JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Current Electricity

Label the corners $1$ to $6$ with current entering at $1$ and leaving at the opposite corner $4$, and let the central node be $O$; every edge and every spoke has resistance $r$. The mirror symmetry about line $1$--$4$ together with the entry/exit antisymmetry fixes $O$ at the mean potential. Taking $V_O=0$, the symmetric corners satisfy $V_2=V_6=q$, $V_3=V_5=-q$ and $V_1=-V_4=p$. Kirchhoff's current law at corner $2$ (joined to $1$, $3$ and $O$) gives $(q-p)+(q+q)+q=0$, so $p=4q$. The injected current is $I=\frac{p-q}{r}+\frac{p-q}{r}+\frac{p}{r}=\frac{3p-2q}{r}=\frac{10q}{r}$, while $V_1-V_4=2p=8q$. Hence $R_{eq}=\frac{2p}{I}=\frac{8q}{10q/r}=\frac{4}{5}r$. The equivalent resistance between opposite corners is $\frac{4}{5}r$.