Question

A null point is obtained at $200\text{ cm}$ on potentiometer wire when cell in secondary circuit is shunted by $5\Omega$. When a resistance of $15\Omega$ is used for shunting, null point moves to $300\text{ cm}$ . The internal resistance of the cell is

• A. $3\Omega$
• B. $4\Omega$
• C. $5\Omega$
• D. $6\Omega$

Hint:

If the shunt resistance triples and the null point increases by 1.5x, the internal resistance often equals the initial shunt.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A potentiometer can be used to determine the internal resistance of a primary cell.
When the cell is shunted by a resistance $R$, the balancing length $l$ corresponds to the terminal potential difference $V$ across the shunt.
The e.m.f. $E$ of the cell corresponds to the balancing length $l_0$ when it is not shunted (open circuit).
Step 2: Key Formula or Approach:
The formula for the internal resistance $r$ is:
\[ r = R \left( \frac{l_0 - l}{l} \right) \]
Since we have two different shunting cases, we can set up two equations with the same unknown $l_0$.
For case 1: $r = R_1 \left( \frac{l_0 - l_1}{l_1} \right)$
For case 2: $r = R_2 \left( \frac{l_0 - l_2}{l_2} \right)$
Equating them gives: $R_1 \left( \frac{l_0 - l_1}{l_1} \right) = R_2 \left( \frac{l_0 - l_2}{l_2} \right)$.
Step 3: Detailed Explanation:
Given values:
$R_1 = 5\ \Omega$, $l_1 = 200\text{ cm}$
$R_2 = 15\ \Omega$, $l_2 = 300\text{ cm}$
Substitute these into the equated formula to find $l_0$:
\[ 5 \left( \frac{l_0 - 200}{200} \right) = 15 \left( \frac{l_0 - 300}{300} \right) \]
Simplify the fractions:
\[ \frac{l_0 - 200}{40} = \frac{l_0 - 300}{20} \]
Multiply both sides by 40:
\[ l_0 - 200 = 2(l_0 - 300) \]
\[ l_0 - 200 = 2l_0 - 600 \]
Rearrange to solve for $l_0$:
\[ 600 - 200 = 2l_0 - l_0 \]
\[ l_0 = 400\text{ cm} \]
Now, substitute $l_0$ back into either of the internal resistance equations (let's use case 1):
\[ r = 5 \left( \frac{400 - 200}{200} \right) \]
\[ r = 5 \left( \frac{200}{200} \right) \]
\[ r = 5 \times 1 = 5\ \Omega \]
Step 4: Final Answer:
The internal resistance of the cell is $5\Omega$.