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 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity (ii) 9.5 Ω 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

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 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of (i) infinity (ii) 9.5 Ω 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 (A) 0.25 Ω (B) 0.95 Ω (C) 0.5 Ω (D) 0.75 Ω

Tardigrade Admin · Accepted Answer

The internal resistance of the cell is r=((l1/l2)-1) R Here, l1=3 m , l2=2.85 m , R=9.5 Ω ∴ r=((3/2.85)-1)(9.5 Ω) =(0.15/2.85) × 9.5 Ω=0.5 Ω

$0.25 Ω$

$0.95 Ω$

$0.5 Ω$

$0.75 Ω$

The internal resistance of the cell is
$r = (\frac{l _{1}}{l _{2}} - 1) R$
Here, $l_{1} = 3 m, l_{2} = 2.85 m, R = 9.5 Ω$
$∴ r = (\frac{3}{2.85} - 1) (9.5Ω)$
$= \frac{0.15}{2.85} \times 9.5 Ω = 0.5 Ω$

0.25Ω

0.95Ω

0.5Ω

0.75Ω

The internal resistance of the cell is r=(l2​l1​​−1)R Here, l1​=3m,l2​=2.85m,R=9.5Ω ∴r=(2.853​−1)(9.5Ω) =2.850.15​×9.5Ω=0.5Ω

$0.25 Ω$

$0.95 Ω$

$0.5 Ω$

$0.75 Ω$

The internal resistance of the cell is
$r = (\frac{l _{1}}{l _{2}} - 1) R$
Here, $l_{1} = 3 m, l_{2} = 2.85 m, R = 9.5 Ω$
$∴ r = (\frac{3}{2.85} - 1) (9.5Ω)$
$= \frac{0.15}{2.85} \times 9.5 Ω = 0.5 Ω$