Puzzle: Controlling two lamps with one SPST switch (AC)

Question

This is the text of the problem:

Draw a circuit with two lamps & one SPST switch.

The lamps must be in opposite states.

New parts are disallowed to add to the circuit, but there is no limit on the type of lamps. We can't replace lamps with LEDs.

Answer 1

Nothing was said about the lamp types, or that they need to be identical. So, assuming that L1 and L2 are both 120V incandescents, and L1 is a 1 watt bulb and L2 is a 100-watt bulb,

^{simulate this circuit – Schematic created using CircuitLab}

when the switch is open, L1 will be lit, and the current through L2 will be so small as to produce insignificant heating and the bulb will be off to the naked eye. When the switch is closed, L2 will receive full voltage and be on, while L1 will be off.

EDIT: A few numbers. A 100-watt bulb at 120v has a resistance of 144 ohms. A 1-watt bulb at 120v has a resistance of 14,400 ohms. With the two lamps in series, the voltage across the 100-watt bulb will be $$V = \frac{144}{144+14400} \times 120 = 1.18 \text{volts}$$ Power dissipated in the 100-watt bulb will be $$P = \frac{V^2}{R} = \frac{1.18^2}{144} = ~.01 \text{watts}$$ In other words, the 100 watt bulb will be running at .01 watts, or 1/10,000 its normal power.

Here's a quick test. Take a 100 watt bulb and drive it with a single battery cell. If you can see the slightest trace of light I'd be very surprised.

And it's actually even more true than the analysis so far. Incandescents have resistances less than 1/10 their operating value when cold. So, since the 100-watt never gets hot using the "hot" resistance, you can expect the voltage across it to be at least 10 times less than the worked numbers, so the power dissipated will be at least 100 times less, or on the order of 100 microwatts.