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Susan Brown

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Viewing 15 posts - 1,741 through 1,755 (of 1,968 total)
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  • in reply to: Midterm question 8 #13966
    Susan Brown
    Keymaster

      In order for loads to be in series, there cannot be any alternate path that electrons can take. Is that the case with the booster and the main, based on what you just said above?

      in reply to: Midterm question 8 #13965
      Susan Brown
      Keymaster

        Oops – my “yes” was to your first response

        in reply to: Midterm question 8 #13964
        Susan Brown
        Keymaster

          Yes! And does the same thing happen if you are the ignitor? And if so, why would any current flow through the main coil?

          in reply to: Midterm question 8 #13961
          Susan Brown
          Keymaster

            Okay – that is where your mistake is.

            There is something you aren’t seeing that affects one of the loads.

            Try doing the Zen trick on the Booster. Become the booster. Imagine you are reaching out for L1 and N with your hands. Can you get there directly or do you have to go through any other loads?

            in reply to: Midterm question 8 #13959
            Susan Brown
            Keymaster

              No- please answer the question I asked. If you don’t know the answer, just say so. I’m trying to step you through the thought process you need to go through to get to the correct answers

              in reply to: midterm question 5 #13957
              Susan Brown
              Keymaster

                It is given clearly in unit 5. The rule of thumb, which is perhaps more important to know, is also shown on the third video. We also asked you a quiz question about it. Please look at that section of Unit 5, and let me know if something isn’t clear to you

                in reply to: Midterm question 8 #13956
                Susan Brown
                Keymaster

                  Voltage drop occurs when current flows though a load.

                  The first question you need to ask yourself is: Do all 4 loads have current flowing through them, as the circuits are shown in the diagram? (With the detector switch closed) Please just answer that question for me

                  in reply to: Midterm: Voltage Drop #13949
                  Susan Brown
                  Keymaster

                    Hi Philip,

                    Yes, that is correct!

                    in reply to: Midterm question 8 #13947
                    Susan Brown
                    Keymaster

                      That is not correct, which is why I asked how you arrived at it. If I know what your thinking is, then I can see where your mistake is and help you better.

                      Do you think that the main coil is in series with some of the other loads? If so, then it would have to share the voltage drop with them. That would mean any loads it was in series with would not be able to drop 120vac.

                      Normally, to calculate voltage drop across loads in series, you would need to know the resistance of each load. Like you did in Question 4 on the midterm. But we don’t give you any resistances for Question 8, and we still say that you can give us a numerical answer for each voltage drop.

                      The key is accurately seeing the layout of the loads, given the state of the circuits as shown in the diagram.

                      A first question to ask yourself is: Do all of the loads receive current?

                      Let me know what you think.

                      in reply to: Midterm question 8 #13945
                      Susan Brown
                      Keymaster

                        and the booster is 30v

                        You mentioned the booster twice. Did you mean to say the main coil here?

                        What is your explanation for the 30vac?

                        in reply to: Midterm issues question #5 #13943
                        Susan Brown
                        Keymaster

                          By the way, if you watch that video I asked you to, he addresses the danger of thinking about “the path of least resistance.”

                          in reply to: Midterm issues question #5 #13942
                          Susan Brown
                          Keymaster

                            That’s the formula, and I can help you figure out how to use it correctly in a little bit. But first…

                            The rule of thumb is just putting into words what the equivalent resistance of parallel loads will be compared to the resistance values of the individual loads. Something like “the equivalent resistance will always be larger than the biggest load”. (that is not necessarily the answer – just an example)

                            This is arguably more important to know, because you can do quick evaluations of circuits using this rule of thumb that will help you as you’re analyzing a schematic.

                            in reply to: Midterm issues question #5 #13939
                            Susan Brown
                            Keymaster

                              First of all, what is the rule of thumb for equivalent resistance of parallel circuits? (It is explicitly stated in the video on equivalent resistance in Mod. 3, unit 5.)

                              in reply to: Fundamentals mid-term exam #13925
                              Susan Brown
                              Keymaster

                                Hi Josiah,

                                Yes – you have to score 90% or higher on EACH exam in the course. The Midterm is perhaps the most important one of all! It really pushes students to help them understand important electrical concepts. I’ll be grading yours soon and will email you.

                                in reply to: Unit 2 Quiz – need help on question 4 #13919
                                Susan Brown
                                Keymaster

                                  Correct. The distance between each pulse (the “square waves”) will vary to do things like control the speed of a motor.

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