Q. The potential difference between the terminals of a cell in open circuit is 2.2V with resistance of 5$\ \Omega$ across the terminals of a cell, the terminal potential difference is 1.8V. the resistance of the cell is
A) $\frac{9}{10}\Omega$
B) $\frac{10}{9}\Omega$
C) $\frac{7}{12}\Omega$
D) $\frac{12}{7}\Omega$
Explanation
(b) Given, $E = 2.2V,\ \ R = 5\Omega$
$V = 1.8V,\ r = ?$
Now current in the circuit
$i = \frac{E}{R + r}$
$or\ \ \ \frac{V}{R} = \frac{E}{R + r}$
$\ \ \ \ \ \frac{1.8}{5} = \frac{2.2}{5 + r}$
$or\ \ \ r = \frac{10}{9}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.