Question

The potential difference between the terminals of a cell in an open circuit is 2.2V. When a resistor of $5\Omega$ is connected across the terminals of the cell, the potential difference between the terminals of the cell is found to be 1.8V. The internal resistance of the cell is

• A. $\frac{7}{12}\Omega$
• B. $\frac{10}{9}\Omega$
• C. $\frac{9}{10}\Omega$
• D. $\frac{12}{7}\Omega$

Answer 1

The correct option is B $\frac{10}{9}\Omega$

Potential difference across the cell in open circuit is the emf of the cell.

Hence, emf $E=2.2V$

When, circuit is closed, potential difference across cell is given by $V=E-Ir$

And,$I=\frac{E}{R+r}$

Hence, $V=E-\frac{Er}{R+r}$

$⟹V=\frac{ER}{R+r}$

$⟹1.8=\frac{2.2\times 5}{5+r}$

$⟹9+1.8r=11$

$⟹r=\frac{2}{1.8}\Omega$

$⟹r=\frac{10}{9}\Omega$

Answer-(B)