JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Current Electricity

By the symmetry of the six-resistor (2Ω each) network in the figure, the arrangement reduces to an equivalent external resistance of $\frac{2}{3}\Omega$. Adding the battery's internal resistance of $\frac{2}{3}\Omega$ in series gives a total circuit resistance

$R_{eq}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\Omega$

The total power delivered by the cell is

$P=\frac{{\epsilon }^2}{R_{eq}}=\frac{(2{)}^2}{4/3}=3\text{W}$