In a Potentiometer experiment the balancing length with a cell is 560 cm, when an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell

Question

In a Potentiometer experiment the balancing length with a cell is 560 cm, when an external resistance of 10 Ω is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell (A) 1 Ω (B) 2 Ω (C) 1.2 Ω (D) 2.1 Ω

Tardigrade Admin · Accepted Answer

Given, l1=560 cm, R=10 Ω l2=l1-60 ⇒ l2=560-60=500 cm ∴ Internal resistance of cell is given by r =((l1-l2/l2)) R=((560-500/500)) 10 =(60/500) × 10=(6/5)=1.2 Ω

Q. In a Potentiometer experiment the balancing length with a cell is $560 c m$ , when an external resistance of $10 Ω$ is connected in parallel to the cell, the balancing length changes by $60 c m$ . Find the internal resistance of the cell

$1 Ω$

$2 Ω$

$1.2 Ω$

$2.1 Ω$

Given, $l_{1} = 560 c m, R = 10 Ω$
$l_{2} = l_{1} - 60$
$\Rightarrow l_{2} = 560 - 60 = 500 c m$
$∴$ Internal resistance of cell is given by
$r = (\frac{l _{1} - l _{2}}{l _{2}}) R = (\frac{560 - 500}{500}) 10$
$= \frac{60}{500} \times 10 = \frac{6}{5} = 1.2 Ω$

Q. In a Potentiometer experiment the balancing length with a cell is 560cm, when an external resistance of 10Ω is connected in parallel to the cell, the balancing length changes by 60cm. Find the internal resistance of the cell

1Ω

2Ω

1.2Ω

2.1Ω