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A potentiometer wire of length 10 m and ...

A potentiometer wire of length `10 m` and resistance `30 Omega` is connected in series with a battery of emf `2.5 V`, internal resistance `5 Omega` and an external resistance `R`. If the fall of potential along the potentiometer wire is `50 mu V mm^(-1)`, then the value of `R` is found to be `23 n Omega`. What is `n`?

Text Solution

Verified by Experts

The correct Answer is:
`(5)`
Potential gradient, `k = (I R_(P))/(L) = (E R_(P))/((R + R_(P) + r)L)`
(where `R_(P)` is resistance of the wire)
or `50 xx 10^(-3) = (2.5 xx 30)/((R + 30 + 5) xx 10)` or R = 115 Omega`
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Knowledge Check

  • A potentiometer wire of length 1m and resistance 10 Omega is connected in series with a cell of emf 2V with internal resistance 1 Omega and a resistance box including a resistance R. If potential difference between the ends of the wire is 1 mV, the value of R is

    A
    `20000 Omega`
    B
    `19989 Omega`
    C
    `10000 Omega`
    D
    `9989 Omega`

  • A potentiometer wire has length 10m and resistance 10Omega . It is connected to battery of EMF 11 volt and internal resistance 1Omega , then the potential gradient in the wire is

    A
    `10 V//m`
    B
    `1V//m`
    C
    `0.1 V//m`
    D
    none

  • A potentiometer wire of length 10m and resistance 20Omega is connected in series with another resistance of 80Omega and a battery of emf 4V . The potential gradient on the wire will be (in mV//cm )

    A
    `0.8`
    B
    `0.16`
    C
    `0.2`
    D
    `0.4`