Current Electricity Ques 115
115. 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: 115.(c)
Solution:
- (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} $
BETA
