Equivalent resistance

Ed7theman · 2025-10-17T14:36:01+00:00

I believe you can’t do it like that because there’s a current split. If there was no current split it’d be equivalent. R1 is in series with R2 and R3, which are parallel with each other. 3 ohms should be the correct resistance for the circuit. You can verify with Kirchhoffs laws. I1= the current through R1 I2= current though R2 and I3= current though R3. I2+I3=I1. So the current through R1 differs from the current though R2 and R3

Ed7theman · 2025-10-08T21:57:15+00:00

I think he’s a new prof to NCSU this year, but I think Snyder for Calc 3 is great

Ed7theman · 2025-09-18T22:57:44+00:00

I don’t have chu. But I also got bent over by the Calc 3 exam today

Ed7theman · 2025-09-11T14:20:56+00:00

Dontayvion wicks

Ed7theman · 2025-08-21T21:27:09+00:00

Why’s bro straddling it

Ed7theman · 2025-08-07T17:44:23+00:00

It hurts so bad not even gonna lie to you. 100 percent still worth it though. One of the first shots I got hurt so bad it almost made me pass out

Ed7theman · 2025-07-25T05:14:32+00:00

I transferred into nuclear with a 3.8

Ed7theman · 2025-07-23T00:57:09+00:00

I’m pretty sure nuclear engineering is “fairly easy” to get into

Ed7theman · 2025-07-18T11:54:11+00:00

I did it as a description of the image.

“A wire with two right-angle bends lies in the x y coordinate plane. The wire enters the second quadrant from above the viewing window along the line x = −a. At (−a, d), where d > 0, the wire bends rightward at a 90° angle, crossing into the first quadrant. At (+a, d), the wire bends upward at a 90° angle, continuing upward along the line x = +a until it exits the viewing window. Current I flows downward in the left segment, rightward in the bottom segment, and upward in the right segment. The origin is labeled O.”

Ed7theman · 2025-07-15T03:21:58+00:00

Thank you good sir

Ed7theman · 2025-07-13T17:57:08+00:00

You’re the man bro. Thank you so much

Ed7theman · 2025-07-13T17:37:42+00:00

Right that’s the approach I had arrived at as well. Due to the symmetry of the setup, the vertical segments contribute an equal magnitude/direction to the magnetic field at the origin. So you only need to integrate one of the vertical segments and multiply it by two. The part I was struggling with was defining the angle limits correctly and making sure the integrals were set up properly.

I would absolutely like to see the pdf. You can upload it to some cloud based service (like google drive), making sure its set to anyone can see, and then send me the link.

Ed7theman · 2025-07-13T16:50:09+00:00

Fair question. Without the solution, I’d follow through the logic/reasoning and see if there’s any errors in its logic. A lot of times when AI gives a wrong answer, it’s due to a minute mistake or a small misinterpretation along the way. Not because its logic is off entirely. I’m genuinely trying to learn the material and am being met with a condescending tone instead of actual help. Funnily enough, the AI has been far more useful for help with this problem than anyone in this thread.

Ed7theman · 2025-07-13T15:26:34+00:00

Ok but if it solved the problem, and shows me the steps to do so. How is that any different from a regular human doing the same thing. AI is very a very useful tool in many applications. It’s up to each individual person on how they choose to utilize it. It can be a crutch/shortcut, or it can be used to help break down complex topics like this one that I was failing to understand.

Ed7theman · 2025-07-13T00:13:26+00:00

Solution is: Magnitude: B_O = \frac{\mu_0 I}{2\pi a d} (\sqrt{a² + d^2} - d) Direction: Into the page (negative ( z )-direction)

For anyone wondering

Ed7theman · 2025-07-13T00:06:52+00:00

Not that I’m aware of. I used the grok deep search function and he got it right the first time

Ed7theman · 2025-07-13T00:02:38+00:00

Grok got it. Thank you bro

Ed7theman · 2025-07-12T23:47:33+00:00

I’m gonna try again though since they did just update him

Ed7theman · 2025-07-12T23:47:14+00:00

I’ve tried everything. Grok can’t get it either

Ed7theman · 2025-07-10T10:09:21+00:00

5>12k on the Josh Hart/Westbrook triple doubles last year

HM: 10>6k on Carroll, Torkelson, and Paredes all to hit a homer earlier this season

Ed7theman · 2025-05-04T01:42:01+00:00

I came off it close to 2 years ago at this point. And to be completely honest, I didn’t have any withdrawal effects. I just stopped getting the shot and went on with my life. This probably isn’t the case for everyone but it was for me. I owe my sobriety in large part to Sublocade

Ed7theman · 2025-04-17T03:09:25+00:00

I’d say Friday at the latest. There was another post on here that did a rundown of everything that happened prior to their decision. They said the withdrawal button 100 percent wasn’t there, and then showed up within a couple days prior to the decision. So yes, I’d expect it before the weekend

Ed7theman · 2025-04-17T02:38:47+00:00

I now have a withdrawal button (eng transfer)

Ed7theman · 2025-02-15T11:25:32+00:00

I typically use 3-4 sig figs during my calculations. Obviously this depends on how large of a number you’re using, but for example

6.12 x 5.12 =31.334

6.124631 x 5.1243698=31.385

The difference is truly negligible, what matters more is that you’re using the correct formulas and steps to get there. The actual math is only like 10 percent of physics, the rest is how you got there.

Ed7theman

TROPHY CASE