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Standard XII

Physics

Electric Bulbs

A potentiomet...

Question

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 emf of $2 V$ and a negligible internal resistance. The potentiometer wire itself is $4 m$ long. When the resistance $R$ is connected across the given cell has value of $(i)$ infinity, $(i i) 9.5 Ω$ the balancing length on the potentiometer wire are found to be $3 m$ and $2.85 m$ respectively. The value of internal resistance of the cell is

0.25 Ω

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0.95 Ω

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0.5 Ω

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0.75 Ω

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Solution

The correct option is C $0.5 Ω$
Given,
$l_{1} = 3 m,$
$l_{2} = 2.85 m$ ,
$R = 9.5 Ω$

Using, $r = (\frac{l_{1}}{l_{2}} - 1) \times R$
$r = (\frac{3}{2.85} - 1) \times 9.5$
$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$

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The correct option is C 0.5 ΩGiven, l1=3 m, l2=2.85 m, R=9.5 Ω Using, r=(l1l2−1)×R r=(32.85−1)×9.5 ∴r=0.152.85×9.5=0.5Ω

The correct option is C $0.5 Ω$
Given,
$l_{1} = 3 m,$
$l_{2} = 2.85 m$ ,
$R = 9.5 Ω$

Using, $r = (\frac{l_{1}}{l_{2}} - 1) \times R$
$r = (\frac{3}{2.85} - 1) \times 9.5$
$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$