An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V it draws a power P. Then:

1
\(\mathbf{P}=\left(\frac{\mathrm{V}}{\mathbf{V}_{0}}\right)^{2} \mathbf{P}_{0}\)
2
\(\mathbf{P}=\left(\frac{\mathbf{V}_{0}}{\mathbf{V}}\right)^{2} \mathbf{P}_{0}\)
3
\(\mathbf{P}=\left(\frac{\mathbf{V}_{0}}{\mathbf{V}}\right) \mathbf{P}_{0}\)
4
\(P=\frac{V}{V_0} P_{0}\)

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