Current Electricity - Result Question 98

103. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire, has an emfof $2.0 V$ and a negligible internal resistance. The potentiometer wire itself is $4 m$ long, When the resistace $R$, connected across the given cell, has values of

(i) infinity (ii) $9.5 \Omega$

The balancing lengths’, on the potentiometer wire are found to be $3 m$ and $2.85 m$, respectively. The value of internal resistance of the cell is

[2014]

(a) $0.25 \Omega$

(b) $0.95 \Omega$

(c) $0.5 \Omega$

(d) $0.75 \Omega$

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Answer:

Correct Answer: 103. (c)

Solution:

  1. (c) Internal resistance of the cell,

$ \begin{aligned} & r=(\frac{E-V}{V}) R=(\frac{\ell_1-\ell_2}{\ell_2}) R \\ & =(\frac{3-2.85}{2.85}) \times(9.5) \Omega=0.5 \Omega \end{aligned} $