Electrical – Effects of rapidly pulsing DC inductive load on solid state relays

With the concerns you state, and especially with having the relay installed in a wall, I guess the answer is:

Don't.

You might add a fuse, but then, you'd have to find a rating in the relay's data sheet telling you if it is approved for usage with a fuse, and if so, what the fuse's rating would have to be. All in all, you would have to do the same qualification for the entire system (relay, fuse, type of outlet socket, possible loads) that the original vendor did for the combinaiton of the relay and fountain.

The relay will not limit the current (amps) to a level that is safe to itself. When overloaded, it will likely act like a short circuit for moderate overload conditions and will fail open (or explode, or burn) for severe or ongoing overload conditions.

The thing to remember is this: Once you exceed any of the ratings, the manufacturer won't guarantee anything they tell you in the data sheet any more and you are on your own.

Also, with having the relay installed in a wall where it can't really get rid of any heat it produces, I recommend using the derating curve for 60...70 °C, which tells you to rather not exceed 2...3 A.

A hacker's answer, however, is:

Yes, your setup might work, ...

... but you would have to make sure all is well and within the specified limits and characteristics.

An old question, but it's a cool topic for beginners to wrap their minds around, so I'll answer it. To answer the last question first, remember that voltage appears across a load, while current is measured through the load.

It may be easier to visualize the phase lag concept if you think of a capacitor rather than an inductor. You're probably familiar with the fact that when you charge a large capacitor, it looks like a short circuit at first. At the instant of connection, current is flowing through the cap, but no voltage appears across it because, hey, it's a short circuit, right? As the cap charges up, the voltage across it rises and the current through it falls. This is all that's meant when people say that "the current leads the voltage" in a capacitor.

With an inductor, we say the voltage leads the current because at the instant of connection the inductor looks like an open circuit. A perfect inductor connected to a voltage source at time=0 will have the whole supply voltage across it, with no current flowing through it. During the 'charging' process the inductor stores energy in its surrounding magnetic field, which cannot happen instantaneously any more than a capacitor can be charged instantaneously. So the voltage "leads" the current in this case.

What's interesting about an inductor is what happens when the source is disconnected. A capacitor will just sit there at the same voltage, slowly losing its charge over a long period of time if there is no load across it. But with an inductor, the magnetic field collapses as soon as the power supply is removed, and this happens quickly. A recently-disconnected inductor will try to maintain the flow of current through the circuit rather than the voltage across itself.... but wait, there is no circuit anymore, because we just opened it.

A perfect inductor would generate an infinite voltage in an attempt to keep the current flowing. Even an imperfect one can turn a few volts into several hundred for a short period of time after disconnection. This is why a zero-crossing switch is not the same thing as a snubber. The snubber's job is to give the inductor a load it can drive when the source is removed altogether -- usually a capacitive one since you don't want it drawing current the rest of the time. It keeps the voltage from rising to levels that could hose semiconductors, burn relay contacts with arcing, or otherwise cause trouble.