In the potentiometer arrangement shown, the driving cell A has emf ε and internal resistance r. The emf of the cell. B is (ε/2) and internal resistance 2 r. The potentiometer wire C D is 100 cm long. If balance is obtained with length C J=l, then

Question

In the potentiometer arrangement shown, the driving cell A has emf ε and internal resistance r. The emf of the cell. B is (ε/2) and internal resistance 2 r. The potentiometer wire C D is 100 cm long. If balance is obtained with length C J=l, then (A) l = 50 cm (B) l > 50 cm (C) l < 50 cm (D) Balance cannot be obtained

Tardigrade Admin · Accepted Answer

As potential gradient k in the potentiometer wire will be less than (ε / 100)( V / cm ) because of presence of r, so for the battery of emf (ε / 2) the balance point l will be greater than 50 cm as shown. l=(ε / 2/k( < ε / 100)) ⇒ l > 50 cm

Q. In the potentiometer arrangement shown, the driving cell $A$ has emf $ε$ and internal resistance $r$ . The emf of the cell. $B$ is $\frac{ε}{2}$ and internal resistance $2 r$ . The potentiometer wire $C D$ is $100 c m$ long. If balance is obtained with length $C J = l$ , then

$l = 50 c m$

$l > 50 c m$

$l < 50 c m$

Balance cannot be obtained

As potential gradient $k$ in the potentiometer wire will be less than $(ε /100) (V / c m)$ because of presence of $r$ , so for the battery of emf $(ε /2)$ the balance point $l$ will be greater than $50 c m$ as shown.
$l = \frac{ε /2}{k ( < ε /100 )} \Rightarrow l > 50 c m$

Q. In the potentiometer arrangement shown, the driving cell A has emf ε and internal resistance r. The emf of the cell. B is 2ε​ and internal resistance 2r. The potentiometer wire CD is 100cm long. If balance is obtained with length CJ=l, then

l=50cm

l>50cm

l<50cm