Add Switch To Extension Cord

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While talking about neutral faults in this answer, I (wrongly) said that neutral faults can cause shock. As mentioned in the comments I am

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I looked at the problem for a while, but I couldn’t wrap my head around it. So I came up with a circuit diagram and asked a question in the chat.

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Being true to my name, I decided to try it. No, I did not catch the basic (neutral) conductor (sorry to disappoint those of you who would like me to go into oblivion).

I ended up using 15W compact fluorescent lamps instead of 60W incandescents, so I only measured 0.125A.

Kirchhoff’s second law states that the total voltage applied to any closed circuit path is always equal to the sum of the voltage drops along the path.

If we apply 120 volts across the loop, the total voltage drop across the loop will also be 120 volts. However, most of the voltage drop is in the first part of the loop (the wire to the switch, the switch, the wire between the switch and the light, the light and the wire from the light to the first turn-on wire. connector). From this point, the same voltage will be applied to the rest of the first loop as is applied to the second loop. The voltage will be so small, it won’t be enough to light the bulb.

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Each wire in the photo is 14 AWG and 6″ long. We can determine the resistance of each bit of wire using Table 8 of Chapter 9 of the National Electrical Code. Solid, 14 AWG, uncoated, copper wire has a resistance of 3.07 ohms per 1000 feet.

Now we can determine the total resistance in the circuit (we will assume 0 ohms across the switch for simplicity).

Note: When you look at the diagrams, you may be dealing with a simple parallel circuit. However, this is actually a short-circuited series circuit. This is why the results can be counter intuitive and confusing.

OK, there seems to be a lot of confusion about this, so let me give a quick overview of the first week or two of an electrical physics course.

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Analyzing circuits in general is quite complex; Even a simple circuit with just a resistor, inductor and capacitor takes about a semester to fully understand. However, most home circuits can be modeled using only resistors, which makes them much easier to understand.

I’m going to assume you’re familiar with electrical concepts common to home wiring: current (amperage), voltage, alternating current, etc.

The symbol is an AC (alternating current) generator, the type that supplies electricity to homes. This would be equivalent to the left and right prongs in the outlet.

This is the symbol for a resistor, which opposes current flow. It can also be used to resist a light bulb, a person touching a wire or a wire.

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This is the wire. We assume it has 0 resistance; If we want to model the resistance of a wire, we add an imaginary resistor to the circuit.

Note that the voltage between any two points on our ideal resistor wire will always be 0, it seems.

That there can never be current between them (or rather, the current is 0/0). We ignore this fact as theoretical, because real wires are always there

The symbol is called “ground,” but it actually represents the neutral wire, not the third shoe that an electrician would call “ground.” If there is more than one ground in the circuit diagram, we assume that they are all connected.

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Analyzing these types of circuits requires knowledge of only three laws: Ohm’s Law, Kirchhoff’s Current Law, and Kirchhoff’s Voltage Law.

Kirchhoff’s Current Law (KCL) “The total current entering a wire-junction or device is equal to the total current leaving the junction/device”.

This seems almost obvious if you think of the water analogy of electricity and replace “current” with “water”. You can also call it “preserving the present”.

It sounds complicated, but it’s really simple; This basically says that if you do all the voltage increases and voltage decreases in a loop, you get 0 when you go back to where you started. This also makes sense if you replace “voltage” with “pressure”. Analogy to water.

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According to KVL, the voltage across R2 = 0. According to Ohm’s law, this means current through R2 = 0, so no light. The diagram you gave in the answer…

There was concern in the comments above that this circuit is not realistic, because it does not model the resistance of the wire. It is true; Let’s try to analyze a more realistic circuit.

First, we find the current through the entire circuit. This can be done with just the three rules above and a lot of algebra

. Since this isn’t an algebra class, I’ll skip this step and just tell you that the equivalent resistance is a little less than 1002Ω, so the generator current is about 120V / 1002Ω ≈ 0.11976A.

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Since the voltage gain on our AC generator is 120V, by KVL that means there is a voltage DROP across it (and the neutral wire).

If the “you” in the picture is really you: According to this answer, the inductance of 60 Hz AC is around 0.4 mA. So even though an electric current is actually flowing through your body, you won’t feel it even if you’re standing in a bath of water.

In the case where “you” are actually an incandescent light, there will be an electric current, but the light it produces will be so dark that you will not see it.

And if “you” have a fluorescent lamp like in your pictures, the ballast will block any current flow at all.

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Finally, let’s compare the above case with the case where the device is turned on and you connect the hot wire to the ground:

In both cases, the current through you will be slightly less than 12 mA. 12mA is enough to cause severe pain and make your muscles contract so you can’t let go of the wire. And if you’re sweating that day or your skin resistance is low, the current will be even higher, possibly enough to kill you.

There may also be some additional resistance at each point that a wire connects to another. (wire nut, outlet etc.)

All three wires connect to the voltage at the outlet because they are different from the breaker box.

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At the outlet near the service entrance, with no load, both neutral and ground have about 0.15VAC.

I also went and measured the voltage to a point on the ground a few feet away from the grounding rod and got the same 0.15VAC measurement.

To prove that the multimeter was working properly, I touched both probes with my finger and got 0VAC.

So you can not only get a shock from the neutral wire, but also from the ground wire.

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For example, sitting on the edge of a pond with a ground pump, and a perimeter that is not based on the same capacity.

I’m pretty sure I’ve done this test before and got a high reading on the grounding rod. They have put up some new transmission lines and poles for the new substation; This would have helped pull the ground conductor closer to 0VAC.

This is also assuming that the owner of the house did not change the outlet and mistakenly switched the hot one for the neutral one.

By clicking “Accept all cookies”, you agree that StackExchange may store cookies on your device and disclose information in accordance with our cookie policy. I wanted to share my recent experience updating the wiring on the SP-20 motor to add ground. (Disclaimer: I am not an electrician although I have consulted with an electrician.)

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My press power cord was cut where it exceeded the press. So, I needed to connect the plug anyway and decided it would be best if it was grounded. First, I checked the existing 18AWG diameter wiring and the motor requires 3.2 amps and 115V. I bought a 16/3 16AWG three wire extension cord. My electrician friend confirmed that I could go a little bigger, but not smaller in diameter.

I cut off the receptor end of the extension cord and ran it through the outlet pipe to the junction box where the switch was. I then peeled back the plastic coating to reveal three internal wires: load, line (hot) and ground. Using spade connectors, I wired in the load and line switches. The ground wire has a hole in the connector that is attached to the junction box. From there I connected the ground wire to the bolt on the body of the printer. I will soon update the connection with a ring connector for better security. Currently, the copper ground wire is simply wrapped around the bolt

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