Twelve wires each having resistance $2 \Omega$ are joined to form a cube. A battery of $6 \mathrm{~V}$ emf is joined across point $a$ and $c$. The voltage difference between $e$ and $f$ is ______V.

Question

Aicharya · Accepted Answer

From symmetry, current through e-b & g-d =0 $\begin{aligned} & \therefore \mathrm{R}_{\text {eq }}=\frac{3}{4} \times \mathrm{R}=\frac{3}{2} \Omega \\ & \therefore \text { Current through battery }=\frac{6 \times 2}{3}=4 \mathrm{~A} \\ & \mathrm{i}_2=\frac{4}{8} \times 2=1 \mathrm{~A} \\ & \therefore \Delta \mathrm{V} \text { across e-f }=\frac{\mathrm{i}_2}{2} \times \mathrm{R}=\frac{1}{2} \times 2=1 \mathrm{~V} \end{aligned}$

Twelve wires each having resistance $2 \Omega$ are joined to form a cube. A battery of $6 \mathrm{~V}$ emf…

Solution