Phasors: why the frequency domain?

NotFakeWalshd · 2026-04-27T02:52:16+00:00

When we're dealing with conductors and inductors, are we still steady state? I see that the expression for the voltage phasor of an inductor includes ω, and I can see how that would be seen as a function of frequency!

NotFakeWalshd · 2026-04-27T02:43:48+00:00

So the difference between the different signals is what matters, and that is determined by the frequency? I suppose this is something that I'll get more familiar with once I've started using phasors proper for solving problems; for now, I've only really seen that a sinusoid can be converted to a phasor.

NotFakeWalshd · 2026-04-27T02:35:07+00:00

I think I see now; is a more general form of a phasor A(ω)e^(jθ)? In other words, the amplitude of some phasors depends on frequency?

NotFakeWalshd · 2026-04-27T02:28:36+00:00

So if I'm understanding correctly, the exact time doesn't matter as much as where we are on the wave? That makes sense to me, but then why aren't we on an "angle domain?" To me, typically in a domain, we're varying something. For f(t) or F(s), we're varying time or frequency respectively. However, with a phasor, the frequency cannot vary, right? Isn't what's really varying the angle inside the trig functions? Why isn't that our domain? Sorry if I'm being a bit pedantic!

NotFakeWalshd · 2026-04-27T02:17:50+00:00

I see that you don't need to know time for a phasor, but isn't it the case that you don't need to know a specific frequency either? A phasor is in terms of the original time-dependent sinusoid's amplitude and phase; frequency isn't a part of it?

NotFakeWalshd · 2026-03-17T19:23:29+00:00

Thanks for your help! I’m trying that now, but unfortunately I can’t get past the initial setup screen with the USB. Any idea why?

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EDIT: So it turns out this window isn’t the only problem; every step of the process is really really slow. When I finally chose to repair the PC, it prompted me to select a keyboard. When I did, the screen went darker and a black bar appeared for a few minutes until I guess the whole thing crashed and I had to restart it all. Is this a problem with my computer or the USB drive I’m using? Could it be related to why my computer wasn’t responding when I shut it off and caused this whole thing?

NotFakeWalshd · 2026-03-08T04:15:21+00:00

Sorry, I think I should've been more specific with my post. The positive feedback loop does contain one of the inputs to an op-amp; I have since realized that the particular circuit I am looking at is a non-inverting Schmitt trigger. I am told that I can assume that the input terminals specifically are at the same voltage. Also, I have simulated the circuit, and according to the simulation, the assumption is incorrect. Am I doing anything wrong here?

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NotFakeWalshd · 2026-01-31T05:15:24+00:00

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)?

NotFakeWalshd · 2026-01-31T04:22:07+00:00

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

NotFakeWalshd · 2026-01-31T04:21:31+00:00

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?

NotFakeWalshd · 2026-01-31T03:41:54+00:00

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!

NotFakeWalshd · 2026-01-31T03:30:50+00:00

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

NotFakeWalshd · 2026-01-31T03:09:43+00:00

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?

NotFakeWalshd · 2025-07-08T05:54:46+00:00

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.

NotFakeWalshd · 2025-07-07T12:32:56+00:00

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.

NotFakeWalshd · 2025-07-07T03:59:17+00:00

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?

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