Question

A light bulb is rated at \[100\,\mathrm{W}\] on a \[220\,\mathrm{V}\] AC supply. What is its resistance?

Options

• \[311\,\Omega\]

• \[220\,\Omega\]

• \[0.454\,\Omega\]

• \[484\,\Omega\]

Answer 1

The supply rating is an RMS voltage, so use \[
\begin{array}{c}
R = \dfrac{V_{\mathrm{rms}}^2}{P}
\end{array}
\]. Substitution gives \[
\begin{array}{c}
R = \dfrac{(220\ \mathrm{V})^2}{(100\ \mathrm{W})} = 484\ \Omega.
\end{array}
\].