Question

With a certain cell, the balance point is obtained at 65 cm from the end of the potentiometer wire. With another cell whose e.m.f differs from that of first by 0.1 V, the balance point is obtained at 60 cm. Find the e.m.f of each cell.

Answer 1

ANSWER

Emf of potentiometer wire, E=2V

Potential of cell, V=1.5V

Balance point obtained when the cell is in open circuit, I

1

=76.3cm

New balance point obtained when a resistor is used in the external circuit, I

2

=64.8cm

External resistor, R=9.5Ω

Now,

r=

V

R(E−V)

=

I

2

R(I

1

−I

2

)

[

V

E

=

I

2

I

1

]

=

64.8

9.5(76.3−64.8)

Ω

=

64.8

9.5×11.5

Ω

=1.686Ω≈1.7Ω