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

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Current Electricity

To measure the internal resistance of a battery, potentiometer is used. For

R = 10Ω

, the balance point is observed at

ℓ = 500 cm

and for

R = 1Ω

the balance point is observed at

ℓ = 400 cm

. The internal resistance of the battery is approximately :

Answer: (D)

With the cell balanced against a resistor

R

, the terminal voltage equals the potential-gradient reading,

V = ϕ ℓ

, and

ε = V (1 + \frac{r}{R})

.

For

R = 10 Ω, ℓ = 500

ε = ϕ (500) (1 + \frac{r}{10})

.
For

R = 1 Ω, ℓ = 400

ε = ϕ (400) (1 + \frac{r}{1})

.
Equating:

500 (1 + \frac{r}{10}) = 400 (1 + r) \Rightarrow 500 + 50 r = 400 + 400 r

100 = 350 r \Rightarrow r = \frac{100}{350} \approx 0.29 \approx 0.3 Ω.

Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option