Question

A potentiometer, a standard cell of emf 5v and of negligible resistance,maintain a steady current through the potentiometer wire of length 5 m. Two primary cells of emfs e, and e are joined in series with same polarity (i) opposite polarity. The combination is connected through a galvanometer and a jockey to the potentiometer. The balancing length in two cases is found to be 350 cm and 50 cm respectively. Find the value of emfs of the two cells.

Answer 1

Answer:

The value of emfs of the two cells are 2 V and 1.5 V.

Explanation:

Let the emf of the first cell is E1

and the emf of the secod cell is E2

At first we will find the potential gradient of the cell

According to the definition = V/L = 50/.050 = 100 volt per meter

Therefore k = 100v/m

Now l1= the balancing length for the galvanometer.

l1= E1+E2

l2= the balancing length for the jockey.

l2 = E1-E2

As per the problem:

l1 = 350 cm

     = 3.5 m

l2 = 50cm

  = 0.5m

Therefore

E 1+E2 = 3.5............. (1)

E1−E2 = 0.5..............(2)

by solving (1) and (2)

2E1 = 4

E1 = 2 V

E2 = 1.5 V

Therefore,  the emf of the first cell is 2 V and   the emf of the second cell is 1.5 V