The figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 2.5 V cell. The balance point of the cell in the open circuit is 75 cm. When a resistor of 10 Ω is used in the external circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. The internal resistance of the cell is:

Question

The figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 2.5 V cell. The balance point of the cell in the open circuit is 75 cm. When a resistor of 10 Ω is used in the external circuit of the cell, the balance point shifts to 65 cm length of potentiometer wire. The internal resistance of the cell is: (A)  2.5 Ω  (B)  2.0 Ω  (C)  1.54 Ω  (D)  1.0 Ω

Tardigrade Admin · Accepted Answer

For a potentiometer, the internal resistance (r) is given by r=R((l1/l2)-1) Given, R=10 Ω, l1=75 cm , l2=65 cm ∴ r=10((75/65)-1) =10 × 0.154=1.54 Ω

$2.5 Ω$

$2.0 Ω$

$1.54 Ω$

$1.0 Ω$

For a potentiometer, the internal resistance $(r)$ is given by
$r = R (\frac{l _{1}}{l _{2}} - 1)$
Given, $R = 10 Ω, l_{1} = 75 c m, l_{2} = 65 c m$
$∴ r = 10 (\frac{75}{65} - 1)$
$= 10 \times 0.154 = 1.54 Ω$

2.5Ω

2.0Ω

1.54Ω

1.0Ω

For a potentiometer, the internal resistance (r) is given by r=R(l2​l1​​−1) Given, R=10Ω,l1​=75cm,l2​=65cm ∴r=10(6575​−1) =10×0.154=1.54Ω

$2.5 Ω$

$2.0 Ω$

$1.54 Ω$

$1.0 Ω$

For a potentiometer, the internal resistance $(r)$ is given by
$r = R (\frac{l _{1}}{l _{2}} - 1)$
Given, $R = 10 Ω, l_{1} = 75 c m, l_{2} = 65 c m$
$∴ r = 10 (\frac{75}{65} - 1)$
$= 10 \times 0.154 = 1.54 Ω$