NEET Exam > NEET Questions > In the potentiometer experiment two cells Of ...
Start Learning for Free
In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is?
Most Upvoted Answer
In the potentiometer experiment two cells Of emf E1&E2 are used in ser...
And E2 are connected in series. The positive terminal of E1 is connected to one end of the potentiometer wire, and the negative terminal of E2 is connected to the other end of the wire. The galvanometer is connected between the positive terminal of E2 and the sliding contact on the potentiometer wire.
When the sliding contact is moved along the wire, the potential difference across the galvanometer changes. By measuring the position of the sliding contact when the galvanometer reading is zero, the emf of E2 can be determined.
To find the emf of E1, the positive terminal of E2 is disconnected from the potentiometer wire and connected to the negative terminal of E1. The galvanometer is connected between the positive terminal of E1 and the sliding contact on the potentiometer wire.
Again, by measuring the position of the sliding contact when the galvanometer reading is zero, the emf of E1 can be determined.
This experiment relies on the principle that the potential difference across a potentiometer wire is directly proportional to the distance along the wire. By measuring the position of the sliding contact when the galvanometer reading is zero, we can determine the emf of each cell.
When the sliding contact is moved along the wire, the potential difference across the galvanometer changes. By measuring the position of the sliding contact when the galvanometer reading is zero, the emf of E2 can be determined.
To find the emf of E1, the positive terminal of E2 is disconnected from the potentiometer wire and connected to the negative terminal of E1. The galvanometer is connected between the positive terminal of E1 and the sliding contact on the potentiometer wire.
Again, by measuring the position of the sliding contact when the galvanometer reading is zero, the emf of E1 can be determined.
This experiment relies on the principle that the potential difference across a potentiometer wire is directly proportional to the distance along the wire. By measuring the position of the sliding contact when the galvanometer reading is zero, we can determine the emf of each cell.
Community Answer
In the potentiometer experiment two cells Of emf E1&E2 are used in ser...
We know that, E1/E2 = L1/L2 . Applying this formula, E1+E2/E1-E2 = L1+L2/L1-L2 [using componendo and dividendo]. We get ratio E1/E2=3:1.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is?
Question Description
In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is?.
In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is?.
Solutions for In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? defined & explained in the simplest way possible. Besides giving the explanation of
In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is?, a detailed solution for In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? has been provided alongside types of In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? theory, EduRev gives you an
ample number of questions to practice In the potentiometer experiment two cells Of emf E1&E2 are used in series & the balance length is found to be 58cm of the wire. If the polarity of E2 is reversed, then the balancing length becomes 29cm. The ratio E1/E2 Of the emf of the two cells is? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup