I found this warning on a little slip of paper, in with an Insignia surge protector:

DO NOT install this device if there is not at least 30 feet(10 meters) or more of wire between the electrical outlet and the service panel.

What could this be about? What are the odds that the device is actually designed to make use of the properties of that 11m+ run of cable?

Also, how the heck is it ok to have a restriction like this? Our service box is optimally located in the center of a small home! Most outlets and appliances are within a 10m run of wire from the box.

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    \$\begingroup\$ Likely this surge protector relies on the presence of certain inductance on the source side, to provide specified protection. And if you failed to comply with 30ft cabling, they will not pay for your damages. \$\endgroup\$ Jan 6, 2017 at 2:42
  • \$\begingroup\$ Well, I guess that that length of wire has two interesting properties: it's free, and the maximum transient voltage that it can handle is massive, compared to any PCB mounted component. I hope this restriction exists for the latter reason, and not only the former. \$\endgroup\$
    – Dan Ross
    Jan 6, 2017 at 2:56
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    \$\begingroup\$ I have a surge protector with the same warning and a 25-foot cord. So the warning is likely boilerplate and not written with consideration of the specific design — or it's not simply the properties of that length of cabling that matter to whatever this is. I hope this question gets a non-speculation answer! \$\endgroup\$
    – Kevin Reid
    Jan 6, 2017 at 3:31
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    \$\begingroup\$ The question is, do they mean 30 ft or round trip, or one way? Cause that's 30 vs 60. \$\endgroup\$
    – Passerby
    Jan 6, 2017 at 5:29
  • \$\begingroup\$ What does it say in the owner manual about this? \$\endgroup\$
    – Andy aka
    Jan 6, 2017 at 11:14

1 Answer 1


To limit the maximum current and voltage in the event of a short circuit.

Extra resistance between the power source and a surge protector limits the maximum peak voltage to the protected equipment. That was good advice.

Close to the breaker panel, the available short-circuit current is considerably more than at the end of a long run of wire. The resistance and the inductance of longer wires sometimes matters.

Years ago, I helped wire a commercial building. All the wire had been pulled, and all that was left was to install the duplex receptacles. Well, I started at the farthest point --at least 75 feet-- from the service entrance. Not knowing which breaker fed what, to shut off the power to each box I simply shorted the wires together to trip the breaker.

Well, that worked very well, until I came to a box that was only a few feet from the panel. When I shorted those wires, the current was so large that the flash temporarily blinded me. When I regained my sight several seconds later, I realized that my gloves were on fire. It did trip the 20A breaker; but when I reset it, I could tell from the feel that the contacts inside the breaker had arced and were nearly welded together.

Here's an excerpt from a Littelfuse PDF.

The apparent 'clamping' of the voltage results from the increased voltage drop (IR) in the source impedance due to the increased current. It should be clearly understood that the device depends on the source impedance to produce the clamping. One is seeing a voltage divider action at work, where the ratio of the divider is not constant but changes. However, if the source impedance is very low, then the ratio is low. The suppressor cannot be effective with zero source impedance (Figure 2) and works best when the voltage divider action can be implemented.

The longer the wires that ultimately power the surge supressor, the greater the source impedance. And the greater the source impedance, the lower the surge voltage will be at the protected equipment.

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    \$\begingroup\$ While this is good information, it lacks any specific technical information to back it up. This is empirical at best. A more scientific answer would be better. \$\endgroup\$
    – Passerby
    Jan 6, 2017 at 5:31
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    \$\begingroup\$ @passerby, This is basic electronics 101.There's no guesswork nor anything empirical about this. The longer the wire, the greater the resistance. The greater the resistance, the less current can flow at a given voltage. On the load side of the of the resistance, the voltage is reduced. Ohm's law (Google it) applies here. A resistor in lieu of the wire would accomplish the very same results. That's just a fact. \$\endgroup\$ Jan 6, 2017 at 6:19
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    \$\begingroup\$ 20m of AWG12 (minimum size for a 20-amp circuit) has a total resistance of 0.1 ohm. It's hard to believe that this would make any significant difference compared to all of the other impedances in the circuit in the type of scenario you're talking about. Maybe it has more to do with the capacitance of the cable limiting the speed of any transients, although adding a small discrete capacitor would have eliminated this external requirement. \$\endgroup\$
    – Dave Tweed
    Jan 6, 2017 at 13:37
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    \$\begingroup\$ @Ali Chen I would be very surprised if a 40M squashed loop (Which is what most house wiring cable looks like) had an inductance anything like 40mH, for PSC calculations it is the resistance that matters. 0.1 ohm on a 230V circuit limits the PSC to less then 2.5kA, get within a meter or so of the inltake and you might easily see ten times that if you are located near the distribution transformer. \$\endgroup\$
    – Dan Mills
    Jan 7, 2017 at 14:36
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    \$\begingroup\$ I should have mentioned in the first line that short circuit mainly referred to the surge suppressors' function during a voltage spike. When they clamp on a HV spike, they temporarily become a short circuit. And that's what protects the connected devices from the voltage surge. \$\endgroup\$ Jan 29, 2017 at 19:45

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