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A cell of emf E and internal resistance ...

A cell of emf `E` and internal resistance `r` connected in the secondary gets balanced against length `l` of potentiometer wire. If a resistance `R` is connected in parallel with the cell, then the new balancing length for the cell will be

A

`((R)/(R-r))`

B

`((R-r)/(R))`

C

`((R)/(r))`

D

`((R)/(R+r))l`

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • A cell of emf (E) and internal resistance (r) is balanced across (l) length of potentiometer wire. If another cell of emf 2E and internal resistance (2r) is connected in parallel to the first cell , then the balancing length will be

    A
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    B
    ` 21//3`
    C
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    D
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    A
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    D
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  • is made of uniform material and cross-sectional area, and it has uniform resistance per unit length. The potential gradient depends upon the current in the wire. A potentiometer with a cell of emf 2 V and internal resistance 0.4 Omega is used across the wire AB . A standard cadmium cell of emf 1.02 V gives a balance point at 66 cm length of wire. The standard cell is then replaced by a cell of unknows emf e (internal resistance r ), and the balance. Point found similarly turns out to be 88 cm length of the wire. The length of potentiometer wire AB is 1 m . If the resistance is connected across the cell e , the balancing length will

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