Question

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \text{ m}$ when shunted by $3 \; \Omega$ resistance and at a length $1.5 \text{ m}$, when cell is shunted by $6 \; \Omega$ resistance. The internal resistance of the cell is

• A. $1 \; \Omega$
• B. $4 \; \Omega$
• C. $2 \; \Omega$
• D. $6 \; \Omega$

Hint:

To verify your solution rapidly without recalculating, find $l_1$. For $r=6 \; \Omega$ and $R=3 \; \Omega$, $l_1 = 1 \cdot (\frac{6}{3}+1) = 3 \text{ m}$. For the second shunt $R'=6 \; \Omega$, $l_1 = 1.5 \cdot (\frac{6}{6}+1) = 3 \text{ m}$. Since $l_1$ matches perfectly, your answer is verified.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem details a potentiometer setup used to measure the internal resistance ($r$) of an electrochemical cell. We are given the balancing lengths on the potentiometer wire for two different external shunt resistances connected across the cell.

Step 2: Key Formula or Approach:
The internal resistance $r$ of a cell measured via a potentiometer is given by:
$$r = R \cdot \left(\frac{l_1}{l_2} - 1\right)$$ where $l_1$ is the open-circuit balancing length (no shunt connected), $l_2$ is the balancing length with shunt resistance $R$.
Rearranging this formula gives a constant expression for $l_1$:
$$\frac{r}{R} + 1 = \frac{l_1}{l_2} \implies l_1 = l_2 \cdot \left(\frac{r}{R} + 1\right)$$

Step 3: Detailed Explanation:
Let's plug in our two experimental conditions:
1.

Case 1: $R = 3 \; \Omega$, $l_2 = 1 \text{ m}$ $$l_1 = 1 \cdot \left(\frac{r}{3} + 1\right) = \frac{r}{3} + 1$$ 2.

Case 2: $R' = 6 \; \Omega$, $l_2' = 1.5 \text{ m}$ $$l_1 = 1.5 \cdot \left(\frac{r}{6} + 1\right) = \frac{3}{2} \cdot \left(\frac{r + 6}{6}\right) = \frac{r + 6}{4}$$ Since the open-circuit length $l_1$ remains constant in both cases, equate the two expressions:
$$\frac{r}{3} + 1 = \frac{r + 6}{4}$$ $$\frac{r + 3}{3} = \frac{r + 6}{4}$$ Cross-multiplying to eliminate fractions:
$$4(r + 3) = 3(r + 6)$$ $$4r + 12 = 3r + 18$$ Subtract $3r$ and $12$ from both sides to isolate $r$:
$$r = 18 - 12 = 6 \; \Omega$$

Step 4: Final Answer:
The internal resistance of the cell is $6 \; \Omega$, which matches option (D).