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

For the branch containing A, B and C.

$R_{e q} = \frac{3 R}{2}$

Current in A : $I_{1} = \frac{E}{R_{eq}} = \frac{2 E}{3 R}$

Current in each of B and C = ${I^{'}}_{1} = \frac{I_{1}}{2} = \frac{E}{3 R}$

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

$R_{e q} = \frac{7 R}{3}$

Current through D and H: $I_{2} = \frac{3 E}{7 R}$

Current through each of E, F and G:

${I^{'}}_{1} = \frac{I_{2}}{3} = \frac{E}{7 R} Since I_{1} > I_{2} > {I^{'}}_{1} > I^{'}$

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

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