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Susan Brown

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Viewing 15 posts - 1,681 through 1,695 (of 1,926 total)
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  • in reply to: 1.3 Fundamentals-Basic Electricity unit 3 #14044
    Susan Brown
    Keymaster

      Hi Matt,

      First ask yourself a question: is a fuse a load? Remember, pretty much any component in a circuit is either a switch or a load. What is a fuse?

      in reply to: Module 3 Unit 3 Ohm`s law #14031
      Susan Brown
      Keymaster

        Hi Bryan,

        The material we put in our courses is what we know, from many years of experience, to be the foundational information and skills that a capable appliance tech needs to have. In order to be able to troubleshoot electrical systems, you not only have to be able to read a schematic and devise a testing strategy, then you have to interpret the measurements you get – which can sometimes be tricky.

        To do this well, you have to have a really good feel for basic circuits and electricity. Ohms Law simply describes how the electrical properties that you need to understand relate to each other. So that’s the main way it’s useful to an appliance tech – to first learn about electricity, then to have something to use when needed to interpret the various measurements you’re taking with your meter(s). There’s no other way to have a solid, functional understanding of how electricity works.

        So, its initial usefulness is as an important part of the learning process. An experienced tech who has a good understanding of circuits and electricity will no longer have to do a lot of actual calculations using Ohm’s Law equations, but they still can come in handy now and then when you encounter an unusual head-scratcher type of a scenario.

        Hope that helps!

        in reply to: dishwasher trick-out tool #14024
        Susan Brown
        Keymaster

          Ah – that is something we picked up at a manufacturer’s training. From Miele, if my memory serves. I don’t know how to obtain one. Are you Miele authorized?

          in reply to: Module 3 Unit 8 Voltage drop #14020
          Susan Brown
          Keymaster

            I’m glad you love this stuff! And you are getting close.

            This is great because we have identified one area of misunderstanding you have about current in a series circuit.

            If you need to, go back and review a little bit of our instruction about current – how electrons flow. In a circuit, if one electron is moving, they are all moving. And all at the same rate. Picture a line of billiard balls that are all touching each other, with no extra space anywhere. If you push the one on the end, they will all move at the same rate. If you stop the one on the front end, they all stop. Electrons behave in that kind of way. What that rate (current flow) is will be determined by the total resistance of the circuit, not by the individual load they are going through at a particular moment in time.

            Again – this is for loads in series, not parallel.

            Look at the equation you gave in 4.A. I = E / (R1+R2). You are saying there that the current is the voltage divided by the total resistance. That math equation is showing what I said above, basically.

            So the circuit current is 6.48 amps, at both R1 AND R2.

            Let me know if that makes sense to you.

            in reply to: Module 3 Unit 8 Voltage drop #14018
            Susan Brown
            Keymaster

              Good to hear back from you!

              Good answers so far.

              3. Does current vary at different points along a series circuit, or will it always be the same?

              4. Based on that answer, how do you calculate current in the circuit?

              in reply to: Module 3 Unit 8 Voltage drop #14016
              Susan Brown
              Keymaster

                Hi Mike,

                Glad to help!

                (Note that I changed the title of your topic to the unit where you scored a 70%.)

                This is THE most common unit in this module for students to struggle with, so don’t feel bad.

                First of all, look at Question 6. The thing that is wrong about the answer you chose is not the number. We were asking for the reading on an ammeter (also sometimes called an amp-meter). Does that mean we were looking for voltage or current?

                Then for 7 and 8, we’ve got two loads in a series. I’m going to step you through this a bit with questions.

                1. What is the source voltage for this circuit?
                2. What do you know about the behavior of current in a series circuit?

                Answer those and then we’ll continue on.

                in reply to: dishwasher trick-out tool #14013
                Susan Brown
                Keymaster

                  Hi David,

                  It’s been awhile since I’ve watched those videos – I’m assuming this is something mentioned there. Are you talking about the tool for Miele dishwashers?

                  in reply to: Basic Refrigerator Troubleshooting Strategies #13996
                  Susan Brown
                  Keymaster

                    Hi there – sorry about that. The recording is available to you here:

                    https://mastersamuraitech.com/webinar-recording-basic-refrigerator-troubleshooting/

                    in reply to: Section 7.2 Swtiches, etc. #13991
                    Susan Brown
                    Keymaster

                      Hi Phil,

                      Here’s what I do. After I’ve watched it through once, I restart the video and pause at each slide to take notes. You can click ahead in the progress bar to advance to the next slide(s) without actually having to listen to all of it again, unless there are parts you didn’t quite catch the first time.

                      Hope that helps!

                      in reply to: Midterm question 7 #13986
                      Susan Brown
                      Keymaster

                        That is correct! The one branch has a failure, of course, and current will stop in that one, which decreases the overall current draw. But the branch parallel to it will be unaffected.

                        in reply to: midterm question 5 #13983
                        Susan Brown
                        Keymaster

                          Yes, that’s correct. (I’m going to hide your answer, by the way.)

                          The equivalent resistance will always be less than the smallest resistance of the individual loads.

                          in reply to: midterm question 5 #13981
                          Susan Brown
                          Keymaster

                            Here it is embedded for you:

                            https://www.youtube.com/embed/iHB3lxdc68E

                             

                            in reply to: midterm question 5 #13980
                            Susan Brown
                            Keymaster

                              That’s why we also want you to know the rule of thumb – not everyone likes to work with fractions.

                              Watch this video. The resistance numbers are not identical, but follow along and see if you can then come up with where your mistake is:

                              https://youtu.be/iHB3lxdc68E

                              in reply to: Midterm question 8 #13979
                              Susan Brown
                              Keymaster

                                You’re welcome! That’s what I’m here for 🙂

                                in reply to: Midterm question 8 #13976
                                Susan Brown
                                Keymaster

                                  its being shunted which means its receiving current just not 120v of current correct each other load is receiving right?

                                  Let’s make sure you are keeping current and voltage straight.

                                  Current (in “amps”) is electrons flowing through a closed circuit.

                                  Voltage (in “volts”) is the pressure, or charge, that gives the electrons the push to want to move in the circuit. You can have a lot of potential voltage, but if there isn’t a complete circuit, current won’t flow.

                                  Voltage drop is a measurable voltage that is created by current flowing through a load.

                                  In these circuits, with the detector switch closed, the main coil is shunted. If there is a path available to electrons with no resistance (no load), they will take that path 100% of the time. That is a shunt. This is what is happening here. And it results in the booster and the ignitor simply being in parallel with each other, and not in series with anything. This is why we can know the voltage drops for all of the loads in this scenario without having to actually calculate them.

                                  By the way, when that detector switch is open, then the main coil is in series with both the ignitor and the booster, and you’d have a series-parallel situation.

                                  Take one more look at the circuits and think it through one more time to make sure you really get it. Being able to see what is going on in a basic schematic is CRITICAL for good troubleshooting.

                                  since the main coil is shunted off the answer for the mail coil will be 0v correct?

                                  Correct

                                Viewing 15 posts - 1,681 through 1,695 (of 1,926 total)