c
$Emf$ of the battery, $E =12\, V$
Internal resistance of the battery, $r=0.4\, \Omega$
Maximum current drawn from the battery $=I$
According to Ohm's law, $E=I r$
$I=\frac{E}{r}$
$=\frac{12}{0.4}=30\, A$
The maximum current drawn from the given battery is $30 \;A$.