Power:

$\overline{){\mathbf{P}}{\mathbf{=}}{{\mathbf{i}}}^{{\mathbf{2}}}{\mathbf{R}}}$, where i is current and R is equivalent resistance, in this case.

Resistance of a bulb can be expressed as:

$\overline{)\begin{array}{rcl}{\mathbf{R}}& {\mathbf{=}}& \frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{P}}\end{array}}$, where V is voltage and P is power.

For resistors connected in series:

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

Two 75.0 W (120 V) lightbulbs are wired in series, then the combination is connected to a 230 V supply. How much power is dissipated by each bulb?

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