Question

In a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35.0cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0cm , what is the emf of the second cell?

Answer 1

Principle

The working principle of a potentiometer is that the potential drop across any portion of a wire of uniform cross-section is directly proportional to the length of that portion, provided a constant current flows through it.

When we find a balance point for a cell, the EMF of the cell is equal to the potential drop across that length of the potentiometer wire.

Mathematically, this can be expressed as:
E ∝ L
or
E = kL

where:
E is the EMF of the cell.
L is the balancing length on the potentiometer wire.
* k is the potential gradient (potential drop per unit length of the wire), which is constant for a given setup.

Given Data

• EMF of the first cell, E₁ = 1.25 V
• Balancing length for the first cell, L₁ = 35.0 cm
• Balancing length for the second cell, L₂ = 63.0 cm
• EMF of the second cell, E₂ = ?

Solution

Step 1: Set up the equations for both cells.

For the first cell:
E₁ = k L₁
1.25 V = k (35.0 cm) ---(Equation 1)

For the second cell:
E₂ = k L₂
E₂ = k (63.0 cm) ---(Equation 2)

Step 2: Find the ratio of the two equations to eliminate the constant 'k'.

Divide Equation 2 by Equation 1:
E₂ / E₁ = (k L₂) / (k L₁)

The potential gradient 'k' cancels out:
E₂ / E₁ = L₂ / L₁

Step 3: Rearrange the formula to solve for E₂.

E₂ = E₁ * (L₂ / L₁)

Step 4: Substitute the given values and calculate E₂.

E₂ = 1.25 V * (63.0 cm / 35.0 cm)

E₂ = 1.25 * (63 / 35)

E₂ = 1.25 * 1.8

E₂ = 2.25 V

Answer

The EMF of the second cell is 2.25 V.