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When people in my classes ask questions, my favorites are the ones that I can't answer off the top of my head, because it makes me think.
We recently had a class in Banff, Alberta, Canada, because somebody has to do it, so it might as well be me. During a break, one of the more curious people in the class (curious as in inquisitive, not as in strange) named Stuart Williamson asked me a great question. It made me think about exploding lamps.
The question had to do with losing the neutral conductor. We had been talking about why the neutral conductor is the only normally current-carrying conductor that is not protected from overloading by a fuse or circuit breaker, which is true. The reason is that the neutral conductor serves as the 0-volt reference. If it's lost, bad things can happen.
Suppose, for example, we have a lighting rig with 16 PAR lamps that are rated 1000 watts at 120 volts. Let's say the power is supplied by a split-phase system with two hot conductors (black and red), a neutral, and a ground, where the phase-to-neutral voltage is 120V and the phase-to-phase voltage is 240V. And suppose we made a mistake circuiting the rig, so we have 10 of them connected from the black leg to neutral, and six from red to neutral. And suppose one of the six lamps burned out during the show (Sheesh, we’re having a bad night, and it’s about to get worse!), and now we have a more unbalanced system.
At the rated voltage, the resistance of the filament in these lamps is about 14.5 ohms (1000W/120V = 8.3A, and 120V/8.3A = 14.5 ohms). Schematically, the group of 10 lamps, each of which is 14.5 ohms, are in parallel with each other, and they have an equivalent resistance of 1.45 ohms [1/(1/14.5 +1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5) = 1.45 ohms). The six lamps (now five after the blown lamp) are in parallel with each other, and they have an equivalent resistance of 2.9 ohms [1/(1/14.5 + 1/14.5 + 1/14.5 + 1/14.5 + 1/14.5) = 2.9 ohms].
When the neutral (0-volt reference) is lost, that creates a voltage-divider circuit between the 1.45 ohm and the 2.9 ohm equivalent resistances with 240V applied to it, which draws 55.2 amps. Not to worry, we still have a quantity of 15,000 watts of load, and 55.2 amps at 240 volts is only 13,248 watts (55.2A x 240V = 13,248W).
On the other hand, the voltage-divider circuit is going to drop some voltage across each of the equivalent resistances in proportion to their impedance. That means we’ll have 80.2 volts applied to the group of 10 lamps (55.2 amps x 1.45 ohms = 80.2V) — which explains the sudden dimming of those lamps — and about 160 volts (Yikes!) applied across the group of five lamps — which explains why they suddenly got much brighter. And because it we're having a really bad day, one of the lamps in the group of five has a weak filament and blows. Now the math changes. That group of lamps now has an equivalent resistance of 3.6 ohms [1/(1/14.5 + 1/14.5 + 1/14.5 1/14.5) = 3.6 ohms], and the voltage-divider circuit only gets worse. The group of 10 lamps still has an equivalent resistance of 1.45 ohms, but the drop in impedance across the other lamps means the current drops to 47.5 amps [240V/(1.45 + 3.6) = 47.A], but now the applied voltage on the group of four lamps rises to 171 volts! You can guess what happens from there. Lather, rinse, repeat. The lamps continue to fall like dominoes (that have been sprinkles with gunpowder and set on fire!) until all the lamps in that group are blown. Then the fireworks stop because that group of fixtures is now an open circuit, which means all the lamps will dowse in a more civilized way, i.e., they fade to black.
This is a simplified version of what would happen in real life because the resistance of the filament will change with the operating temperature, which changes with the current. But the process will follow the same script except the numbers would change slightly. And if it was a 3-phase system instead of a split-phase system, then it would be a more complex circuit, but the same thing would happen.
Say what you will about LEDs, but an all-LED rig would likely never have this problem because most of them will operate on any voltage from 90V to 250V, nor does the impedance change when the emitters fail.
