Question

The emf of a cell is balanced at 52 cm length of the potentiometer wire. If 5 ohm resistance is inserted from
the resistance box connected with the cell, then balancing length obtained is 40 cm The internal resistance of
the cell will be​

Answer 1

Answer:

1.5 Ω

Given,

R=5Ω

l

1

=52cm,l

2

=40cm

To find,

Internal resistance of cell, r=?

r=R(

l

2

l

1

−1)=5(

40

52

−1)=5(

40

52−40

)=5(

40

12

)=5(

10

3

)=

10

15

=1.5Ω

⇒r=1.5Ω

Therefore, the internal resistance of the cell is 1.5Ω

Answer 2

The internal resistance of the cell comes out to be 1.5 ohm.

Step-by-step procedure

Given:

The balancing length at 5 ohms resistance; $L_{1}$= 52cm

The resistance; R= 5 ohm

When the resistance of 5 ohms is inserted, the balancing length becomes; $L_{2}$= 40cm

To find: Internal resistance

Solution:

We know from the formula to calculate the internal resistance of a cell:

r= R($L_{1}$ ÷$L_{2}$ -1)

r= 5(52÷40 - 1)

r= 5(52-40÷40)

r= 5(12÷40)

r= 15÷10

Thus, the internal resistance; r= 1.5 ohm

Result:

The internal resistance of the cell comes out to be 1.5 ohm.

(#SPJ3)