Question

In a potentiometer experiment, a wire of length 10 m and resistance $5\,\Omega$ is connected to a cell of emf 2.2 V. If the potential difference between two points separated by a distance of 660 cm on potentiometer wire is 1.1 V, then the internal resistance of the cell is

• A. $1.6\,\Omega$
• B. $1.4\,\Omega$
• C. $1.2\,\Omega$
• D. $1\,\Omega$

Hint:

Potential Gradient $k = \frac{V_{wire}}{L} = \left( \frac{E}{R_{wire} + r} \right) \frac{R_{wire}}{L}$. Use the data given for a segment to find $k$ first.

Answer 1

The Correct Option is A

Step 1: Calculate Potential Gradient ($k$):
We are given that the potential difference across a length $l = 660 \text{ cm} = 6.6 \text{ m}$ is $V = 1.1 \text{ V}$. Potential gradient $k = \frac{V}{l} = \frac{1.1}{6.6} = \frac{1}{6} \text{ V/m}$.
Step 2: Calculate Current in the Wire ($I$):
The resistance of the potentiometer wire is $R_p = 5\,\Omega$ and length $L = 10 \text{ m}$. Resistance per unit length $\lambda = \frac{5}{10} = 0.5\,\Omega/\text{m}$. Since $k = I \lambda$, we have: \[ \frac{1}{6} = I \times 0.5 \implies I = \frac{1}{6 \times 0.5} = \frac{1}{3} \text{ A} \]
Step 3: Calculate Internal Resistance ($r$):
The current $I$ is supplied by the cell of emf $E = 2.2 \text{ V}$ connected in series with the wire. Let $r$ be the internal resistance. Total resistance of the circuit = $R_p + r = 5 + r$. Using Ohm's law: $I = \frac{E}{R_p + r}$. \[ \frac{1}{3} = \frac{2.2}{5 + r} \] \[ 5 + r = 6.6 \] \[ r = 6.6 - 5 = 1.6\,\Omega \]