Calculate battery current and equivalent resistance of the network shown in figure.

In the given circuit, all the resistors are in parallel across the cell.

$\frac{1}{\mathrm{R}}=\frac{1}{8}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\Rightarrow \mathrm{R}=\frac{48}{30}=8/5\mathrm{\Omega }$

$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{24}{8/5}=15\mathrm{A}$