The bulbs act as resistors.

Assuming that each bulb has a resistance of R, we can get the equivalent resistance for the different cases.

Remember,

Equivalent resistance for resistors in parallel:

$\overline{)\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{eq}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{n}}}}$

Equivalent resistance for resistors in series:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}{{\mathbf{R}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{n}}}}$

From lowest to highest, rank the overall value of the resistance across the circuit from the point labeled j to the point labeled k.

a. A, B, C

b. A, C, B

c. B, A, C

d. B, C, A

e. C, A, B

f. C, B, A

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