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 is 4 m long- When the resistance R-9-5 - is connected across a secondary cell the balancing length on the potentiometer wire was found to be 2-85 m- If the balancing length for the emf of the secondary cell is 3 m- the value of internal resistance of the cell is

Given: l_1=3 m, l_2=2.85 m R=9.5 Ω Using the relation, r=(l_1l_2 1)× R r=(32.85 1)× 9.5 ∴ r=0.152.85× 9.5=0.5 Ω Therefore, option (A) is correct. ...

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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 is $4 m$ long. When the resistance $R = 9.5 Ω$ is connected across a secondary cell the balancing length on the potentiometer wire was found to be $2.85 m$ . If the balancing length for the emf of the secondary cell is $3 m$ , the value of internal resistance of the cell is

0.5 Ω

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0.95 Ω

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0.25 Ω

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0.75 Ω

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Solution

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

Using the relation,

$r = (\frac{l_{1}}{l_{2}} - 1) \times R$

$r = (\frac{3}{2.85} - 1) \times 9.5$

$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$

Therefore, option $(A)$ is correct.

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The correct option is A 0.5 ΩGiven: l1=3 m, l2=2.85 m R=9.5 Ω Using the relation, r=(l1l2−1)×R r=(32.85−1)×9.5 ∴r=0.152.85×9.5=0.5 Ω Therefore, option (A) is correct.

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

Using the relation,

$r = (\frac{l_{1}}{l_{2}} - 1) \times R$

$r = (\frac{3}{2.85} - 1) \times 9.5$

$∴ r = \frac{0.15}{2.85} \times 9.5 = 0.5 Ω$

Therefore, option $(A)$ is correct.