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.