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Susan Brown

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Viewing 15 posts - 211 through 225 (of 1,968 total)
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  • in reply to: Using Schematics to Troubleshoot Appliances, Part 1 #25581
    Susan Brown
    Keymaster

      What a coincidence – I was just asking the Samurai this question myself yesterday!

      It is definitely not neutral, neither is it line voltage. But it is “hot”. In other words, you would measure some voltage in between the two loads (with respect to neutral).

      As an example, if the two loads had the same resistance, then you would measure 60v wrt N in between the loads, because each one would drop 60v.

      Make sense?

      in reply to: Module 6 Unit 7 #25579
      Susan Brown
      Keymaster

        Hi Anthony,
        Just to make sure we understand your question – are you asking if the explanation (“Both the main and sub PCBs would be replaced because 1) the Fast Track says to replace the PCB *assembly* which means BOTH boards, if they meant just the main PCB, they would have said that”) comes from somewhere on the Fast Track, or just from our general knowledge of these things?

        in reply to: Basic Electricity MOD 4. Unit 6 #25578
        Susan Brown
        Keymaster

          Hi Mitchell,

          Most of our videos are hosted on Vimeo, but a few, like these, are YouTube videos. They work fine for me right now, so I assume you were having some temporary issue with those.

          See the Video Playback section on this page for more suggestions to try, if they still aren’t playing for you.

          https://my.mastersamuraitech.com/appliance-repair-course-support/student-resource-page/#video-playback

          Let me know if you continue to have this issue, and I’ll make sure you get the video links another way.

          in reply to: module 9 unit 2 quiz question #15 #25571
          Susan Brown
          Keymaster

            Yes, 2 is the correct answer. It is great to see you working on understanding the questions you missed, even though you “passed.” Not all students do that.

            in reply to: Can i move back and forth between courses? #25569
            Susan Brown
            Keymaster

              Hi Michael,
              Questions like that you can always email me about, FYI. But this is fine, too!

              MST Courses are like parallel circuits – what happens in one does not affect the others 😀

              The only thing you risk by taking a break is needing to spend more time refreshing for the Final Exam. However – that is actually a better method in terms of retaining the info. And hopefully you’ve been taking good notes to help you with this.

              in reply to: Voltage drop #25563
              Susan Brown
              Keymaster

                Hi Dylan,

                I’ll start by just telling you how I came to understand it, to see if it helps.

                All voltage is a difference in charge between two points. For our purposes (dealing with circuits in appliances), there are two categories of voltage.

                1. Voltage potential – the difference in charge between two points where there is no current flowing, but there is the potential for current to flow if there is a closed circuit between those two points. A common measurement of this type of voltage is putting one probe at one side of a load, and the other at a good neutral point. Or, measuring across an open switch.

                2. Voltage drop – the difference in charge from one side of a load to the other, created by current flowing through the load. In other words, you will only measure voltage drop in an active circuit.

                “Dissipation” sounds more like an idea from power transmission, where you can have some losses due to small amounts of resistance over very long stretches of power lines. Small amounts of the power are transformed into heat and lost to the environment. This is not something we deal with in appliance circuits, which are very small compared to transmission lines.

                Here’s something else to keep in mind. In circuits, there are always two sides to every power supply. Either L and N, or L1 and L2. These two sides meet at loads.

                Switches, on the other hand, are always on just one side of the power supply (otherwise they would create a short).

                Let me know if this helps, or if you have any follow up questions.

                • This reply was modified 1 year, 2 months ago by Susan Brown.
                in reply to: Midterm Exam for Core Q#8 #25559
                Susan Brown
                Keymaster

                  Hi Douglas,

                  That is correct. The closed detector switch is a [answer hidden]
                  Good job!
                  Note – I will need to hide these answers.

                  • This reply was modified 1 year, 2 months ago by Susan Brown.
                  in reply to: Module 4 – Unit 5 quiz question #18 parallel circuit #25555
                  Susan Brown
                  Keymaster

                    Great! Keep up the good work.

                    in reply to: Module 4 – Unit 5 quiz question #18 parallel circuit #25553
                    Susan Brown
                    Keymaster

                      Hi Edwin,

                      I love to see students playing around with these calculations in order to understand!

                      Your mistake is in how you figured out the current in each circuit – trying to use proportions. Remember, current is inversely proportional to resistance. Did you notice that you were getting higher current flows in the circuits with higher resistances? That is not how it works. I = E/R

                      The voltage supply to each parallel circuit is 120v. This fact is true regardless of the state of any circuits in parallel to it. So the voltage drop across these single loads will always be 120v if they are active.

                      The resistance of each load doesn’t change – it is a fixed quantity. So, how can the current in each circuit change due to the failure of a circuit in parallel?

                      Let’s look at the calculations.

                      Given your scenario, with three parallel circuits with R1 = 100 ohms, R2 = 200 ohms, and R3 = 300 ohms, Req = 55 ohms, Itotal = 2.2 amps.

                      I1 = 120/100 = 1.2 amps
                      I2 = 120/200 = 0.6 amps
                      I3 = 120/300 = 0.4 amps

                      Totals up to 2.2 amps.
                      Notice that I1 is 3 times bigger than I3, which makes sense because R1 is three times *smaller* than R3. (Current is inversely proportional to Resistance.)

                      If you remove R2, you get Req = 75 ohms, Itotal = 1.6 amps
                      You can see from above that I1 + I3 = 1.2 + 0.4 = 1.6 amps

                      Same thing if you only have R2 and R3 active, It = 1 amp, which is what you get if you add I2 + I3.

                      Does that make sense?

                      in reply to: Mod 6 unit 3 #25550
                      Susan Brown
                      Keymaster

                        Hi Anthony,

                        The correct answer is “serial data transfer”, which is a type of digital data communication. You are not actually wrong, but we want the better, more specific terminology for the answer here.

                        We realize the distinction is pretty subtle, so we decided to take away “digital” as one of the answer choices.

                        in reply to: Midterm exam ( question #8 ) #25543
                        Susan Brown
                        Keymaster

                          Hi Tiago,
                          All you need to know is that the detector switch is closed. Like any other closed switch, it acts just like a wire, with no resistance. So the question you ask yourself is: if you do the “Zen trick” on the Ignitor or the Booster, how will you reach N? Through the closed switch, the main coil, or both? Review the first video in Unit 5 for a refresher on the Zen trick and the effect of a pathway like this closed switch on circuits.

                          in reply to: Module 3 unit 4 #25537
                          Susan Brown
                          Keymaster

                            Hi Travis,

                            The Advanced Courses assume that students have taken the Core course, so there may be questions that draw on fundamental topics like electricity and circuits.

                            But most of the time quiz questions are drawn from the material in the unit at hand.

                            For example, the question “What are two common operating voltages for door locks in computer-controlled washers? (Mark both correct answers)”
                            In the first video, we show different examples. At least one is using 120v, and the other 12vdc. (5vdc is just data, not operation).

                            For any of the questions you are stumped on, you can ask us for guidance. Just post it here.

                            Susan Brown
                            Keymaster

                              Question #9: The heater relay on the heater PCB switches _____ to the heater.

                              in reply to: Mod 8 unit 1 #25513
                              Susan Brown
                              Keymaster

                                Hi William,

                                The first sentence of the question has important technical terminology: You’re testing a micro switch in a live, 12 V DC circuit. That tells you all you need to know – the switch is in a circuit which will have at least one load.

                                in reply to: Module 7 – Unit 3 #25506
                                Susan Brown
                                Keymaster

                                  Okay – just did that for you.

                                Viewing 15 posts - 211 through 225 (of 1,968 total)