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A battery is connected to a potentiomete...

A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`.
Calculate the internal resistance of the cell.

A

`4 Omega`

B

`2 Omega`

C

`5 Omega`

D

`1 Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
`r = (l_(1) - l_(2))/(l_(2))R = (84 - 70)/(70) xx 5 = 1 Omega`
`1 = (l_(1) - l_(3))/(l_(3)) xx 4`, put `l_(1) = 84 cm` get `l_(3) = 67.2 cm`
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Knowledge Check

  • A battery is connected to a potentiometer and a balance point is obtained at 84 cm along the wire. When its terminals are connected by a 5 Omega resistor, the balance point changes to 70 cm . Find the new position of the balance point when 5 Omega resistor is changed by 4 Omega resistor.

    A
    `26.5 cm`
    B
    `52 cm`
    C
    `67.2 cm`
    D
    `83.3 cm`

  • For a cell of e.m.f. 2 V , a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2Omega resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is

    A
    `0.25 Omega`
    B
    `0.50 Omega`
    C
    `0.80 Omega`
    D
    `1.00 Omega`

  • The emf of a standard cells is balanced at 150 cm length of a potentiometer wire. When this cell is shunted by a 2 ohm resistance, the null point is obtained at 100 cm. the value of internal reistance of the cell is :

    A
    0.1ohm
    B
    1ohm
    C
    2ohm
    D
    0.5ohm