Browse by Stream
QnA
Home

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The bubs are made and the resistance of the filament increases with the increase in temperature. If at room temperature, 80 <img alt="\small \omega" src="https://entrancecorner.oncodecogs

The bubs are made and the resistance of the filament increases with the increase in temperature. If at room temperature, 80 \small \omega, 60 \small \omega and 20 \small \omega  bulbs have filament resistance \small R_{80},\ R_{60} and \small R_{20}. respectively, the relation between these resistances are:-

 

Option: 1

\frac{1}{R_{80}}=\frac{1}{R_{60}}+\frac{1}{R_{20}}


Option: 2

R_{100}>R_{60}>R_{40}


Option: 3

R_{100}=R_{60}+R_{20}


Option: 4

\frac{1}{R_{100}}>\frac{1}{R_{20}}>\frac{1}{R_{60}}


Answers (1)

best_answer

P=\frac{V^2}{R} 

Since same voltage is applied across all bulbs.

\\So, \\80 \omega>60 \omega>20 \omega

\frac{V^2}{R_{80}}>\frac{V^2}{R_{60}}>\frac{V_2}{R_{20}}

\Rightarrow \frac{1}{R_{80}}>\frac{1}{R_{60}}>\frac{1}{R_{20}}

80 \omega=60 \omega+20\omega \text { [At room temperature}]

So,

\frac{V^2}{R_{80}}=\frac{8^2}{R_{60}}+\frac{V^ 2}{R_{20}}

\frac{1}{R_{80}}=\frac{1}{R_{60}}+\frac{1}{R_{20}}

 

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel
anam-khan

CSAB 2025 counselling dates out for special, supernumerary rounds through JEE Mains

Latest Question