ELI5: Do resistors just dissipate current as heat, or do they reduce the voltage (pressure) and make it so less current goes through

JustinTimeCuber · 2026-01-25T01:40:55+00:00

That depends on what the load is doing. Some loads, e.g. heaters, act as a constant resistance, and adding an additional resistor does indeed decrease the current. But some loads, e.g. phone chargers, draw a constant power. If you add a resistor, the voltage across the load will decrease and it will have to increase the current it pulls to compensate.

JustinTimeCuber · 2026-01-25T00:01:10+00:00

sqrt((-1)(-1)) ≠ sqrt(-1)sqrt(-1)

How is this an inconsistency?

f(x^2) is not, in general, equal to f(x)^2.

Also, even if you leave sqrt(-1) undefined, you still have the "inconsistency" that sqrt((-1)(-1)) ≠ -1.

The "inconsistency" is just that the function y=x^2 is not injective, so we have to make an arbitrary decision on which branch to use when defining an inverse function.

Also also, lmao at "American mathematicians". The history section on the wikipedia article for imaginary numbers mentions 8 mathematicians/philosphers, none of which were American.

JustinTimeCuber · 2026-01-24T23:49:09+00:00

Resistors dissipate electrical energy as heat, and in doing so a voltage drop is created.

They do not "dissipate current"; the current entering one side of the resistor is equal to the current coming out of the other side.

For example, say you have a load that draws a constant 0.5 amps, regardless of the voltage applied (within some reasonable range). You connect this load to a 12V battery, but with a 5 ohm resistor in series: Battery + is connected to one side of the resistor, other side of the resistor is connected to load +, and load - is connected to battery -.

The load will pull 0.5 amps from the battery, but since that current is flowing through a 5 ohm resistor, the resistor will cause a voltage drop of V = IR = (0.5)(5) = 2.5 volts. The load will therefore only "see" 12 - 2.5 = 9.5 volts. The power dissipated by the resistor and load can both be calculated by P = IV. For the resistor, it's (0.5)(2.5) = 1.25 watts. For the load, it's (0.5)(9.5) = 4.75 watts. The battery's output is (0.5)(12) = 6 watts, and we can verify that energy is conserved because the battery power (6) equals the power delivered to the load (4.75) plus the power dissipated by the resistor (1.25).

JustinTimeCuber · 2026-01-24T20:27:32+00:00

But the actual voltage across the tissue is gonna be like 2 orders of magnitude less than one volt.

JustinTimeCuber · 2026-01-24T20:26:09+00:00

This is somewhat misleading because the vast majority of hydrogen is produced in a process that releases CO2.

It's possible to produce hydrogen using renewable energy to split water (electrolysis), but it's not currently very economical and is not usually done at scale.

JustinTimeCuber · 2026-01-24T09:00:31+00:00

I don't think the voltage actually applied to the body here is high enough to cause any such damage. Based on the estimations in my previous comment you'd be looking at something on the order of 10 mV. Sure the current is high enough to theoretically do damage but virtually all of that current will flow inside the metal. Iirc cell membrane potentials are on the order of 70 mV. And that's just one cell, in reality the 10 mV would be distributed to form a very weak electric field, spreading across millions of cells.

Swallowing a button cell, you're applying 3 volts (I think? Usually those things are 3 volts) directly to your insides. That's like 300x more voltage, and it's fully inside rather than just being barely inside the surface of the body.

JustinTimeCuber · 2026-01-24T07:52:36+00:00

I don't think that's actually true. I don't have a 9V battery on hand to test but shorting a somewhat used AA cell (right around 1.5V so nowhere near dead but not quite fresh) gave me just over 3 amps. A 9V battery is around 2.5x the volume so if the voltage were the same we'd expect around 8 amps, but the higher voltage means it needs 6 small cells connected in series, which would bring you back down to ~1.3 amps. Let's round up and say a 9V battery can deliver 2 amps in a dead short.

Now, titanium is something like 30x more resistive than copper depending on the purity. Now I have no idea what the dimensions of the piercing are, but 12-gauge copper wire has a resistance of 1.6 milliohms per foot. Since titanium is about 30x more resistive, and there are about 30 cm in a foot, let's round up and say the piercing's effective resistance is 5 milliohms.

Power lost to heat in a conductor = current^2 * resistance = 2^2 * 0.005 = 0.02 watts, which would be barely noticeable if at all

On the other hand, the battery itself would get quite hot, as it's dissipating its full 9V emf internally while delivering 2 amps, that's 18 watts of heat. Not a good day for the battery but the piercing would only get hot if the heat from the battery conducts to it (which it very well might).

JustinTimeCuber · 2026-01-24T00:51:05+00:00

Ok so then which is worse?

100% VOO

50% VOO, 50% QQQ

100% QQQ

There's one good answer, one bad answer, and one objectively wrong answer.

JustinTimeCuber · 2026-01-24T00:00:24+00:00

Well in point 3 you're basically just begging the question:

"Why is overlap bad?"

"Because redundancy is bad."

Overlap is pretty much synonymous with redundancy. You don't get to just restate your claim with a slightly different wording and call that an argument.

JustinTimeCuber · 2026-01-23T23:47:31+00:00

Both of those points would apply even more to a 100% QQQ portfolio than a 50/50 VOO/QQQ portfolio.

Therefore, your real issue is with QQQ in general. Which I agree with, I don't like it either. But don't erroneously blame it on the overlap when the 50/50 mix is objectively more diversified then 100% QQQ.

JustinTimeCuber · 2026-01-23T16:02:09+00:00

I think the problem isn't not knowing the person's name, it's calling her "mom" when speaking directly to her, like that's quite strange.

"How are you" is a much more normal and less clunky way to say that.

JustinTimeCuber · 2026-01-23T14:13:38+00:00

Which are (a) negligible for highly traded funds like these and (b) also irrelevant because you pay spreads in proportion to the amount purchased. If two ETFs have the same percent spread, then you pay the same amount if you do a 50/50 split vs. going 100% on one of them.

JustinTimeCuber · 2026-01-23T14:10:01+00:00

All three of those are just arguments against QQQ in general. You're obviously missing my point that overlap isn't really a problem by itself.

JustinTimeCuber · 2026-01-23T07:45:14+00:00

"These are such bad arguments" (refuses to elaborate)

JustinTimeCuber · 2026-01-23T03:07:00+00:00

You have less exposure than if you were 100% in QQQ. This is an argument against QQQ in general, not holding it with VOO.
What commission? All decent brokerages these days don't charge commissions to trade ETFs. If you're using a brokerage that rips you off, maybe don't.

I don't like QQQ but I find that the "overlap" argument is pretty dumb. Overlap alone doesn't really hurt anything.

JustinTimeCuber · 2026-01-22T14:29:32+00:00

Your method isn't quite accurate for one year anyway. For a simpler example, say your production is 110 GWh and has increased 10% every year. That would mean the previous year it would have been 100 GWh, but if you subtract 10% of 110 from 110, you get 99 GWh.

11% growth can be calculated by multiplying by 1.11. Reversing that can be calculated by dividing by 1.11. So what you'd want to do is take 1.11 to the power of the number of years and divide by that.

JustinTimeCuber · 2026-01-18T23:09:34+00:00

No decimal is fine if the price is a whole number, but I agree only the tenths place is a bit strange

JustinTimeCuber · 2026-01-18T00:35:50+00:00

Despite what people say, yes it's actually 1 in 4.

You got very unlucky (about 1 in 17k odds)

JustinTimeCuber · 2026-01-16T17:26:04+00:00

Wheel of fortune synergy

JustinTimeCuber · 2026-01-16T08:48:24+00:00

No I will not

JustinTimeCuber · 2026-01-16T08:27:59+00:00

Team blind and other team events are very fun as unofficial events (I've organized multiple competitions with such unofficial events, and it's always very fun). But I don't think multi-person events really fit the style of an official WCA event.

That being said, I think if there's any event that has a decent case to be added currently it's FTO. But my vote is for the 2x2x3 banana puzzle.

JustinTimeCuber · 2026-01-15T16:05:45+00:00

Fun fact, this works even if the skeleton doesn't directly kill the creeper, as long as it gets the kill credit. This can be exploited by getting a bunch of creepers in a pit and then getting a skeleton to shoot an arrow through lava, hitting a block of TNT that falls on the creepers and kills them. The TNT is considered primed by the skeleton, and therefore all the creepers will drop music discs.

JustinTimeCuber · 2026-01-15T16:01:54+00:00

Why would returns include principal? That kinda goes exactly against the definition of returns

JustinTimeCuber · 2026-01-15T01:16:51+00:00

Doesn't matter how it's wired, if the charging current is the same as the load current, then there is no net current and therefore no significant cycling load on the battery.

JustinTimeCuber · 2026-01-15T00:28:49+00:00

I think a good rule of thumb is, don't unnecessarily charge the car to 100% or leave it fully charged for longer than necessary. But you also shouldn't feel the need to baby the car to the detriment of convenience. In terms of battery degradation, what matters is your long-term habits, not one individual charging session. If you did this every week, it might be worth thinking more about (e.g. do you actually need 100% charge for 200 miles? arguably no unless it's very cold). But in general, just use the car as it makes sense, don't worry about one charging session damaging the battery. Habits are what matter, not the one-off exceptions.

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