A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ϵ and internal resistance r . A cell C having EMF ϵ /2 and internal resistance 3r is connected. The length AJ , at which the galvanometer, as shown in the figure, shows no deflection is

Question

A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ϵ and internal resistance r . A cell C having EMF ϵ /2 and internal resistance 3r is connected. The length AJ , at which the galvanometer, as shown in the figure, shows no deflection is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8bugwncznmothb3i.jpg" /> (A) (5/12)L (B) (11/24)L (C) (11/12)L (D) (13/24)L

Tardigrade Admin · Accepted Answer

Potential difference across AJ=IRA J , =(ϵ /13 r)⋅ ((12 r/L) ⋅ x) For no current in galvanometer, (ϵ /2)= potential difference across AJ (ϵ /2)=(12 ϵ /13 L)x x=(13L/24)

$\frac{5}{12} L$

$\frac{11}{24} L$

$\frac{11}{12} L$

$\frac{13}{24} L$

Potential difference across $A J = I R_{A J}$ ,
$= \frac{ϵ}{13 r} \cdot (\frac{12 r}{L} \cdot x)$
For no current in galvanometer,
$\frac{ϵ}{2} =$ potential difference across $A J$
$\frac{ϵ}{2} = \frac{12 ϵ}{13 L} x$
$x = \frac{13 L}{24}$

Q. A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ϵ and internal resistance r . A cell C having EMF ϵ/2 and internal resistance 3r is connected. The length AJ , at which the galvanometer, as shown in the figure, shows no deflection is

125​L

2411​L

1211​L

2413​L

Potential difference across AJ=IRAJ​ , =13rϵ​⋅(L12r​⋅x) For no current in galvanometer, 2ϵ​= potential difference across AJ 2ϵ​=13L12ϵ​x x=2413L​

$\frac{5}{12} L$

$\frac{11}{24} L$

$\frac{11}{12} L$

$\frac{13}{24} L$

Potential difference across $A J = I R_{A J}$ ,
$= \frac{ϵ}{13 r} \cdot (\frac{12 r}{L} \cdot x)$
For no current in galvanometer,
$\frac{ϵ}{2} =$ potential difference across $A J$
$\frac{ϵ}{2} = \frac{12 ϵ}{13 L} x$
$x = \frac{13 L}{24}$