Question

For a cell terminal potential difference is $2.2\text{V}$ when circuit is open and reduces to $1.8\text{V}$ when cell is connected to a resistance of $R=5\Omega ,$ then determine internal resistance of cell.

• A. $\frac{10}{9}\Omega$
• B. $\frac{9}{10}\Omega$
• C. $\frac{11}{9}\Omega$
• D. $\frac{5}{9}\Omega$

Answer 1

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

Emf of the cell is equal to open terminal potential difference.
$\epsilon =2.2\text{V}$

current in the circuit, $i=\frac{2.2}{5+r}$
Potential difference across battery is equal to potential difference across SR resistance.
$\therefore V=iR$
$\Rightarrow 1.8=\left(\frac{2.2}{5+r}\right)5$
$\Rightarrow 9+1.8r=11$
$\Rightarrow 1.8r=2$
$\therefore r=\frac{10}{9}\Omega$