Question

In a potentiometer circuit a cell of EMF $1.5\, V$ gives balance point at $36\, cm$ length of wire. If another cell of EMF $2.5\, V$ replaces the first cell, then at what length of the wire, the balance point occurs ?

• A. 60 cm
• B. 21.6 cm
• C. 64 cm
• D. 62 cm

Answer 1

The Correct Option is A

To solve this problem, we need to understand the working principle of a potentiometer. The potentiometer compares the EMF of a test cell against a known potential difference, not drawing any current from the test cell. The balance point in a potentiometer occurs when the potential drop across the length of the wire is equal to the EMF of the cell.

The formula used in a potentiometer is:

\(V = \frac{E}{L}\cdot x\)

Where:

• \(V\) is the potential difference across the wire at the balance point.
• \(E\) is the EMF of the cell.
• \(L\) is the total length of the wire.
• \(x\) is the balancing length for the given EMF.

Given:

• The EMF of the first cell, \(E_1 = 1.5 \, V\)
• The balance point for the first cell, \(x_1 = 36\, cm\)
• The EMF of the second cell, \(E_2 = 2.5 \, V\)

We need to find the balancing length \(x_2\) for the second cell:

Since the same wire and similar connections are used, the potential gradient \(\left(\frac{V}{L}\right)\) remains constant. Hence, we can write:

\(\frac{E_1}{x_1} = \frac{E_2}{x_2}\)

Substitute the given values into the above equation:

\(\frac{1.5}{36} = \frac{2.5}{x_2}\)

Solve for \(x_2\):

\(x_2 = \frac{2.5 \times 36}{1.5}\)

\(x_2 = 60 \, cm\)

Thus, the balance point for the second cell of EMF \(2.5 \, V\) occurs at \(60 \, cm\) of the wire.