As shown in the figure, a potentiometer wire of resistance 2 0 Ω  and length 300cm is connected with resistance box (R.B) and a standard cell of emf 4 V. For a resistance ‘R’ of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of ‘R’ is __________ Ω .

0 4 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago

    R A B = 2 0 Ω

    L = 300 cm

    Null point is at 60 cm from A, so

    2 0 m V = 4 R + 2 0 × 2 0 × ( 6 0 3 0 0 ) × 1 0 3

    2 0 = 1 6 R + 2 0 × 1 0 3

    R = 1 5 6 0 0 2 0 = 7 8 0 Ω

Similar Questions for you

A
alok kumar singh

Kindly go through the solution

 

A
alok kumar singh

It is balanced wheatstone bridge

 

A
alok kumar singh

Charge remains same

 

A
alok kumar singh

Kindly go through the solution

 

A
alok kumar singh

It is balanced wheatstone bridge

 

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post