Assuming all bulbs are identical, rank the brightnesses of the bulbs, from brightest to dimmest.

 

Solution:

For the branch containing A, B and C.

${\mathrm{R}}_{eq}=\frac{3R}{2}$

Current in A : ${\mathrm{I}}_1=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{2\mathrm{E}}{3\mathrm{R}}$

Current in each of B and C = ${I^{\text{'}}}_1=\frac{I_1}{2}=\frac{E}{3R}$

For the branch containing D, E, F, G and H:

${\mathrm{R}}_{eq}=\frac{7R}{3}$

Current through D and H: $I_2=\frac{3E}{7R}$

Current through each of E, F and G:

${{\mathrm{I}}^{\text{'}}}_1=\frac{{\mathrm{I}}_2}{3}=\frac{\mathrm{E}}{7\mathrm{R}}\mathrm{Since}{\mathrm{I}}_1>{\mathrm{I}}_2>{{\mathrm{I}}^{\text{'}}}_1>{\mathrm{I}}^{\text{'}}$

A > D = H > B = C > E = F = G