Given: emf,\(E=12\;V\)
Internal resistance \(r=0.4 \ \Omega\)
Output voltage,
\(V=E-Ir\)
\(r=0.4\; \Omega\)
and for maximum output voltage external resistance, \(R\) will be \(0\).
As per Ohm’s Law,
\(V=\mathit{IR}\)
On substituting the given values, we have
\(E=I(R+r)\)
\(I=\frac E{R+r}\)
\(I=\frac{12}{0+0.4}\)
\(I=30A\)
Hence, the maximum current that can be drawn from the battery is\(30\;A\).