (and V->0 for any individual slice as the differential ->0). It is one of those (un?)fortunate happenstances (like the Bohr hydrogen atom model) where the system in the limit sums to something we already know and leads to a bit of misunderstanding when you confuse two things which appear to be related but really aren't...
- mindslight 9y agoThe V in V^2/R would be the voltage drop along the wire, not the total system voltage.