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Susan Brown

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  • in reply to: Advance refrigerator unit 5 Video 3 #25289
    Susan Brown
    Keymaster

      I found it – it is actually in Unit 6.

      Unfortunately, that is an older video and we don’t have the schematic for it (or even the model number so we could look it up). So we can’t answer your question definitively. There are no quiz questions related to that, however, and we’ve decided to remove the video from the unit.

      • This reply was modified 1 year, 3 months ago by Susan Brown.
      in reply to: Advance refrigerator unit 5 Video 3 #25285
      Susan Brown
      Keymaster

        Hi Ray – would you please clarify which Module and Unit you are talking about? Thanks!

        in reply to: Module 3 unit 6 question 3 #25271
        Susan Brown
        Keymaster

          Hi Terrance – you got the answers correct on your recent attempt on that quiz – do they make sense to you, or do you still have a question about it?

          in reply to: CS3 LOI #25269
          Susan Brown
          Keymaster

            Hi Matt – we interacted via email about this – are you all set or do you still have questions?

            in reply to: Module 6 unit 4 #25265
            Susan Brown
            Keymaster

              Hi Cooper,
              Did you rewatch the final video in Unit 4, from 46:20 – 50:20. That’s where Samurai reviews when to use them and when not to, and what they look like. Then he goes through a troubleshooting scenario where I know he at least demonstrates jumping a switch.

              If you think about the function of a switch – to complete a circuit (a closed switch is like a wire) – then you can see how a jumper just fills in for that function.

              Similarly, a load needs power to do work. A cheater cord provides power directly to the load to help you troubleshoot – does the load work when you provide power with a cheater cord? If so, then you know the load is not faulty, and vice versa.

              in reply to: Mod 8 unit 1 #25258
              Susan Brown
              Keymaster

                Hi Andrea,

                I just thought I would chime in to review a few basics, to see if this helps you.

                Whether we are talking about an AC or a DC circuit, the following basic concepts of voltage and current are the same.

                The presence of voltage gives us the *potential* for current to flow. If we have a complete (closed) circuit along with voltage, we will have current.

                We cannot have current in one part of a circuit and not another. (You can have circuits in parallel with each other where one has current and another doesn’t, but that is not what we’re talking about here. We’re just talking about a single circuit.)

                Measuring voltage at an outlet or across an open switch are both examples of measuring voltage potential. Measuring voltage across a working load (current going through it) is measuring voltage drop.

                A live test just means that the machine is plugged in and we are turning it on. Voltage is being applied, but if there is an open somewhere, then no current will be flowing.

                Does that help?

                in reply to: Module 8 Unit 2 Schematic Review #25254
                Susan Brown
                Keymaster

                  Hi Brian,

                  I can do this for you now (although it is unit 3, not 2, that needs the reset). Also – Mod. 7, unit 4 is under 80%. You need to use your second attempt to improve that before you move on.

                  For future reference, please use the Quiz & Exam Reset Request form in the “Campus Support” menu when you need a reset. Thanks!

                  in reply to: Refrigerator ice maker basics #25242
                  Susan Brown
                  Keymaster

                    if I had picked a temperature between 8F and 17F I would have been correct, instead of picking one outside of this range as I thought that what the question was asking?

                    Correct.

                    Every model of refrigerator out there has a particular temperature at which the ice maker will stop functioning. If you made a list of what those temperatures are for all the various models, you would see that they all will be somewhere in that range. But for any individual refrigerator, it is a single number. Does that make sense?

                    in reply to: Redo property of refrigerants unit 4 #25236
                    Susan Brown
                    Keymaster

                      Hi Ted – I can set you back to Unit 4. FYI, it will clear the two quizzes you took after that, so you’ll need to retake them. Make notes if you want to, then let me know when you’re ready for the reset.

                      FYI, for future reference, there is a Quiz & Exam Reset Request form in the Campus Support menu. That’s the best way to get a reset.

                      in reply to: Module exam question 17 #25227
                      Susan Brown
                      Keymaster

                        Sorry – I’m not sure what you mean. What would be 32?

                        The total resistance of the circuit is 32 ohms.

                        Did you rewatch the videos in Unit 4? There’s one where I do a some different calculations on loads in series, including the heat generated. And then there’s another where Samurai is finding the heat generated by a loose connection (with slightly different numbers for the resistances)

                        in reply to: Module exam question 17 #25225
                        Susan Brown
                        Keymaster

                          To just find the heat generated by the 4-ohm loose connection, it would have to be [the voltage dropped across the loose connection] squared / 4.

                          You can do it that way, but I think it’s easier to find the current in the circuit, then use I squared x 4.

                          in reply to: Module exam question 17 #25222
                          Susan Brown
                          Keymaster

                            P = E squared / R is a correct formula, but if you are going to use this, E has to be the voltage dropped across the R that you are interested in. Since there are two resistances in this circuit, you can’t use 240 for E and 4 ohms for R. Does that make sense?

                            We suggest using P = I squared x R, and step through a similar calculation in Unit 4 of Basic Electricity.

                            in reply to: Non-invasive method for checking refrigerator #25217
                            Susan Brown
                            Keymaster

                              To measure pressures, you have to tap into the sealed system – that’s what we mean by “invasive”. Temperature and amps don’t require that.

                              in reply to: Non-invasive method for checking refrigerator #25215
                              Susan Brown
                              Keymaster

                                Hi Anthony,

                                This question is referring to the actual measurements we would make to evaluate the sealed system. The point of the non-invasive techniques are to find ways to know the pressure without directly measuring it, which can only be done invasively. The Danfoss app technique you refer to is taking the direct measurement of temperature then combining that with thermodynamic knowledge to determine pressure.

                                It’s kind of like how you can look at a glass of ice water and know that its temperature is right around 32 degrees F. You didn’t actually stick a thermometer in it and directly measure it. But the visual information (ice in a glass with water) plus your knowledge of the characteristics of ice let you determine pretty accurately what the temperature is.

                                Does that make sense?

                                in reply to: Voltage drop #25212
                                Susan Brown
                                Keymaster

                                  Hi David,

                                  Voltage drop is the voltage you measure across a working load. You can calculate the expected voltage drop using E = I x R or E = P/I, depending on what information you have.

                                  So the amount of voltage drop will depend on the amount of current flowing through it. And this is influenced by the presence of any other loads in the circuit or things like variable speed systems.

                                  The Samurai recommends that you watch these Workshops at Appliantology if you haven’t already to help you strengthen your understanding of this important topic!

                                  Voltage, Voltage Drop, Loads & Switches, Jumpers & Cheaters

                                  Schematic Analysis Using Ohm’s Law

                                  Voltage Measurements and More

                                  • This reply was modified 1 year, 5 months ago by Susan Brown.
                                Viewing 15 posts - 196 through 210 (of 1,901 total)