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

Answer: (D)

$0.3 \, \Omega$

Answer 2

Let the potential gradient be λ.

For R= 10 Ω:

i × 10 = λ × 500 = ε − i rs

500 λ = ε − 50 λ rs

For R= 1 Ω:

i' × 1 = λ × 400 = ε − i' rs

400 λ = ε − 400 λ rs

Subtracting these equations:

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

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