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Susan Brown

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  • in reply to: Current and voltage #18674
    Susan Brown
    Keymaster

      There’s a lot of complexity to the relationships between voltage, current, resistance, and power, but we’re trying to present it as simply as possible – just what you need to know to help you be a strong electrical troubleshooter.

      Which Ohm’s Law formula you use depends on what you are dealing with.

      For example, in a circuit, the current is I = E/R.

      Higher resistance will mean lower current. Inversely proportional.
      Higher voltage will mean higher current. Directly proportional.

      The way the formula is written shows you that.

      If you want to calculate the voltage drop across a load, and you know the circuit current and resistance of the load, then you’d use E = I x R.

      When you are talking about power (watts), then you might be using P = I x E. For example, if you know that a single load in a 120v circuit is supposed to generate 120 watts of heat, then you would calculate that there should be 1 amp of current in that circuit.

      This module takes more effort than most to master, but it will really help you down the road.

      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

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