## 149. Math Puzzle $10 / 77$

Peter connected two 6 V lamps in series to a 12 V battery, one lamp rated at 18 W and the other at 3 W. One of the lamps burned out. Which one was it? Why did the lamp burn out?
Through both lamps connected in series, the same current $I$ flows. The internal resistances of the lamps are:

Lamp $6 \mathrm{~V} / 18 \mathrm{~W}: R_{1}=\frac{36 \mathrm{~V}^{2}}{18 \mathrm{~W}}=2 \Omega$

Lamp $6 \mathrm{~V} / 3 \mathrm{~W}: R_{2}=\frac{36 \mathrm{~V}^{2}}{3 \mathrm{~W}}=12 \Omega$

The current $I$ is: $I=\frac{12 \mathrm{~V}}{(12+2) \Omega}=\frac{12}{14} \mathrm{~A}$.

It causes the voltage drops across the lamp $6 \mathrm{~V} / 18 \mathrm{~W}$: $U_{1}=R_{1} \cdot I=1.71 \mathrm{~V}$.

across the lamp $6 \mathrm{~V} / 3 \mathrm{~W}$: $U_{2}=R_{2} \cdot I=10.29 \mathrm{~V}$.

The lamp $6 \mathrm{~V} / 3 \mathrm{~W}$ is overloaded and burns out.