In the given potentiometer circuit arrangement, the balancing length A C is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, (ε1/ε2) is :

Question

In the given potentiometer circuit arrangement, the balancing length A C is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, (ε1/ε2) is : (A) (5/3) (B) (8/5) (C) (4/3) (D) (3/5)

Tardigrade Admin · Accepted Answer

E1=k l1 ...(i) E1+E2=k l2 ... (ii) (E1/E1+E2)=(l1/l2)=(250/400)=(5/8) 8 E1=5 E1+5 E2 3 E1=5 E2 (E1/E2)=(5/3)

$\frac{5}{3}$

$\frac{8}{5}$

$\frac{4}{3}$

$\frac{3}{5}$

$E_{1} = k l_{1}$ ...(i)
$E_{1} + E_{2} = k l_{2}$ ... (ii)
$\frac{E _{1}}{E _{1} + E _{2}} = \frac{l _{1}}{l _{2}} = \frac{250}{400} = \frac{5}{8}$
$8 E_{1} = 5 E_{1} + 5 E_{2}$
$3 E_{1} = 5 E_{2}$
$\frac{E _{1}}{E _{2}} = \frac{5}{3}$

Q. In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400cm. The ratio of the emf of two cells, ε2​ε1​​ is :

35​

58​

34​

53​

E1​=kl1​ ...(i) E1​+E2​=kl2​ ... (ii) E1​+E2​E1​​=l2​l1​​=400250​=85​ 8E1​=5E1​+5E2​ 3E1​=5E2​ E2​E1​​=35​

$\frac{5}{3}$

$\frac{8}{5}$

$\frac{4}{3}$

$\frac{3}{5}$

$E_{1} = k l_{1}$ ...(i)
$E_{1} + E_{2} = k l_{2}$ ... (ii)
$\frac{E _{1}}{E _{1} + E _{2}} = \frac{l _{1}}{l _{2}} = \frac{250}{400} = \frac{5}{8}$
$8 E_{1} = 5 E_{1} + 5 E_{2}$
$3 E_{1} = 5 E_{2}$
$\frac{E _{1}}{E _{2}} = \frac{5}{3}$