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I have an Outlet Tester like this:

enter image description here

It is used to see the mains electric connection, it has the next states:

I opened it and then I reversed engineer it. The diagrams I obtained are these:

enter image description here

I would like to make the circuit analysis (Kirchoff's Laws) but I don't know how to apply them in this case, I don't see a well defined voltage source, like in other cases:

enter image description here

I mean, yes, there is a Hot (+) and a Neutral (-), but, what about ground? .. Let's suppose that Hot is 127 Vrms and 60 Hz, How do I apply KVL and KCL here?

Dau
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2 Answers2

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First of all, analyzing an AC circuit with purely real ohmic impedances is just as straight forward as analyzing a DC circuit, you just let neutral be ground, and use the RMS voltages and currents as if they were DC values.

In your case you have four basically different circuits, because you have three circuits where one of the terminals are disconnected, these are all unique, and three circuits where all three terminals are connected to some voltage source with the same reference. This means that we need to solve four different circuits individually to find the solution for each configuration.

I will leave "Open Line", "Open Neutral" and "Open Earth" for you to solve, this should be easy now that you know that you need to solve all these three individually.

However it might not be obvious to you how you can solve the remaining three in one go, so I will go ahead and give you a hand.

schematic

simulate this circuit – Schematic created using CircuitLab

In the schematic above I have made an abstraction, and avoided thinking of the three terminals as being connected to either line, neutral or earth, and just thought of them all as connected to arbitrary voltage sources V1, V2 and V3, this allowes me to solve the circuit ones, and then substitute the voltage values afterwards to find the final solutions.

When we solve circuits with multiple sources we use the principal of superposition, which to put it in short terms, means that we calculate the voltages and currents that we are interrested in as a function of each source individually, and then we add those results together in the end to get the final solution with all the sources acting on the circuit. This is always valid to do if your cirucit is linear.

For voltage sources, when they are not part of the particular solution being calculated, they are shorted. So to calculate the solution we get with only V1 we do as follows.

schematic

simulate this circuit

Let's reference the currents in all the resistors in the direction going from the top to the bottom of the diagrams, and let's reference the voltages accordingly, going from the top down, I will call the currents in the resistors I1, I2 and I3, to denote the currents in resistors R1, R2 and R3.

In the diagram above, where we only considder V1, the currents are as follows;

$$I_1=\frac{V_1}{R_1}$$ $$I_2=\frac{V_1}{R_2}$$ $$I_3=0$$

Now the diagram drawing tool is really cumbersome, so I wan't draw all the diagrams, but the soluting when considdering V2 and V3 are:

$$I_1=-\frac{V_2}{R_1}$$ $$I_2=0$$ $$I_3=\frac{V_2}{R_3}$$

$$I_1=0$$ $$I_2=-\frac{V_3}{R_2}$$ $$I_3=-\frac{V_3}{R_3}$$

Now we add the solutions together:

$$I_1=\frac{V_1}{R_1}-\frac{V_2}{R_1}$$ $$I_2=\frac{V_1}{R_2}-\frac{V_3}{R_2}$$ $$I_3=\frac{V_2}{R_3}-\frac{V_3}{R_3}$$

Above is the final solution of the KVL/KCL analyzis, for the three circuits where all terminals are connected.

Now to find the particular solutions for the three individual configurations we just substitute V1, V2 and V3 with the voltages of the terminals that they are connected to, for this purpose we considder neutral and earth to both be at 0V, and line to be at Vl.

Line/Earth Reversed

$$I_1=\frac{0}{R_1}-\frac{0}{R_1}=0$$ $$I_2=\frac{0}{R_2}-\frac{V_l}{R_2}=-\frac{V_l}{R_2}$$ $$I_3=\frac{0}{R_3}-\frac{V_l}{R_3}=-\frac{V_l}{R_3}$$

Line/Neutral Reversed

$$I_1=\frac{0}{R_1}-\frac{V_l}{R_1}=-\frac{V_l}{R_1}$$ $$I_2=\frac{0}{R_2}-\frac{0}{R_2}=0$$ $$I_3=\frac{V_l}{R_3}-\frac{0}{R_3}=\frac{V_l}{R_3}$$

Correct

$$I_1=\frac{V_l}{R_1}-\frac{0}{R_1}=\frac{V_l}{R_1}$$ $$I_2=\frac{V_l}{R_2}-\frac{0}{R_2}=\frac{V_l}{R_2}$$ $$I_3=\frac{0}{R_3}-\frac{0}{R_3}=0$$

Since we are dealing with AC we don't care if the sign is positive or negative, as this just means that the AC waveform is phase shifted 180 deg, and we only care about the RMS values in this case.

If we match the solutions above with your original truth-table, we see that they match up, indeed the solution gives us currents in the lamps that we are expecting.

Now as I sayed I will leave it for you to solve the remaining three where one of the terminals are disconnected, for these you need to do the same thing that I did, but just leaving the disconnected terminals floating throughout the calculations, so for example when Line is disconnected, R1 and R2, and their lamps, will be in series.

Vinzent
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  • There is one problem .. The IV Curve of Neon Lamps, is not linear, it has negative Resistance .. How does this affect in the analysis? – Dau Jan 17 '20 at 18:33
  • This is NOT a Linear Circuit, the resistance of the Neon Lamps is not constant – Dau Jan 17 '20 at 19:26
  • No you are right, it is not strictly a linear circuit, but almost nothing in this world is. In engineering we are used to assuming that circuits are linear, because that is usually the only way we can solve them, but it is also almost always reasonable to linearize a circuit, as most circuits behave very close to linearily. In this instance it doesn't really matter at all, because although you include the non-linear characteristics in your solution, you will still come to the same result as I did, although you will have captured the current more accurately. – Vinzent Jan 17 '20 at 19:32
  • ..It doesn't matter that the lamps are not linear, they are close enough to linear when the circuit it operating in steady state. – Vinzent Jan 17 '20 at 19:34
  • Ok, ... Another thing, In the experiment, I obtained that the current i1 and i2 is 0.7 mA .. Theoretically, if I consider the Lamp Resistance as 0, I obtained 1.867 mA .. So I guess this means that I need take into consideration the Neon Lamp Resistance, unfortunately, my multimeter is unable to measure it ... Nevertheless, I guess the Analysis is the same ... – Dau Jan 17 '20 at 20:01
  • Yes if you really want to model the circuit accurately you need to take the resistance of the lamps into account, but you just add them to the values of the resistors, it doesn't change the whole picture regardless, you still get the same lamps lighting up and that's what's important. – Vinzent Jan 17 '20 at 20:18
  • All right, I think now I understand it, more or less .. Just one thing, Why is it that the The Superposition Principle doesn't work with non Linear Circuits? Is there a way to prove that? – Dau Jan 17 '20 at 20:23
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    If you considder the case of a bjt transistor which is a non linear component, its current to voltage ratio is \$\ I_E = I_{ES} \ln ( \frac{V_{BE}}{V_{T}}+1) \$ If you raise the base emitter voltage from zero to 0.2V it doesn't cause the same change in current as raising the voltage from 0.2V to 0.4V, ie. by the same amount, so individual sources can not be separated out of the equations as they all have an effect on the absolute voltage across the base emitter. And yes it is easy to find the proof of the superposition principle online, it's a harry bit of math. – Vinzent Jan 17 '20 at 21:37
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You will have to do six separate analysises, one for correct wiring, and one for each fault condition, making the appropriate power and ground connections for each condition.

For the "Correct" state, you apply power between Line and Neutral, and connect Ground to Neutral.

For "Open Ground", connect power between Line and Neutral, and leave Ground unconnected.

Remember that Ground should be connected to Neutral in the main breaker panel.

Peter Bennett
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    "You will have to do six separate analysises" Nope, only four are needed. – Vinzent Jan 16 '20 at 20:02
  • Assuming that the neon under the red lens is the same, symmetry implies that there are really only two distinct cases - when two lines are connected across the supply and one open, and when two are commoned. – Phil G Jan 16 '20 at 20:52