Series and Parallel Circuits: Which Bulb Glows BRIGHTER?

[@TheScienceCube]

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ELECTRIC CURRENT PLAYLIST: kzbin.info/aero/PLQgL7RKnNg_yKfpNYK6KEOchLAsDhp4Lp

[@matt32videos30]

Thanks , I have finished electricity , at this time I am just doing some exercises , my next chapter is magnetism :)

[@TheScienceCube]

Great to know! Do check out thesciencecube.com/. It would probably have useful resources, particularly the PDF summary notes. Best wishes!

[@nickkang1917]

Can you elaborate on the parallel piece, feels like there's missing info on why you just used 2 for R without calculating R equivalence in parallel

[@TheScienceCube]

Good question Nick! You can do both ways. What I have done is stayed with the actual situation where there are 2 bulbs and I have found power delivered by each. If you find the equivalent resistance, it'll come to 1ohm and therefore the current through that 1 ohm resistor would be 8 amperes (I = V/R = 8/1)...and then if you calculate the power, it will be I^2 R or 8^2 * 1 = 64 watts which is the same as I calculated. Does that answer your question?

[@Khan-pk7tr]

​@@TheScienceCube thank you sir❣️

[@TheScienceCube]

@@Khan-pk7tr Where are you from?

[@nataliadabek2566]

very nicely explained , thank you!!!

[@TheScienceCube]

Glad you liked Natalia :) Can you suggest any one thing that can make these videos more useful for students?

[@TheScienceCube]

...also do check-out this lesson I uploaded this week kzbin.info/www/bejne/mnyzq4uMiqh7nKc

[@TheScienceCube]

Hi! Please also join my Telegram Channel t.me/TheScienceCube_Community for PDF lesson summaries and other physics solutions. You can always leave if you wish to, but trust me, you won't want to miss this! 😉

[@satheeshkumark4]

EXCELLENT CLARITY SIR

[@TheScienceCube]

So glad you liked it! I am trying to grow this channel. Can I request you to share this video with your friends. Thanks!

[@TheScienceCube]

Hi! Please also join my Telegram Channel t.me/TheScienceCube_Community for PDF lesson summaries and other physics solutions. You can always leave if you wish to, but trust me, you won't want to miss this! 😉

[@Sandhya-l3d]

thank u sm for this!!!!!!!!!!!

[@TheScienceCube]

@quantizedenergy so happy to this lesson helped!! What's sm?

[@Sandhya-l3d]

@@TheScienceCube sm means so much haha :D

[@narayanankutty8497]

Why are we using P = V²/R instead of P = i²R ?? is it because source voltage is same for all?

[@TheScienceCube]

You can use either. The answer will be the same

[@aparnasubramanian1203]

Sir , if number of identical bulbs are increased in series and parallel , will there be any change in brightness?in which circuit?

[@TheScienceCube]

Hi Aparna! First, I totally overlooked your question. Sorry for that. Yes the power in bulbs in series will reduce but that in parallel will not reduce. I suggest you do the calculation the way I have done taking 4 resistors of 2 ohms each in both situations, using the same logic I have used

[@aparnasubramanian1203]

@@TheScienceCube thanks sir

[@TheScienceCube]

@@aparnasubramanian1203 let me know if you got it.

[@goyaldeekshant]

nice

[@TheScienceCube]

Are you prepping for an exam?

[@PranavKumar-dd2xo]

i have a doubt if i connect 3 same bulbs in series will all the 3 glow with same intensity of the 1st one will b brightest. doubt is will the voltage equally divide bcz I read somewhere that the last bulb will glow the least

[@TheScienceCube]

The way you should see the situation is that you have introduced another resistor in the circuit. This means you are blocking free flow of electrons. So the current goes down across the series. However I will still be the same for each bulb since electrons have only one path to pass through. So lower current means lower glow for each of the 3 bulbs, but equal

[@sowmyasrithiramdas482]

shouldn't effective resistance in parallel be 1

[@TheScienceCube]

That is right but that's not what we are after.

[@sowmyasrithiramdas482]

@@TheScienceCube that's right, I was clarifying my doubt because I have an exam tomorrow by the way thank you for such a clear and informative explanation,I appreciate your efforts

[@TheScienceCube]

@sowmyasrithiramdas482 Tanks a bunch Sowmya! 😊 Join me on Telegram t.me/TheScienceCube_Community for PDF lesson summaries and other physics solutions that'll hopefully make you go "Eureka!" Link: 🔬📚💡 Let's geek out on Physics! 💫 PS: You can always leave if you wish to, but trust me, you won't want to miss this! 😉

[@bwk8327]

R between d c in parallel should be 1 as follow 1/R=1/r1+1/r2

[@TheScienceCube]

That is correct

[@merykjenkins3274]

We find this out soon...skip

[@TheScienceCube]

Umm... is there a question?

[@silentnerd1112]

Sir if we use P=VI Than the answer is coming 32 W? Why P=VI P=8*4(in case of series) =32 Please help

[@TheScienceCube]

Have you taken V = 8V in series?

[@silentnerd1112]

Sir actually i found my mistake... I doubled the current to 4 it should had been 2 only that would give 8x2=16. Sorry for wasting your time Thank you for giving your time

[@TheScienceCube]

@@silentnerd1112 That is correct. You should also try to calculate individually and then add to see if you get the same answer.

[@silentnerd1112]

Yes sir tried it both ways power of individual bulb is 8W since the voltage will get divided equally among the bulb as they have the same resistance That would mean P=4*2=8 Than 8+8=16

[@TheScienceCube]

@@silentnerd1112 Fantastic! I am really very happy that you figured out on your own :) Where are you from?