For batteries in series,

$\overline{){{\mathbf{V}}}_{\mathbf{e}\mathbf{q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{{\mathbf{V}}}_{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{{\mathbf{V}}}_{{\mathbf{3}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{{\mathbf{V}}}_{{\mathbf{4}}}}$

Ohm's law:

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{i}}{\mathbf{R}}}$

V_{eq} = V_{1} + V_{2} + V_{3} + V_{4} = 4.5 + 4.5 + 4.5 + 4.5 = **18 V**

You have a circuit consisting of four 4.5 V D-cell batteries in series, some wires, and a light bulb. The bulb is lit and the current flowing through the bulb is 0.50 A. You may assume that all of the voltage drop occurs in the light bulb. What is the resistance of the bulb?

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