Solved example: Power dissipated in bulbs

[@dharanishree2401]

when I feel like I have no one, i come here because you make things easy. thankyou😢

[@mohemmedilham3854]

Same here🙌

[@dynaspinner64]

The 2nd bulb glows brighter since it has a high resistance. If it has high resistance, that means more heat will be produced resulting in a stronger glow than the 1st bulb. Although as others mentioned, the 2nd bulb will get fused because the power generated exceeds its power rating. So basically the power rating is not what matters to determine whether or not a bulb glows really bright or not, it is the resistance of the filament. This is my answer. I'm thinking my reasoning is probably correct unless proven otherwise.

[@sanjudevi7633]

Can you also give a theoretical explanation for the same bulbs connected in parallel? The resistance if the bulbs will still be same and according to P=V²/R the power dissipated in the bulb with lower resistance will be high means it'll glow brighter. But this doesn't fits in the definition that more resistance means more power yn Please explain this

[@siyab0303]

​​@@sanjudevi7633 If according to equations, then, When they are connected in series, the current remains same but voltage changes. This is why we consider the eqn P=I²R. According to this the more the resistance, the more the power would be. When they are connected in parallel, voltage remains the same but current changes. This is why we consider the eqn P=V²/R. According to this the more the resistance the less the power dissipated will be. I hope this helps

[@Rodion_raskalnikov__]

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[@adeypurohit01]

Finally I got it ! Life is chilled now 😄

[@muskanagrawal7428]

Thanks u actually talked about the concepts behind these problems ...

[@AliSafdarSaeed]

Sir very well described.I think the answer of your last question is simply that the bulb with low power rating has high resistance.If resistance is high ,the bulb will glow brighter.If my answer is wrong,kindly tell me the right one.

[@ultimateaaj2921]

No in parallel connection, the one with lower resistance has a higher power dissipated.

[@Cheatrunner]

​@@ultimateaaj2921but it is not parallel connection, it is series connection

[@siyab0303]

​@@ultimateaaj2921 If according to equations, then, When they are connected in series, the current remains same but voltage changes. This is why we consider the eqn P=I²R. According to this the more the resistance, the more the power would be. When they are connected in parallel, voltage remains the same but current changes. This is why we consider the eqn P=V²/R. According to this the more the resistance the less the power dissipated will be. I hope this helps

[@quratulainraza1415]

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[@vestoic_o77]

awesome,i learned a tricky concept so easily

[@allendora6709]

Best teacher 😍😍😍...

[@sushantraj604]

Man! You are justttttttt awesome. Love you❤❤

[@andrewjustin256]

Would you please clarify this to me that for a bulb we know it has power of 50W and voltage of 10V, therefore we can calculate how much current will pass through it from the equation P= I∆V, and we find it to be 50/10= 5Amps, which is going to be the same if we were to have found this from your answer of having 2ohm resistance, turing out to be P= I^2R ==} I =√(50/2) ==} I=5Amps. However, after this, we know P=I ∆V, and by solving this for new power we get P= 5Amp× (∆V of the battery, ie. 5V) ==} P= 25W, which is twice of 12.5W. Likewise, if I solve this with P=I^2R, I get (5)^2×2, which is 50 Watts. It appears that every answer of mine is multiplied by 2, like 12.5× 2=25×2=50. Please tell me what I did wrong here!

[@abhishekvishwakarma252]

Best teacher..

[@lumeshpawar4568]

Nicely explained. 👍👍 Thank u.

[@GovardhanReddy-os7cx]

Good content sir

[@cluBMallu]

Nice class. Thankyou

[@andrewjustin256]

In the second problem, you solves the final power with P=I^2×R; however, if I solve this with other equation, I am receiving different answers. Like, for instance, P=V^2/R => (20 volt of first bulb)^2/2 => 200watts. Please tell me what is happening here.

[@liketh]

Bro they are in series so voltage is different for both resisto. I - 4A R1-2ohm so voltage is Ir which is 8volts and second one voltage is 32volts

[@liketh]

Now use v2/R

[@liketh]

We get 64/2 32 watts and 32*32/8 which is 4 into 32 which is 128 watts

[@amrelsheikh5770]

The last question Solution is So easy >>> we notice that the rating of the two Bulbs have the same Voltage (20 V) >> and the same Voltage mean if we have a circuit of 20 V for the Bulbs the first will be 200 and the other one will be 50 >> and we can get this case if we connect them parallel so we have the same Voltage ,and by the way in Parallel Circuits if we have 20 V and raise this to 40 V (20*2) the Power will multiply by 4 directly so the 1st bulb will be 800 W ,and2nd be 200 W directly and the current will multiply by 2 only like voltage >>> But If we Put them as series we still have the same current but different Voltages So the High Voltage will cause the High heat and the High Glow .. Hope that answer your question

[@adityarishi8451]

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[@aviaasia4114]

such a lifesaver!! tysm

[@sudhansukumarmaharana4457]

Don't the 80watt bulb get fused? Because power consumption is now 128watt

[@AlagarPandi-lr9fo]

Superb

[@tejaswaniaratipalli8090]

awesome explanation

[@shushpeople]

Thanks!! This has helped a lot 😃

[@MathGyanBySK]

Mast

[@RelevantDad]

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[@sonukumar.7501]

THANK YOU BEST VEDIO

[@rosemarychacko3690]

Thanks sir

[@ChinitheLovebird]

Thankyousomuch sir!

[@abhijeetsrivastava5614]

Thank you so much sir

[@sreenivasansreya6029]

Thanks.

[@JamesKalelemba-b7w]

What if you calculate the current first and then use it to calculate power? How's it

[@sandipbhunia4426]

That's nice

[@shadanali5858]

Thanks

[@abhishekbatoy5532]

Thanks 👍

[@nisa_rosario]

Thank you sir but how about parallel circuit?

[@soulwanderer2143]

Thank you...😁

[@rajkarmavat8163]

❤️❤️❤️❤️❤️

[@awontyy]

why don't they just say what's the resistance on the electronic devices? 🤷🏽‍♀️ why do we have to calculate it !

[@ayman9422]

🇪🇬👍🏻

[@priteshkhilari1918]

Sir plz elaborate why 2nd bulb of less rating glows out to be brighter when connectee in series?🤔

[@HappySM]

Bcz from calculation we got: R1=2 ohms & R2=8 ohms & i=4A So, Power dissipated in bulb 1, P1=32W Power dissipated in bulb 2,P2=128W Since the power dissipated in bulb 2 is more than that of bulb 1, It glows brighter. It's not always the case that the bulb whose power rating is higher glows brighter.....you have to find it with thw data given..

[@sirajalam1521]

The power dissipated in second bulb comes out to be 128 watt, which is greater than the rated power across it,so second bulb will get fused. Second bulb will not glow at all, Only the first bulb will glow in this case.

[@Ratulphysics]

How a 20 w bulb can produce 128 w !!!!!!!!!

[@Ratulphysics]

50 w to 128 w impossible.... Great mistake here

[@petermarkwood9077]

@@Ratulphysics the only mistake is he took a battery which provides 32V across the 8ohm register (bulb) where the max voltage we can provide is 20V according to the voltage rating of the bulb...otherwise voltage difference over 20 volts will damage the bulb...but our mathematical equation is not able to predict the highest voltage...so that's why due to this mistake we got 128W from 50W bulb which will never be possible in real life. Hope it will help you🙂🙂

[@Ratulphysics]

@@petermarkwood9077 yes I understood it clearly but the problem is that this mistake makes a wrong conception Among many students that's why I commented here. And everything fine but we must consider the Voltage case here also with power in every aspect.since no light can be bright than it's actual limit.. Thank you for the reply

[@ultimateaaj2921]

​@RA Physics - Ishtiak Kabir Ratul The actual problem here is that there are two bulbs. So the voltage is split. If you calculate the voltage of each bulb, you will find that the 50W bulb gets a lot more voltage as it has higher resistance. So Naturally, it will dissipate more power.

[@Learnwithme-d2b]

But can it dissipated more power than it's actual rating?​@@ultimateaaj2921

[@malekqaddarah]

Hello sir pls do like the second example but the bulbs are in parallel

[@mounivarma1840]

Parallel problems?

[@freddykambata421]

I am still thinking why the one with less power glows more

[@tiredcat6653]

@Ahmad Almusharraf omg I love you for the explanation. Tysm💕💕💕💕

[@nikeshbabu2875]

In the last problem why sir we aree not using V2/R?

[@KhanAcademyIndiaEnglish]

You can use it and I encourage you to try. But, we would first have to find the V across each resistor. 40V is the total voltage across both resistors.

[@nikeshbabu2875]

@@KhanAcademyIndiaEnglish oh... Okay sir thanks a lot 👍

[@Abhishekrayo]

I could understand logically why bulb with low power rating glows brighter ? Please explain

[@adityarishi8451]

Thats 80ohm not 8ohm

[@salenagomez3240]

Sir in 2nd question we can answer without finding current in series because current in series are same

[@anugrahahaha1908]

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[@salenagomez3240]

@@anugrahahaha1908 😣

[@ishika6802]

Hey it's selena...

[@ch.jamilu6626]

But if u use v^2/r formula here v is not constant in series. In series I is same. Then you get correct value while using i^2r

[@martinkykta3784]

Your answer for the power dissipated in the 10V, 50W bulb operating at 5V is incorrect. The resistance at 10V and 5V is not the same because of the temperature dependence of the tungsten filament resistance. The filament is cooler at 5V. The correct power is determined as follows: P2=P1*(V2/V1)^1.55 P2=50*(5/10)^1.55=17.1W

[@Sami-el8jk]

Where does the 1.55 comes from

[@kaiyeeyu4788]

How about the parallel one?

[@samiabdelghaffar4877]

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[@heetsonani8371]

3:00

[@shivamgulati1838]

Is it of 10th

[@pholososello1299]

Those accents

[@sindhugummadilli3662]

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[@bhavanimuthuraman9808]

Haven’t they asked to find the power dissipated

[@electronicslecturesbygulsh3799]

When 2nd bulb is having rated power of 50watt how 128 Watt is paper dissipated sir

[@jamilakhatoonazad2243]

Actually the second bulb will fuse due to higher power through it than rated...hope it helps

[@ruwafariha3723]

As the voltage is low in the 1st bulb

[@martinkykta3784]

Cccccc

[@faizanishfaq4503]

what about power in parallel?

[@worldaroundivana5398]

Why in second question power dissipation according to the formula P=V^2/R is different than P= I^2/R..

[@bobnob4393]

Because V=I/R. You can just substitute. They're are 3 different ways you can arrange the power equation with these two equations

[@HappySM]

You are wrong here. From both the ways, you get P1 and P2 to be 32W and 128W (resp) only. Let me explain this to you.. From calculation we got: R1=2ohm R2=8ohm... Now... Total volatage across circuit is 40 volts & We know that in series circuit V is proportional to R ( since in V=IR, I is costant for both resistances and so V is proportional to R ) So V1÷V2 = R1÷R2...........equation 1 V1 + V2 = 40volts V2 = 40 - V1 volts.........equation 2 Putting eqn 2 in eqn 1, we get V1= 8volts and therefore V2= 32 volts And therefore Power diss. in bulb 1= 32W Power diss. in bulb 2= 128W EXACT SAME VALUE as we got from I^2.R formula. Hope it helps☺️

[@shreyashsaxena6248]

hello, could anyone help figure out why the bulb with a lower power rating glows brighter?

[@arunikaraghuvanshi3412]

@@shreyashsaxena6248 the resistance of the second bulb is higher. Which means the heat generated in the second bulb will be higher. So power dissipated would be higher.

[@_N_o._.one_]

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[@ohmygoddw859]

your observation is extraordinary

[@_N_o._.one_]

@@ohmygoddw859 Thanks. At least someone pays respect to my abilities😂😂🤣🤣🤣🤣

[@khursheedshaik9540]

Sir, is the electrical energy available to the bulb to convert into the light and heat energy comes from the moving charges? means by their collisions? plzzzz zzz replyyyyyyy sirrrrr

[@dr4568]

Pagal bana rahe hai ye ........ 50 W power ke bulb me 128 W kaha se aya ? 50 W se jara sa bhi uppar gaya to phut jayega wo bulb ! bachhe ki jaan loge kya ?

[@Abhishekrayo]

Hmm. Same I am Amazed

[@Abhishekrayo]

Do You Know Where The Mistake is ?

[@bonzoking1909]

fake accent

[@shivamgulati1838]

Is it of 10th