Question

A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf's is :
A) 3:4
B) 5:4
C) 5:1
D) 3:2

Answer 1

D) 3:2

In series : E1 + E2 = K(50)

In parallel : E1 - E2 = K(10)

E1 + E2 / E1 - E2 = 5/1

=> E1/E2 = 3/2

The emf's are in the ratio of 3:2.

Answer 2

Answer: (d). The ratio of emf is 3:2.

Explanation:

Given that,

Length l = 100 cm

The

The emf when the two cells are connected in series then the balance point is 50 cm.

$E_{1}+E_{2}=50$....(I)

The emf when the two cells are connected in opposite direction then the balance point is 10 cm.

$E_{1}-E_{2}=10$....(II)

From equation (I) and (II)

Using elimination method we get,

$E_{1}= 30$

$E_{2}= 20$

Now, we will find the ratio of E₁ and E₂

$\dfrac{E_{1}}{E_{2}}=\dfrac{3}{2}$

Hence, The ratio of emf is 3:2.