Question

A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of $10\, \Omega.$ Its internal resistance is

• A. 2.0 ohm
• B. zero
• C. 1.0 ohm
• D. 0.5 ohm

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of potentiometer and how internal resistance of a cell is measured using it. The potentiometer is a device used to measure the electromotive force (emf) of a cell, potential difference, and internal resistance without drawing any current from the cell.

Given:

• A cell is balanced at 110 cm of the potentiometer wire when it is not short-circuited.
• The same cell is balanced at 100 cm of the potentiometer wire when short-circuited through a resistance of 10\, \Omega.

We have to calculate the internal resistance r of the cell.

Use the formula for internal resistance:

\frac{E}{V} = \frac{L_1}{L_2} = \frac{R + r}{R}

Where:

• E is the emf of the cell.
• V is the potential difference across the potentiometer wire when the circuit is closed.
• L_1 = 110 \, \text{cm} (Length without short circuit)
• L_2 = 100 \, \text{cm} (Length with short circuit)
• R = 10\, \Omega.

Substitute the given values into the formula:

\frac{110}{100} = \frac{10 + r}{10}

Simplifying this equation:

1.1 = 1 + \frac{r}{10}

Subtract 1 from both sides:

0.1 = \frac{r}{10}

Multiplying both sides by 10 gives:

r = 1\, \Omega.

Thus, the internal resistance of the cell is 1.0 ohm. Therefore, the correct answer is:

1.0 ohm