Get Answers to all your Questions

header-bg qa

A set of 'n' equal resistors, of value 'R' each, are
connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is

 

  • Option 1)

    20

  • Option 2)

    11

  • Option 3)

    10

  • Option 4)

    9

 

Answers (1)

best_answer

As we have learned

Main current / current from each cell -

i=frac{nE}{R+nr}

- wherein

n- identical cells which connected in series

 

 

Main current -

i= frac{E}{R+frac{r}{n}}

-

 

 

 

I= \frac{E}{R+nR}    And 10I= \frac{E}{R/n+R}

10I= \frac{nE}{R+nR}= nI

n= 10

 

 

 

 

 

 

 


Option 1)

20

This is incorrect

Option 2)

11

This is incorrect

Option 3)

10

This is correct

Option 4)

9

This is incorrect

Posted by

Avinash

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks