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 $30 \;A$.

$(A)$ The current through $P Q$ is zero.
$(B)$ $I_1=3 A$.
$(C)$ The potential at $S$ is less than that at $Q$.
$(D)$ $I _2=2 A$.


