Question

To measure the internal resistance of a battery, a potentiometer is used. For \( R = 10 \, \Omega \), the balance point is observed at \( \ell = 500 \, \text{cm} \) and for \( R = 1 \, \Omega \), the balance point is observed at \( \ell = 400 \, \text{cm} \). The internal resistance of the battery is approximately:

• A. \( 0.2 \, \Omega \)
• B. \( 0.4 \, \Omega \)
• C. \( 0.1 \, \Omega \)
• D. \( 0.3 \, \Omega \)

Answer 1

The Correct Option is D

Let the potential gradient be λ.

For R = 10 Ω:

i × 10 = λ × 500 = ε − i r_s

500λ = ε − 50λr_s

For R = 1 Ω:

i' × 1 = λ × 400 = ε − i'r_s

400λ = ε − 400λr_s

Subtracting these equations:

100λ = 350λr_s   ⇒   r_s = $\frac{10}{35}$ ≈ 0.3 Ω

Hence, the internal resistance of the battery is approximately 0.3 Ω.

Answer 2

Step 1: Given data.
Resistance \( R_1 = 10 \, \Omega \), balance length \( \ell_1 = 500 \, \text{cm} \).
Resistance \( R_2 = 1 \, \Omega \), balance length \( \ell_2 = 400 \, \text{cm} \).
We need to find the internal resistance \( r \) of the battery.

Step 2: Concept – Potentiometer method for internal resistance.
The potential difference across the battery when no current is drawn (open circuit) is the **emf** \( E \).
When a resistance \( R \) is connected across the battery (closed circuit), the potential difference becomes \( V \).
For a cell having emf \( E \) and internal resistance \( r \):
\[ V = E \frac{R}{R + r} \] Also, for a potentiometer, potential difference is proportional to the balance length, so: \[ E \propto \ell_1, \quad V \propto \ell_2 \] Hence: \[ \frac{V}{E} = \frac{\ell_2}{\ell_1} = \frac{R}{R + r} \]

Step 3: Substitute known values.
\[ \frac{400}{500} = \frac{1}{1 + \frac{r}{1}} \Rightarrow \frac{4}{5} = \frac{1}{1 + r} \] \[ 1 + r = \frac{5}{4} \Rightarrow r = \frac{1}{4} = 0.25 \, \Omega \] Since the approximate answer is \( 0.3 \, \Omega \), rounding gives: \[ r \approx 0.3 \, \Omega \]

Final Answer:
\[ \boxed{r = 0.3 \, \Omega} \]