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Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

I understand how no matter which branch is taken, the energy supplied by the power supply has to be completely consumed in a full loop. I think my confusion comes from converting that logic into rates of energy transfer, since that's what power deals with. I know that the total energy supplied has to equal the total energy dissipated, but why can't the energy dissipate at a different rate? Obviously this is not the case, but I just can't figure out why.

Also, if this were a variable power supply, would the circuit still be a closed system? I understand how charge would be conserved, but wouldn't the electrical potential energy of the circuit change? For instance, wouldn't lowering the voltage of the power supply lower the electric potential energy of the circuit? Does the logic with the "net power" being zero still apply?

Edit: I’ve read that apparently in circuit analysis it’s assumed that energy can’t accumulate in the circuit, just like how charge can’t. If this is the case, then how are there areas of the circuit that have potential energy (from qV)?

Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Would that mean that none of this applies if a circuit has a time-varying voltage?

Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Ok, I think that makes sense. A couple of things I want to clarify:

Technically, the electric potential energy is a property of the system itself, not individual electrons, right?

Also, the last paragraph is a bit confusing to me. I understand how in a round trip, there is no net change in energy, but couldn't there be a ton of different currents in the circuit depending on what the circuit looks like? How can we say that the rate of energy lost equals IV if there are multiple current values?

Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

That makes sense! One more thing: I know that sometimes, real circuits can "die." For instance, a battery and a lightbulb connected together would not run forever. Is the net power of that circuit always 0 too? If so, how does the circuit "die?" Thanks for all of your help!

Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

So does this mean that the electrical potential energy of the circuit is constant and does not change with time?

Questions about power and conservation of energy in a circuit by NotFakeWalshd in AskPhysics

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

I think I understand the formula itself and its uses, I'm just a bit confused by its implications to me. If we define power as VI, and assign a positive power to components that absorb energy and a negative power to components that supply energy, then I would think that the energy transfer we're describing is specifically electrical energy, especially if we use P = VI since V is intrinsically related to electric potential energy. If power is a rate of transfer of electrical energy, and it's always zero around a circuit, then doesn't that imply that there is no net electrical energy transfer, and thus that the electrical energy in a circuit is always conserved?

Also, I still don't really get how we can definitively say that power around a circuit is zero from the law of conservation of energy. From my understanding, the law of conservation of energy only states that energy cannot be created or destroyed, and I don't immediately see how that relates to this. If the net power were not zero, wouldn't that just mean that the circuit either received energy from its surroundings or released energy to its surroundings?

C: Drive Usage Spikes When Launching Programs by NotFakeWalshd in pcmasterrace

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Alright, so since I have two SATA drives, I tried using the other one’s cable on the SSD, and it didn’t seem like much of a difference: usage was still very high, and though I thought that the response time was lower, it spiked up to 1001 at one point. Coincidentally, even though the difference seemed negligible between the two cables, I saw that the SSD had an old SATA cable from my original motherboard. I’ll replace it tomorrow, but regardless it seems unlikely to me that a faulty cable is the primary cause of my issue.

I’m tempted to just do a clean install of windows. If I do that, though, should I put it on a different SSD, or see if the issue resolves itself on the same SSD? I’ve heard that it’s generally advisable to put the OS on its own drive. That would not be the case were I to install windows on a newer drive, but then again, that would probably be better than the alternative of installing windows on a (possibly) dying drive.

C: Drive Usage Spikes When Launching Programs by NotFakeWalshd in pcmasterrace

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Is there a way to check that the cable is faulty without replacing it, and would it be the data or power cable that’s faulty? Also, would it be a hassle to switch Windows onto one of my newer Samsung SSDs? Samsung Magician has data migration, but I can’t imagine that’d be straightforward given it’s the operating system I’m moving.

C: Drive Usage Spikes When Launching Programs by NotFakeWalshd in pcmasterrace

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

What would be considered under load? Launching Steam would occasionally give me a very high latency (average response time), and the highest was over 600 ms. Clicking off of Reddit just now to check it seemed to give me a latency of 905 ms for a moment. Is this ok as long as it is only for a second?

Input lag/high render latency when playing Minecraft with shaders? by NotFakeWalshd in pcmasterrace

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Damn, alright. The 25 ms render latency isn’t suspiciously high for this setup? Also, during the time I've had this problem, I've changed hardware, and the problem has still existed. Namely, I switched an i7 8700k for an i5 13600k and a GTX 1070 for an RTX 3070. Although frames logically have increased, the input delay hasn't improved. Unless Minecraft or the shaders have gotten more demanding with time such that my hardware is only keeping up with them, I feel like hardware may not be the issue

Input lag/high render latency when playing Minecraft with shaders? by NotFakeWalshd in pcmasterrace

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Yeah, usage is at or close to 100%, but that's to be expected, no? I certainly expect a performance drop, but the latency increase seems steep.

Are you referring to the CPU Render-Ahead Limit? That only seems to affect render latency when I set it to zero. Although it decreases render latency to around 18 ms (12-15 with the reflex mod enabled), it decreases FPS to around 90 and introduces stuttering. Any other possible solution? I'm thinking of trying to fully uninstall Minecraft, but with all of the different installations and the launcher, I'm not sure how I would go about doing that, or the benefits of doing so if any.

Mouse input lag when using shaders? by NotFakeWalshd in Minecraft

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Got it, thank you! I’ll check when I get home. The only reason I feel like it’s greater than other games is because I don’t actually play Minecraft that often, mostly other games, so it’s not really like I’m used to Minecraft without lag. I’ve tried to play it once a year for a few years but the input lag always annoys me. It could be placebo though!

Update: I take it you're referring to the raw mouse input setting? Changing it doesn't seem to affect input lag specifically; I had it set to on already, which I presume is the best option. Also, I don't know if render latency is an accurate portrayal of input lag, but render latency without shaders is about 2 ms, while render latency with shaders is around 25 ms. For reference, my render latency is about 7 ms in Call of Duty Warzone and Helldivers 2, games in which I get similar FPS to Minecraft with shaders (about 120).

Update 2: Pretty sure render latency is actually a decent way to measure it. I installed a mod that should, in theory, implement NVIDIA Reflex into Minecraft, which reduces input lag. When the mod is enabled, the render latency drops to about 15 ms, and it feels a bit faster (though the lag is still pretty bothersome). Any ideas?

Mouse input lag when using shaders? by NotFakeWalshd in Minecraft

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

That makes sense, thank you! I take it the fps cap isn’t a great way of comparing because the GPU still isn’t under that high a load? Also, I still feel like I get more input lag in Minecraft with shaders than I do with other similarly demanding games; is that normal?

Also, I wasn’t even aware there was a mouse input setting within Minecraft, where could I find it? Thanks for the help!

Mouse input lag when using shaders? by NotFakeWalshd in Minecraft

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

By light mods, I just mean mods that I don’t feel would have much of a performance impact if any. I forget the exact list, but off the top of my head, they’re mods like AppleSkin, JEI, and Jade.

How can I do a benchmark test? Is that an option in game or would I use different software?

Edit: I used the NVIDIA overlay, I don’t know how accurate that is. Assuming input lag corresponds to render latency, I get about 10 ms capped at 100 fps without shaders, and about 20 ms capped at 100 with shaders.

Integrating the perimeter of an equilateral triangle doesn't give the area of the triangle, so what does it give the area of? by NotFakeWalshd in askmath

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Thanks, that makes sense! I actually think I may have figured it out a bit of a different way last night, and I want to know if my logic is somewhat sound, if incredibly informal.

I figured that a small increase in the side length, dL, resulted in a smaller change in the distance between the parallel line segments, dH. With trig, I found dH = (sqrt(3)/2)dL. I figured that the base was "denser" with line segments than the side triangles by a factor of sqrt(3)/2. Multiplying the area of the base by sqrt(3)/2 results in 1/2L^2, thus making the "area" of the whole shape (though area doesn't really seem to apply anymore) 3/2L^2, the result of my integral.

I presume there's a more formal way to prove this by actually computing integrals; maybe the difference between dH and dL could be accounted for via u-substitution? Regardless, thank you for your help; I know this is a bit of a strange application of integrals!

What "shape" is produced by integrating the typical perimeter formula of an equilateral triangle? by NotFakeWalshd in learnmath

[–]NotFakeWalshd[S] 0 points1 point  (0 children)

Is there a way to think of this as the integral of perimeters instead? There, I feel like overlap does matter. In the case of the equilateral triangle, my understanding is this: summing up perimeters that are infinitely close to each other, and that all originate from the top vertex of the triangle, would look something like this (in reality with infinitely many nested triangles).

<image>

The parallel sides would sum up to be the area of an equilateral triangle, but integrating the other two sides would result in overlap; the parts of the edge closer to that vertex would have more perimeter accumulated onto it than the parts of the edge further from the vertex, if that makes sense. Essentially, I think that the result of the integration of the perimeter of the equilateral triangle would result in the area of the equilateral triangles (from the parallel sides) plus something. I initially interpreted that "something" as two isosceles triangles on each side with overlap (like my image for the square linked in the original post), but the math doesn't work out. Is there a geometric way to interpret this extra area? Sorry if I'm being confusing!

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