thermal conduction

[wise magnet]

how to solve this . i designed the question on my own inspired from infinite resistors ladder problems.

consider lower rods to be highly conductive . i.e zero thermal resistance
(like a wire in circuit)

[oak trench]

<@&1227988359910260737>

Note for OP

+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

[wise magnet]

<@&1226380963228160000> pls ping phy one

[cobalt hound]

<@&1227987967939838003>

[wise magnet]

i have solved infinite ladder with same repeating units but here the resistance is decreasing continuously by a factor of 2

[wise magnet]

we can do that

damn

[marble flint]

no, hes a mod lol

[wise magnet]

oh lol

[cobalt hound]

https://tenor.com/view/gumball-discord-mod-gif-22325292

me fr ;-;

this is just sad

anyways i have no idea how to do this

seems very intresting tho

solution mile to ping karna please @wise magnet

[fringe merlin]

so here the repeating unit, instead of being R, would be R/2 right

[fringe merlin]

actually wait lol

i = V/R, and R = l/KA

i.e. i proportional to K, which just keeps increasing here

i think you've made an infinite current setup

[wise magnet]

not exactly

[wise magnet]

infinite ladder ye

[wise magnet]

lowkey true 💀

[fringe merlin]

no like the question you've made

you asked for net heat flow

instead of equivalent resistance right

[wise magnet]

yeah that deltaT/thermal resistance

[fringe merlin]

oh im tripping wait

lemme actually use pen and paper

[wise magnet]

so in short u have to solve infinite ladder only

[fringe merlin]

this isnt the same method as infinite ladder

this doesnt have a repeating unit, just a pattern

[wise magnet]

yeah it looks like infinite ladder but it isnt really

[ember jacinth]

neeche waali rods bhi k 2k 4k 8k..... follow kar rahi hain kya?

[wise magnet]

forgot to mention...

consider lower rods to be highly conductive . i.e zero thermal resistance

(like a wire in circuit)

[ember jacinth]

wouldn't the majority of the heat will flow through the lower rods then?

heat current should become infinite

[vital hamlet]

Infinite na? Series diverges

[pastel yarrow]

@frank zodiac pls? look into that error

we are getting rate limited

must be some function which sends requests every second tbh

[snow girder]

Lower rod conductivity is k,2k,4k...?

[vital hamlet]

What is your basis for considering that the rest will be x/2?

[snow girder]

We are scaling all resistances to half their value. Also, I considered T1 and T2 at upper and lower rod otherwise the result diverges.

[snow girder]

If purely conducting rods are used, same approach to find the value ig.

[wise magnet]

they have high conductivity so zero thermal resistance

[wise magnet]

this tho. pls clarify this

[snow girder]

If you consider the point at T2 to be at infinity, converting to the electrical analogy, you can see the points are short circuited due to them being connected by conducting wires, hence effective current tends to infinity.

[snow girder]

Ah, then same proceedure, just the lower portion as conducting wires.

[wise magnet]

i was inspired from this problem

problem is in my question the resistance is getting halved . so idk the answer...

[ember jacinth]

Isme starting ke dono ends different potential pe honge

Na

[snow girder]

This is what is done here. (Approach for both will be the same)

[wise magnet]

i dont understand r/2 wala thing here

r is net resistance or resistance of that resistor

[snow girder]

Let the net resistance of the entiire thing be x. Now you are scaling down each resistor by a factor by 2. So the effective resistance of the entire portion after the verticle r/2 resistance is x, but scaled down by 2. (r=L/k.A)