Susan Brown

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  • in reply to: Ohms #18673
    Susan Brown
    Keymaster

      Yes, if loads are in series, the total resistance of that circuit is just the sum of the resistances of the loads.

      That calculation is correct.

      For Questions 7 and 8, review the video at the end of Unit 3 about the heat generated by the loose connection. Very similar scenario.

      in reply to: Current and voltage #18668
      Susan Brown
      Keymaster

        stumped?

        Nope! We get asked this question periodically. You happened to ask this on a Saturday night… we work a lot, but we do take breaks here and there!

        in reply to: Current and voltage #18667
        Susan Brown
        Keymaster

          The power company is supplying power.

          P = I x E.

          So, if your goal is to deliver a certain amount of wattage, then you can see that a higher voltage will mean a lower current.

          If you want to have 1000 watts, for example, you could have 100 volts and 10 amps, or you could have 200 volts and 5 amps, etc.

          When you are looking at the voltage drop in a circuit, that’s when we use E = I x R, and current is directly proportional to voltage.

          in reply to: CS2: Load of Interest Quiz #18664
          Susan Brown
          Keymaster

            Yes, surface control. The L2-H2 contacts have to be closed as well for current to go through the element.

            in reply to: quiz reset? #18663
            Susan Brown
            Keymaster

              1. We don’t base Certification on the overall course grade, which is an average.
              2. Certification is based on EACH individual unit quiz being 80% or higher. Yours are all 90% or higher at this point, so you are fine so far.
              3. The exams in the course (Module, Midterm, Final) have to be 90% or higher on EACH one.

              So far, so good!

              in reply to: CS2: Load of Interest Quiz #18656
              Susan Brown
              Keymaster

                For the element to get hot, it needs both L1 and L2. If the L1 to H1 contacts are stuck closed, does that mean that the element would also get L2?

                in reply to: Just curious #18655
                Susan Brown
                Keymaster

                  Good question!

                  The electrons will move IF they are in a conductive material (AND have a complete circuit from a higher charge to a lower one).

                  Copper wire (like most metals) is very conductive. The electrons move pretty readily in it. Other materials are not at all conductive, such as rubber.

                  If you only touch the insulated part of the wire, that protects you. That insulation prevents electron flow. If you touch a bit of bare wire that is live, however, you can get shocked.

                  in reply to: I disagree with Refrigerator Mid-Term Answer #18653
                  Susan Brown
                  Keymaster

                    Hi Steve,

                    Good job working to understand all of the questions and answers, even if you passed the exam!

                    You’ve got most of the correct idea – the condenser does remove enough heat to condense the refrigerant into a liquid. However, it remains a hot liquid, not cool.

                    in reply to: req #18633
                    Susan Brown
                    Keymaster

                      Hi Shaya,

                      This video may help you see how to calculate equivalent resistance in parallel circuits:

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

                       

                      in reply to: Module 3 , unit 2 Auto dry analysis #18602
                      Susan Brown
                      Keymaster

                        No worries!

                        in reply to: Module 3 unit 2 #18600
                        Susan Brown
                        Keymaster

                          Hi Kris,

                          14,400 watts should definitely sound wrong to you – think of how hot a 100 watt lightbulb feels!

                          Here’s why your calculation didn’t work: there is another load in that circuit – the element.

                          In the P = E^2/R formula, E is the voltage dropped across the load. It is only equal to the supply voltage if there is just one load in the circuit. In the loose connection scenario, you have two resistances in series.

                          So, if you want to use this formula for P, you have to calculate the voltage drop across the loose connection first.

                          Alternatively, and this is what we show in the loose connection video in Unit 3, you can use P = I^2 x R. This means you have to calculate the circuit current first, but that’s pretty straightforward (I = E/R, where E is source voltage and R is total resistance in circuit).

                          Susan Brown
                          Keymaster

                            The Final Exam covers material from the entire course. It’s a good time to do a quick review of each unit to refresh your memory, and skim through your quizzes. If there are any lingering areas that you don’t feel strong on, ask us for help!

                            in reply to: question 18 #18555
                            Susan Brown
                            Keymaster

                              Hi Ed,
                              I just emailed you about this…

                              in reply to: Final Exam Q. 7 #18487
                              Susan Brown
                              Keymaster

                                A short or an open? These terms have precise meanings.

                                From Mod 3, unit 1:

                                Short: A condition that creates a circuit that offers no resistance to the current flowing through it. A direct short will cause a fuse to blow, or possibly start an electrical fire.

                                Whereas an open is just that – some break in the circuit (open switch, broken wire, for example) that prevents current.

                                If the fault is on the L1 side, how do you explain this measurement?

                                You use your meter to test for 120 V AC from one of the terminals of the light to neutral and you read 120 V AC on your meter

                                in reply to: Fundamentals Overall Question #18478
                                Susan Brown
                                Keymaster

                                  Hi Darren,
                                  The Final Exam covers material from the entire course.
                                  🙂

                                Viewing 15 posts - 1,351 through 1,365 (of 2,007 total)