When a 6.0 V battery is being charged by a current of 2.0 A , the...

During the charging of a battery, the terminal potential difference $V$ is given by the equation:

$V = E + Ir$

where $E$ is the emf of the battery, $I$ is the charging current, and $r$ is the internal resistance.

Given values are:
$V = 7.2 \text{ V}$
$E = 6.0 \text{ V}$
$I = 2.0 \text{ A}$

Substituting the given values into the equation:

$7.2 = 6.0 + 2.0r$

$2.0r = 1.2$

$r = \dfrac{1.2}{2.0} = 0.6 \ \Omega$