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Electric Charges And Fields

Question
CBSEENPH12040053
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A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is

  • 1.0 Ω

  • 0.5 Ω

  • 2.0 Ω

  • zero

Solution

A.

1.0 Ω

This is a problem is a base on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential differences, then
E/V = l1/l2
where l1 and l2 are the length of potentiometer wire with and without short-circuited through a resistance

Since comma space straight E over straight V space equals space fraction numerator straight R plus straight r over denominator straight R end fraction space left square bracket because space straight E space equals straight I space left parenthesis straight R plus straight r right parenthesis space and space straight V space equals IR right square bracket therefore comma space fraction numerator straight R plus straight r over denominator straight R end fraction space equals straight l subscript 1 over straight l subscript 2 or space 1 space plus space straight r over straight R space equals space 110 over 100 straight r over straight R space equals space 10 over 100 straight r equals space 1 over 10 space straight x space 10 space equals space 1 straight capital omega

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point? 

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