Two lamps, one rated 100 W at 220 V

solution:
Resistance of first lamp,

$R_{1}=\frac{V^{2}}{P}=\frac{(220)^{2}}{100}$

resistance of the second lamp,

$\mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{60}$

Since the two lamps are connected in parallel, the equivalent resistance is given by

$\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{100}{(220)^{2}}+\frac{60}{(220)^{2}}=\frac{160}{(220)^{2}}$

or $\mathrm{R}_{\mathrm{P}}=\frac{(220)^{2}}{160}=\mathbf{3 0 2 . 5} \boldsymbol{\Omega}$

Current drawn from the line, i.e.,

$I=\frac{V}{R_{p}}=\frac{220 V}{302.5 \Omega}=0.727 \mathrm{~A}$