In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52\, cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \Omega$ , a balance is found when the cell is connected across $40\, cm$ of the wire. Find the internal resistance of the cell.

Question

An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

Answer 1

To solve this problem, we will be using the principle of a potentiometer and the concept of balancing length to calculate the internal resistance of the cell.

In a potentiometer experiment, the balance point is achieved when no current flows through the galvanometer, indicating that the potential difference across the unknown cell is equal to the potential drop across the potentiometer wire.

The initial condition is described as:

E = k \times 52\, \text{cm} (Equation 1)

Where E is the EMF of the cell, k is the potential gradient along the wire.

When the cell is shunted with a resistance R_s = 5 \Omega, the balancing length changes to 40\, \text{cm}:

V_{out} = k \times 40\, \text{cm} (Equation 2)

By definition, the output voltage when the cell is shunted can be expressed using the internal resistance r:

V_{out} = \frac{E}{1 + \frac{r}{R_s}}

Equating the potential drop (from Equation 2) with this expression:

k \times 40 = \frac{E}{1 + \frac{r}{5}}

Substitute Equation 1 (E = k \times 52):

k \times 40 = \frac{k \times 52}{1 + \frac{r}{5}}

By cancelling k and simplifying:

40 \times (1 + \frac{r}{5}) = 52

Expand and solve for r:

40 + \frac{40r}{5} = 52
40 + 8r = 52
8r = 12
r = \frac{12}{8} = 1.5 \Omega

This calculation reveals that the internal resistance of the cell is 1.5 \Omega. Hence, the correct option is:

1.5 \Omega

Answer 2