Susan Brown

[Susan Brown]

Ack! How does neutral “send” hot voltage? Neutral is at ground potential. Have you reviewed the material in Unit 6 while working on this question?

Also, this is not a neutral line. We’re dealing with an L1-L2 circuit, but using a neutral point for some of the electrical measurements as reference. In other words, for measurements 2 and 3 we’ve got the red probe of our meter on either L1 or L2, and the black probe on a known-good neutral point. That neutral point is not part of this particular circuit.

[Susan Brown]

If there is an open on L2, where is that 120vac coming from in Measurement #2? (L2 wrt N)

[Susan Brown]

Which measurement makes you think there’s an open on L2?

[Susan Brown]

Both of those Figures have the same type of measurements. We just get a different reading at Meter 3 (L1 wrt N) in Figure 2 compared to Figure 1.

This is a series circuit. We just concluded above that there is an open somewhere in the circuit. Do we have current anywhere?

[Susan Brown]

Remember that current is electrons moving through a wire at a certain rate.

Parallel circuits share a power supply. The electrons (current) coming from that power supply is what we mean by “total circuit current”.

Total circuit current equals the sum of the currents in the parallel branches.

When one branch fails, its current goes to zero, but the other one is unaffected. So, less current is needed from the power supply.

[Susan Brown]

You are in the right place for that! 🙂

[Susan Brown]

Also, we have a two types of voltage measurements in this problem. There are the measurements across the element (L1 wrt L2) where we would expect a voltage drop reading IF current were flowing through the element.

(Remember, only current “flows”. Voltage is just “present.”)

We also do some measurements from L1 and L2 wrt N, just using a neutral point somewhere as reference to see if we have any potential voltage.

[Susan Brown]

So, since there is no current (despite the presence of voltage), we know there is an open in the circuit.

The element has continuity, so what would you conclude?

[Susan Brown]

P.S. We haven’t taught a lot about what “voltage drop” is yet at this point in the course. We’ll be teaching that more as we go along!

[Susan Brown]

There are several different equations for P. How you use them depends on the scenario and what you are trying to find.

The “E” in the equations is the voltage drop of the resistance that you are trying to find the heat for. (NOTE: we haven’t yet taught a lot about voltage drop, but you’ll keep learning more about this as we go along.)

Our scenario is two resistances in series with each other. If you want to just find the heat generated by the entire circuit, then you could do
P = E squared over R
Where E is the source voltage (because the total voltage drop will = source voltage) and R is the total resistance in the circuit.

But if you only want the heat generated by one of the resistances, then E has to be the voltage drop across just that resistance, and of course R = the ohms of that load.

That’s doable for this scenario, but the technique we show in the video at the end of the unit is easier. Find the circuit current first (using the total resistance in the circuit, because that’s what determines current), then P = I squared * R (where R is the resistance of the load you are interested in)

[Susan Brown]

Hi Steven,

I moved your questions to their own topic, just to keep things a little tidier here in the Forums 🙂

For your first question:
There are several different equations for P. How you use them depends on the scenario and what you are trying to find.

The “E” in the equations is the voltage drop of the resistance that you are trying to find the heat for.

Our scenario is two resistances in series with each other. If you want to just find the heat generated by the entire circuit, then you could do
P = E squared over R
Where E is the source voltage (because the total voltage drop will = source voltage) and R is the total resistance in the circuit.

But if you only want the heat generated by one of the resistances, then E has to be the voltage drop across just that resistance, and of course R = the ohms of that load.

That’s doable for question 11 in the Module 3 quiz, but I think the technique we show in the video at the end of the unit is easier. Find the circuit current first (using the total resistance in the circuit, because that’s what determines current), then P = I squared * R (where R is the resistance of the load you are interested in)

For your second question:

We are assuming that the element is the only load in this circuit. We give you E (voltage) and P (watts). We want you to find R (what the resistance of that element must be). Which equation will you use?

[Susan Brown]

Hi Darren,
Always nice to hear from you!
No, we don’t have any plans for a dishwasher course. There just really isn’t enough unique technology beyond what we already teach in the other courses. Same thing with water heaters (very little demand for that – I think that’s mostly the domain of plumbers).

We’re good, thanks! Hope your side effects were minimal.

[Susan Brown]

that’s correct – there is no current flowing in the circuit. We know that there’s some voltage present, so what is wrong with the circuit?

(Think about what is necessary for current flow – voltage and a ____ circuit)

[Susan Brown]

It’s a little easier to help you with this if I know what your answer is for Part 1. Based on Figure 1, what can you say about the general nature of the fault in the circuit?

[Susan Brown]

Hi Jeremy,

Read the section near the end of Unit 7 about Loading vs. Non-loading Meters

https://my.mastersamuraitech.com/module-3/basic-electricity-electrical-measurements-in-appliance-repair/

Short answer: you want to use a loading meter or a DMM that has the LoZ function to measure AC voltage.

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