Question

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of $5\Omega$ the balancing length is $250 \text{ cm}$. When the cell is shunted by $20\Omega$, the balancing length of potentiometer wire is $400 \text{ cm}$. The internal resistance, of the cell is

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

Hint:

Balancing length $l_2$ is always less than $l_1$ because terminal voltage is less than EMF.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A potentiometer balances the potential difference across a cell against a length of its resistive wire.
When a cell is shunted with an external resistor $R$, the potentiometer balances the terminal voltage $V$ of the cell, not its full e.m.f. $E$.
The terminal voltage $V$ depends on the external resistance $R$ and the cell's internal resistance $r$.
Step 2: Key Formula or Approach:
The balancing condition gives terminal voltage $V \propto l$. Let $V = kl$, where $k$ is potential gradient.
Terminal voltage is $V = E \frac{R}{R+r}$.
So, $kl = E \frac{R}{R+r}$.
We can create a ratio of the balancing lengths for two different shunt resistances to eliminate constants $k$ and $E$.
Step 3: Detailed Explanation:
For the first case, shunt resistance $R_1 = 5\Omega$ and balancing length $l_1 = 250 \text{ cm}$. \[ k l_1 = E \frac{R_1}{R_1+r} \implies k(250) = E \frac{5}{5+r} \quad \text{--- (Equation 1)} \] For the second case, shunt resistance $R_2 = 20\Omega$ and balancing length $l_2 = 400 \text{ cm}$. \[ k l_2 = E \frac{R_2}{R_2+r} \implies k(400) = E \frac{20}{20+r} \quad \text{--- (Equation 2)} \] Divide Equation 1 by Equation 2: \[ \frac{250}{400} = \frac{\frac{5}{5+r}}{\frac{20}{20+r}} \] Simplify the fractions: \[ \frac{5}{8} = \frac{5}{20} \cdot \frac{20+r}{5+r} \] \[ \frac{5}{8} = \frac{1}{4} \cdot \frac{20+r}{5+r} \] Multiply both sides by 4 to simplify further: \[ 4 \times \frac{5}{8} = \frac{20+r}{5+r} \] \[ \frac{5}{2} = \frac{20+r}{5+r} \] Cross-multiply to solve for $r$: \[ 5(5+r) = 2(20+r) \] \[ 25 + 5r = 40 + 2r \] Bring the terms involving $r$ to one side and constants to the other: \[ 5r - 2r = 40 - 25 \] \[ 3r = 15 \] \[ r = 5\Omega \] Step 4: Final Answer:
The internal resistance of the cell is $5\Omega$.