Q120 / 200 · NEET 2002 · Physics
For a cell, terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open. If it reduces to $1.8 \mathrm{~V}$ when the cell is connected to a resistance of $5 \Omega$. The internal resistance of cell $(r)$ is then:
- Correct answerA$\frac{10}{9} \Omega$
- B$\frac{9}{10} \Omega$
- C$\frac{11}{9} \Omega$
- D$\frac{5}{9} \Omega$
Solution
Terminal potential difference
$$
\begin{aligned}
& V=E-I r \\
& \therefore V=E-\left[\frac{E}{R-r}\right] r=\frac{E R}{R+r}
\end{aligned}
$$
From given condition $=E=2.2$ and where $R=5$ then potential $V=1.8 \mathrm{~V}$ Therefore $1.8=\frac{2.2 \times 5}{5+r}$
$\Rightarrow r=\frac{10}{9} \Omega$
Hence, the correct answer is option: 'A'.