A cell, shunted by a 8 Ω resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 Ω resistance. The value of internal resistance of the cell will be Ω.

Question

A cell, shunted by a 8 Ω resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 Ω resistance. The value of internal resistance of the cell will be Ω. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(V1/V2)=(3/2)=(E-i1 r/E-i2 r) =(E-(E/8+r) × r/E-(E)4+r × r) (3/2)=(8(4+r)/4(8+r)) 24+3 r=16+4 r r=8 Ω

Q. A cell, shunted by a $8 Ω$ resistance, is balanced across a potentiometer wire of length $3 m$ . The balancing length is $2 m$ when the cell is shunted by $4 Ω$ resistance. The value of internal resistance of the cell will be _______ $Ω$ .

$\frac{V _{1}}{V _{2}} = \frac{3}{2} = \frac{E - i _{1} r}{E - i _{2} r}$
$= \frac{E - \frac{E}{8 + r} \times r}{E - \frac{E}{4 + r} \times r}$
$\frac{3}{2} = \frac{8 ( 4 + r )}{4 ( 8 + r )}$
$24 + 3 r = 16 + 4 r$
$r = 8 Ω$

Q. A cell, shunted by a 8Ω resistance, is balanced across a potentiometer wire of length 3m. The balancing length is 2m when the cell is shunted by 4Ω resistance. The value of internal resistance of the cell will be _______Ω.

V2​V1​​=23​=E−i2​rE−i1​r​ =E−4+rE​×rE−8+rE​×r​ 23​=4(8+r)8(4+r)​ 24+3r=16+4r r=8Ω

Q. A cell, shunted by a $8 Ω$ resistance, is balanced across a potentiometer wire of length $3 m$ . The balancing length is $2 m$ when the cell is shunted by $4 Ω$ resistance. The value of internal resistance of the cell will be _______ $Ω$ .

$\frac{V _{1}}{V _{2}} = \frac{3}{2} = \frac{E - i _{1} r}{E - i _{2} r}$
$= \frac{E - \frac{E}{8 + r} \times r}{E - \frac{E}{4 + r} \times r}$
$\frac{3}{2} = \frac{8 ( 4 + r )}{4 ( 8 + r )}$
$24 + 3 r = 16 + 4 r$
$r = 8 Ω$