When talking about neutral faults in this answer, I (incorrectly) said that a neutral fault could lead to a shock hazard. It was pointed out in the comments that I was dead wrong, and that the scenario that I presented was incorrect.
I looked at the problem for a bit, but just couldn't wrap my head around it. So I came up with a circuit diagram, and posed a question in chat.
So given this diagram, will R2 light up?
Best Answer
As it turns out, no. R2 will not light up.
The Experiment
Being true to my username, I decided to test this out. No, I didn't grab hold of a grounded (neutral) conductor (Sorry to disappoint those of you who would have liked to see me get zapped into oblivion).
I started by wiring up a sample circuit.
Then I added an extension cord, so I could plug it into a receptacle.
When I flipped the switch, the second light did not light!
Finally I took some readings with my ammeter, and this is what I found.
Notes:
I ended up using 15W compact flourescent lamps instead of 60W incandescents, which is why I measured only 0.125A.
The Explanation
Kirchhoff's second law says that the total voltage applied to any closed circuit path, is always equal to the sum of the voltage drops in that path.
If you look at the circuit, there is a "closed circuit path" which I've highlighted in red below.
If we apply 120 volts to the loop, the total voltage drop in the loop will also be 120 volts. However, the majority of the voltage drop occurs in the first part of the loop (the wire to the switch, the switch, the wire between the switch and light, the light, and the wire from the light to the first twist-on wire connector). From this point, the same voltage will be applied to the remainder of the first loop as is applied to the second loop. This voltage will be so small, it will not be enough to light the bulb.
Maths
Each length of wire in the photo is 14 AWG, and 6" long. We can determine the resistance of each bit of wire using Table 8, from chapter 9 of the National Electrical Code. Solid, 14 AWG, uncoated, copper wire has a resistance of 3.07 ohms per 1000 feet.
3.07 ohms per kFT. / 1000 FT. = 0.00307 ohms per foot0.00307 / 2 = 0.001535 ohms per half foot (6")We can determine the resistance of a 15W compact fluorescent lamp (CFL), using Ohm's law Resistance (R) equals voltage (E) squared divided by power (P).
R = 120V ^2 / 15W = 14400V / 15W = 960 ohmsIf we plug those values into a diagram, we'll end up with something like this...
Now we can determine the total resistance in the circuit (we'll assume 0 ohms in the switch for simplicity).
Ra = R1 + R2 + R3 + R4 + R5 + R6 = 960.007675 ohmsRb = R7 + R8 + R9 = 960.00307 ohmsRt = (Ra * Rb) / (Ra + Rb) = 921610.31522356225 / 1920.010745 = 480.0026862472388142806982051551Next we'll calculate the total current.
Ia = Et / Ra = 120V / 960.007675 ohms = 0.12499900065903118951627131522672 amperesIb = Et / Rb = 120V / 960.00307 ohms = 0.12499960026169499645454258807735 amperesIt = Ia + Ib = 0.12499900065903118951627131522672 amperes + 0.12499960026169499645454258807735 amperes = 0.24999860092072618597081390330407 amperesThen we can calculate the voltage drop across each component.
DeltaR1 = R1 * Ia = 0.00019187346601161287590747646887302 voltsDeltaR2 = R2 * Ia = 0.00019187346601161287590747646887302 voltsDeltaR3 = R3 * Ia = 119.99904063266994193562046261765 voltsDeltaR4 = R4 * Ia = 0.00019187346601161287590747646887302 voltsTotal Voltage Drop after R4 = DeltaR1 + DeltaR2 + DeltaR3 + DeltaR4 = 119.99961625306797677424818504706 voltsVoltage after R4 = 120V - 119.99961625306797677424818504706V = 0.00038374693202322575181495294VNOTE: While looking at the diagrams, it may seem as though you're dealing with a simple parallel circuit. However, it's actually a short-circuited series circuit. This is why the results may be counter intuitive, and confusing.