Physics - Current Electricity Question with Solution | TestHub

The diagonal wires (with jump-overs at the crossings) short the corners of each cell, so all seven resistors sit between the same two equipotential node groups - they are all in parallel.

Equivalent resistance: $\frac{1}{R}=\frac{1}{8}+\frac{1}{6}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{12}+\frac{1}{8}=1$, so $R_{eq}=1\Omega$ and $i_1=\frac{24}{1}=24\mathrm{A}$ - (A) is correct.

Each resistor carries 24 V, giving branch currents 3, 4, 6, 4, 2, 2, 3 A; applying Kirchhoff's current law along the labelled paths then gives $i_2=2\mathrm{A}$, $i_3=15\mathrm{A}$ and $i_4=6\mathrm{A}$ - (B), (C), (D) all correct.