Question 325 of 380
To measure the internal resistance of a battery, potentiometer is used. For , the balance point is observed at and for the balance point is observed at . The internal resistance of the battery is approximately :
Show answer and solution
Answer: (B)
Let potential gradient be .
$\begin{aligned}
& \therefore \mathrm{i} \times 10=\lambda \times 500=\varepsilon-\mathrm{ir}_{\mathrm{s}} \\
& \Rightarrow 500 \lambda=\varepsilon-50 \lambda \mathrm{r}_{\mathrm{s}}
\end{aligned}$
Also,
$\begin{aligned}
& \mathrm{i}^{\prime} \times 1=\lambda \times 400=\varepsilon-\mathrm{i}^{\prime} \mathrm{r}_{\mathrm{s}} \\
& \Rightarrow 400 \lambda=\varepsilon-400 \lambda \mathrm{r}_5 \\
& \therefore 100 \lambda=350 \lambda \mathrm{r}_{\mathrm{s}} \Rightarrow \mathrm{r}_{\mathrm{s}}=\frac{10}{35} \approx 0.3 \Omega
\end{aligned}$
$\begin{aligned}
& \therefore \mathrm{i} \times 10=\lambda \times 500=\varepsilon-\mathrm{ir}_{\mathrm{s}} \\
& \Rightarrow 500 \lambda=\varepsilon-50 \lambda \mathrm{r}_{\mathrm{s}}
\end{aligned}$
Also,
$\begin{aligned}
& \mathrm{i}^{\prime} \times 1=\lambda \times 400=\varepsilon-\mathrm{i}^{\prime} \mathrm{r}_{\mathrm{s}} \\
& \Rightarrow 400 \lambda=\varepsilon-400 \lambda \mathrm{r}_5 \\
& \therefore 100 \lambda=350 \lambda \mathrm{r}_{\mathrm{s}} \Rightarrow \mathrm{r}_{\mathrm{s}}=\frac{10}{35} \approx 0.3 \Omega
\end{aligned}$