## 93. Math Puzzle $2 / 73$

A light bulb rated for $60 \mathrm{~V}$ and $30 \mathrm{~W}$ is to be connected to a $120 \mathrm{~V}$ alternating current $(50 \mathrm{~Hz})$ with the same power output by inserting a capacitor in series. What capacitance must this capacitor have?
$$
R=\frac{U_{1}^{2}}{P}=120 \Omega \quad ; \quad Z=\frac{U_{2}}{I}=240 \Omega
$$

from $\frac{1}{\omega C}=X_{c}=\sqrt{Z^{2}-R^{2}}$ we get

$$
C=\frac{1}{2 \pi f X_{c}} \approx 1.53 \mu \mathrm{F}
$$