In a potentiometer experiment, the balancing length of 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.The internal resistance of a cell is

Question

In a potentiometer experiment, the balancing length of 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.The internal resistance of a cell is (A) 1.4 Ω (B) 1.6 Ω (C) 0.12 Ω (D) 1.2 Ω

Tardigrade Admin · Accepted Answer

Consider a cell ofe mf E and balancing length

l_{1}

E = k l_{1}

The potential difference is balanced by length

l_{2}

.

V = k l_{2}

The internal resistance of the cell is given by

r = (\frac{E - V}{V}) R = (\frac{E}{V} - 1) R = (\frac{l _{1}}{l _{2}} - 1) R

= (\frac{560}{560 - 60} - 1) 10 = (\frac{56}{50} - 1) 10

= \frac{6}{5} = 1.2Ω

Q. In a potentiometer experiment, the balancing length of 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 $60 c m$ .The internal resistance of a cell is

$1.4 Ω$

$1.6 Ω$

$0.12 Ω$

$1.2 Ω$

Q. In a potentiometer experiment, the balancing length of a cell is 560cm. When an external resistance of 10Ω is connected in parallel to the cell, the balancing length changes by 60cm.The internal resistance of a cell is

1.4Ω

1.6Ω

0.12Ω

1.2Ω

Consider a cell ofe mf E and balancing lengthl1​ E=kl1​ The potential difference is balanced by length l2​. V=kl2​ The internal resistance of the cell is given by r=(VE−V​)R=(VE​−1)R=(l2​l1​​−1)R =(560−60560​−1)10=(5056​−1)10 =56​=1.2Ω

Q. In a potentiometer experiment, the balancing length of 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 $60 c m$ .The internal resistance of a cell is

$1.4 Ω$

$1.6 Ω$

$0.12 Ω$

$1.2 Ω$