Question

state the working of principle of a potentiometer. Draw a circuit diagram to compare the emfs of two primary cell. Derive the formula used​

Answer 1

Explanation:

a) Working principle of potentiometer

When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Application of Potentiometer for comparing emf's of two cells:

The following figure shows an application of the potentiometer to compare the emf of two cell of emf of two cells of emf E

1

andE

2

E

1

,E

2

are the emf of the two cells

1,2,3 form a two way key.

When 1 and 3 are connected E

1

is connected to the galvanometer (G)

Jokey is moved to N

1

which is at a distance li from A to find the balancing length

Applying loop rule to AN

1

G31A

ϕl

1

+0−E

1

=0

Similarly, for E

2

balanced against l

2

(AN)

2

ϕl

2

+0−E

2

=0

E

2

E

1

=

l

2

l

1

---- 1

thus we can compare the emf's of any two sources.Generally one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Equation (1)

(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared .

(ii)The positive ends of all cells are not connected to the same end of the wire .

solution