CURRENT ELECTRICITY- 13 | BULB & POWER IN SERIES AND PARALLEL COMBINATION |

[@sayanghosh1698]

Sir GOD ho aap..... Aapke jaisa koi nhi padhata ... Comparison nhi karta hu sir lekin fir bhi aa jata hai.... Kya kare..... Ek mamuli sa Ramlal hoke yhi duaa karta hu ki ishwar aapko itni lambi umar de jitna aaj tak koi nhi jiya hai......
AGAR YE BIHARI APNE JEEVAN ME NA HOTA,,,,
TO NEET DENE KE LIYE MAJDURI ME MARNA PARTA GOTA..
LOTS OF LOVE TO YOU MR SIR ❤️❤️❤️❤️

[@pareshkumarkhatua8950]

I want to take my utmost gratitude to my mr sir.....Worlds best teacher ..❤

[@jyotiswami4632]

sir u are the best 💯🥳😍🙏

[@ravikumarraj]

Amazing lecture ❤❤❤

[@Silver9634]

This was an awesome lecture. It was really detailed and all my doubts were cleared. Thankyou so much sir 🤍

[@topspotphysics8416]

Awesome lecture. I m from pakistan. Really appreciate your help

[@sabanaz7223]

Aap jaisa koi na 😊

[@husan_5911]

Sir pehle apne series ko do alag alag formula se nikal. Aur pehle vae apne inversely proportional Kiya aur dusre Mae directly proportional kar diya ????

[@kartikfuke5113]

All doubts cleared 😇

[@sukhdev9411]

Sir bura mat manana lekin maine power aur bulb ka concept alakh sir se b padha tha unacademy se b padha tha pichle saal....lekin mujhe kabhi samjh hi ni aaya tha....
Aaur aab toh bss ye sab yu paani jaisa nirmal lag raha hai....well akhirkaar physics m toh m kaha hi skta hu aab..."kaha pada tha m chakkar me koi nahi mr k takkar m"

[@VivekKumar-zy6ve]

Amazing

[@NepaliQuanta]

sir at 29:12 why we can't use net power =I^2×Rnet =I^2(R1+R2)=I^2R1 + I^2R2 =P1+P2

[@hnsandhu4746]

MR sir OP

[@MirselimaliAli]

Legend❤

[@Sabeel-infactron]

Concepts clear

[@MamtaYadav-qo1eo]

👍😲

[@ArshwaH5172]

Thanks sir ❤

[@shamilshaheen3966]

Sir is always bulb connected in series always have same voltage rating

[@nobodyelse104]

🎉

[@pukhrajmeena5511]

2:42 pr board pr 1 laser light kisne notice kiya

[@ajeetyadav4637]

At 41:41 since when key is in open conditions then all bulbs are in series,then how Peq =v2/3R,rather it should be less than v2/9R....??

[@JoydeepBasumatary-kx3wz]

No , sir is right , it must be
V²/3R
Like :-
Since key is open so Req = R+R+R =3R
And we know that the formula of POWER is, P= V²/R
So putting value of R we get---
PoWER = V²/3R .
Aya ab samajh ??

[@wandering-magician]

sir one doubt
parallel me Req =1/r1+1/r2....hota hai
par Pconsumed nikalte wakt Vs²/Req me Req ko 1/r1+1/r2 ...kyu nahi liya gaya
aisa lete to Pconsumed=Vs²/1/r1+1/r2....banta formula
what mistake am i making sir???

[@udayshankar454]

Parallel mein 1/Req=1/r1+1/r2+1/r3 hota hai but you wrote R(req)= 1/r1+1/r2+1/r3

[@jeevanshvijay357]

Thank u sir

[@harshrajput4015]

Mr God🙏🙏🙏🙏

[@SajalYadav-t2f]

P= VI

[@SajalYadav-t2f]

❤

[@vanshikasaxena5446]

thank u so much sir

[@sunilchoudhary9090]

Nice sir

[@ShivamKumar-xf1mh]

P rated zyada toh P consumec kam in series