Question

For a cell, the terminal potential difference is $2.2V$ when circuit is open and reduces to $1.8V$ when cell is connected to a resistance $R=5\Omega$ . The internal resistance $\left(\right.r\left.\right)$ of the cell is

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

Answer 1

Answer: (A)

When circuit is open then it would be EMF of cell so
$E=2.2V$
When connected with an external resistance $5\Omega$ we will measure terminal potential difference so

$I=\frac{E}{R + r}$
$V_{A}-V_{B}=E-I\left(\right.r\left.\right)=I\left(\right.R\left.\right)=\frac{E R}{R + r}$ .....(i)
So , $V_{A}-V_{B}=1.8VoltR=5\Omega$
$E=2.2V$
from equation (i)
$1.8=\frac{\left(\right. 2 . 2 \left.\right) \left(\right. 5 \left.\right)}{5 + r}$
$9+1.8r=11$
$1.8r=2$
$r=\frac{20}{18}=\frac{10}{9}\Omega$