/sci/ - Science & Math » Thread #12192859

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/sci/ brainlet test

Anonymous Sun 04 Oct 17:16:01 2020 No.12192859 View Reply Original Report

Quoted By: >>12192896 >>12193340 >>12193537 >>12193555 >>12193608 >>12193641 >>12193771 >>12194184 >>12196631

which light glows brighter /sci/?

Anonymous

Anonymous Sun 04 Oct 2020 17:24:07 No.12192896 Report

Quoted By: >>12192906

>>12192859
60W

ee here

Anonymous

Anonymous Sun 04 Oct 2020 17:26:21 No.12192906 Report

Quoted By: >>12192924 >>12192967

>>12192896
Why? Wouldn't the 100 W bulb have more current flowing through it since they're the same voltage?

t. Not ee

Anonymous

Anonymous Sun 04 Oct 2020 17:29:38 No.12192919 Report

Quoted By: >>12192940

60W

t. also EE

Anonymous

Anonymous Sun 04 Oct 2020 17:31:17 No.12192924 Report

Quoted By:

>>12192906
>Wouldn't the 100 W bulb have more current flowing through it
They have the same current flowing through it, they are in series.

Anonymous

Anonymous Sun 04 Oct 2020 17:35:15 No.12192940 Report

Quoted By: >>12192949 >>12192964 >>12192976

>>12192919

Why?

Assuming these are incandescent bulbs, the 60W bulb has a higher resistance. That is why it uses less energy. Making the assumption that they both have the same efficiency (light to heat ratio) the current is the same across both (series) therefore, the lower resistance bulb has a lower voltage across it and dissipates less energy; hence it gives off less light.

Anonymous

Anonymous Sun 04 Oct 2020 17:37:14 No.12192949 Report

Quoted By: >>12192964

>>12192940

(cont) that is why lower wattage bulbs are dimmer even when they are on the same series circuit in your home.

Anonymous

Anonymous Sun 04 Oct 2020 17:43:16 No.12192964 Report

Quoted By:

>>12192949
>>12192940
Y'all chuds are actually heckin pathetic. Leave science to actual educated leftists please.

Anonymous

Anonymous Sun 04 Oct 2020 17:44:43 No.12192967 Report

Quoted By: >>12194184

>>12192906
R(L1) = Umax(L1)^2/Pmax(L1)
Same R(L2).

Req = R(L1) + R(L2)

I = U/Req

P(L1) = R(L1) * I^2
Same for P(L2)

Replace with value and P(L1) > P(L2).
Or : when P is high so R is low. So there will be more voltage accross the one that have low P.

Anonymous

Anonymous Sun 04 Oct 2020 17:48:11 No.12192976 Report

Quoted By: >>12193041 >>12193536

>>12192940
R1 = ( 230 V )^2 / ( 60W ) = 882 Ohms
R2 = ( 230 V )^2 / ( 100 W ) = 529 Ohms
I total = ( 230 V ) / ( 882 + 529) = 0.163 Amps

0.163 Amps * 882 Ohms = 143 Volts across the 60W bulb
0.163 Amps * 529 Ohms = 82 Volts across the 100W bulb

Anonymous

Anonymous Sun 04 Oct 2020 18:12:02 No.12193041 Report

Quoted By: >>12193055

>>12192976
This is quite misleading then right, putting the voltage above each bulb

Anonymous

Anonymous Sun 04 Oct 2020 18:15:08 No.12193055 Report

Quoted By: >>12193613

>>12193041
thats part of the brainlet test. if you cant ascertain what the values mean you are a brainlet.

Anonymous

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Anonymous Sun 04 Oct 2020 19:49:10 No.12193340 Report

Quoted By:

>>12192859
/sci/ demands new puzzles. this one is from last week

Anonymous

Anonymous Sun 04 Oct 2020 20:59:37 No.12193536 Report

Quoted By: >>12193594 >>12194180

>>12192976
You forgot to square the I brainlet

Anonymous

Anonymous Sun 04 Oct 2020 20:59:43 No.12193537 Report

Quoted By:

>>12192859

The solution can't be known. Incandescent light bulbs are very inefficient. Most of the energy that's dissipated is in the form of heat, so you'd have to know an efficiency to at least make a guess which one is brighter and not just hotter.

Anonymous

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Anonymous Sun 04 Oct 2020 21:09:29 No.12193555 Report

Quoted By:

>>12192859
Nobody knows, electricity is magic

Anonymous

Anonymous Sun 04 Oct 2020 21:24:21 No.12193589 Report

Quoted By:

again?

Anonymous

Anonymous Sun 04 Oct 2020 21:25:23 No.12193594 Report

Quoted By:

>>12193536
you are the actual branlet

Anonymous

Anonymous Sun 04 Oct 2020 21:26:50 No.12193599 Report

Quoted By:

Hint: if you are doing any calculations it is a fail.

Anonymous

Anonymous Sun 04 Oct 2020 21:30:42 No.12193608 Report

Quoted By:

>>12192859
Your question willfully doesn't state that the watts is the power rating of the bulb (which is the power used in a 120V parallel connection to the bulb with i^2*R as power) not the power dissipated by the bulb in that circuit.
I'll leave you to figure out the rest of why i^2 is more important than R and how it corresponds to the two other circuits you didn't draw.

Anonymous

Anonymous Sun 04 Oct 2020 21:32:01 No.12193613 Report

Quoted By: >>12193763

>>12193055
82
60
285
40
100
65
If you can't figure out what these values are you're a brainlet

Anonymous

Anonymous Sun 04 Oct 2020 21:43:24 No.12193641 Report

Quoted By: >>12193644

>>12192859
actual ee here, not just a larping mathfag
unironically impossible to answer without knowing the resistances of each bulb

Anonymous

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Anonymous Sun 04 Oct 2020 21:44:28 No.12193644 Report

Quoted By: >>12193656

>>12193641

Anonymous

Anonymous Sun 04 Oct 2020 21:46:01 No.12193649 Report

Quoted By: >>12194753

Actual EE here. You need to take inductance into account too, because the filament is shaped like a coil so there is a lot of reactive power that is returned back to the grid. Two lighbulb in series is a pretty serious offense since you are lagging now so expect a nasty call from your utility company

Anonymous

Anonymous Sun 04 Oct 2020 21:48:45 No.12193656 Report

Quoted By: >>12193660 >>12193763

>>12193644
completely different configuration

Anonymous

Anonymous Sun 04 Oct 2020 21:49:39 No.12193660 Report

Quoted By: >>12193668

>>12193656
anon are you autistic or something

Anonymous

Anonymous Sun 04 Oct 2020 21:51:39 No.12193668 Report

Quoted By: >>12193683 >>12193763

>>12193660
Each lightbulb is getting 230 volts in your new configuration

Anonymous

Anonymous Sun 04 Oct 2020 21:55:27 No.12193683 Report

Quoted By: >>12193694 >>12193696

>>12193668
have you considered that analyzing the unrelated problems may give you insight into solving the problem at hand?

Anonymous

Anonymous Sun 04 Oct 2020 21:59:32 No.12193694 Report

Quoted By:

>>12193683
no they are two different configurations.

Anonymous

Anonymous Sun 04 Oct 2020 22:01:29 No.12193696 Report

Quoted By:

>>12193683
have you considered that analyzing the unrelated problems may give you insight into solving the problem at hand?

Anonymous

Anonymous Sun 04 Oct 2020 22:20:27 No.12193763 Report

Quoted By: >>12193765

>>12193613
>>12193656
>>12193668
you have all failed the brainlet quiz sad!

Anonymous

Anonymous Sun 04 Oct 2020 22:21:19 No.12193765 Report

Quoted By:

>>12193763
wrong
we cannot know what proportion of the voltage each bulb is receiving
therefore we cannot know how much power each bulb is dissipating
therefore we cannot know which bulb is brighter
go back to doing proofs

Anonymous

Anonymous Sun 04 Oct 2020 22:24:49 No.12193771 Report

Quoted By:

>>12192859
The one on the right because the electrons go that way first

Anonymous

Anonymous Mon 05 Oct 2020 01:15:24 No.12194180 Report

Quoted By:

>>12193536
Why the fuck would I square the current

Anonymous

Anonymous Mon 05 Oct 2020 01:17:14 No.12194184 Report

Quoted By:

>>12192859
in series, you're going to get brighter glow with lower rated wattage. derivation is probably somewhere ITT
here: >>12192967

Anonymous

Anonymous Mon 05 Oct 2020 05:24:28 No.12194739 Report

Quoted By:

Not enough information is provided, this is dumb.

Anonymous

Anonymous Mon 05 Oct 2020 05:28:33 No.12194753 Report

Quoted By:

>>12193649
You haven't calculated anything in years.

Anonymous

Anonymous Mon 05 Oct 2020 18:41:49 No.12196597 Report

Quoted By:

its 50/50. They either glow brighter or they dont.

Anonymous

Anonymous Mon 05 Oct 2020 18:55:01 No.12196631 Report

Quoted By:

>>12192859
idk, analog is a bitch

CE here

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