Susan Brown

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  • in reply to: 2nd attempt at the basic electricity exam #26164
    Susan Brown
    Keymaster

      Hi Ethan,
      First of all – you indicated in your student info that you want to earn Certification for the course. Do you recall the requirements for that?

      You must earn 80% or higher on EACH unit quiz, and
      90% or higher on EACH exam in the course.

      So, not only would you need to retake the exam anyway, but there are two unit quizzes that are below 80%. In order to retake those, you would need to get set back to Mod 4, unit 5.

      Let me know!

      in reply to: MidTerm – Retakes #26147
      Susan Brown
      Keymaster

        Hi Michael – I reset it so you can start fresh and answer all the questions.

        in reply to: Question 8 Midterm Exam #26141
        Susan Brown
        Keymaster

          You got it!!!

          (FYI, I’ll hide this answer so we don’t just give it away to other students. They need to experience their own “a-ha!” moment 🙂

          in reply to: First Quiz attempt Wifi Issues #26140
          Susan Brown
          Keymaster

            Hi Victor – I reset that quiz for you.

            in reply to: Midterm Question #8 #26129
            Susan Brown
            Keymaster

              Hi Kenneth,
              Right, because if you only have one load in a circuit, the voltage drop will equal the source voltage.

              The key here is the impact of the closed detector switch on the circuits with the Ignitor, Booster, and Main.

              If you do the “Zen trick” on the Ignitor or the Booster, how do you “reach” N? Through the closed switch, through the Main, or both?

              in reply to: Equivalent Resistance vs. Total Resistance #26125
              Susan Brown
              Keymaster

                Hi Chris,

                “1 over” is the same as “1 divided by”. So, for example, 1/10 is “1 divided by 10”. When you do that on your calculator, you should get 0.1 (“one tenth”)

                1/2 should result in 0.5 (“five tenths”, which is the same as one half)

                So, let’s say you have two 10-ohm resistances in parallel.

                The calculation would be
                1/(1/10 + 1/10) = 1/(0.1 + 0.1) = 1/(0.2) = 1 divided by 0.2 on your calculator = 5 ohms.

                Do you get that when you do it on your calculator?

                What do you get if you try it again but with two 20-ohm resistances in parallel? Let me know!

                in reply to: Final Exam test 2 question 8 #26120
                Susan Brown
                Keymaster

                  It would be similar to a refrigerator – the point is that it’s a cold, damp environment. So the choice of connector and other precaution are important. You had a quiz question about this in that unit.

                  in reply to: Module 3 Unit 2 #26101
                  Susan Brown
                  Keymaster

                    Hi Raja,

                    This is a little Basic Electricity practice.

                    In order to get current to flow through a load, you need a voltage and a complete circuit.

                    And by voltage, we mean a voltage difference. All voltage is expressed as the difference in charge between two points. A classic reading of voltage potential is measuring from L1 with respect to Neutral (where you use a known-good neutral point as reference). Reading voltage drop is when you are measuring across a load (a component with resistance that does work – element, pump, bulb, etc.). Current flowing through a load produces voltage drop.

                    So – if you have L1 on both sides of a load, there is no difference in voltage that will drive current through that circuit. So, no current.

                    in reply to: Midterm Question #8 #25894
                    Susan Brown
                    Keymaster

                      If there is only one load in a circuit, do you know its voltage drop? (think about what Kirchhoff’s law teaches)

                      in reply to: Midterm Question #8 #25892
                      Susan Brown
                      Keymaster

                        Hi Troy,

                        No calculations are needed to answer Question 8, and you definitely don’t need to make something up. The key is seeing the impact that the closed detector switch has on the circuits.

                        Let’s back up a bit – if you look just at the Safety, do you know what its voltage drop is?

                        in reply to: Reset Advance Troubleshooting Unit 3 #25887
                        Susan Brown
                        Keymaster

                          Hi Brian,
                          I reset you. FYI – it’s best to use the Quiz & Exam Reset Request form when needed (in the “Campus Support” menu).

                          ~ Susan

                          in reply to: current and resistance #25851
                          Susan Brown
                          Keymaster

                            That’s correct

                            in reply to: current and resistance #25836
                            Susan Brown
                            Keymaster

                              Also – go to the Core course, Mod 4, unit 4, and watch the second video. I start with a circuit with a single 60 ohm load, then add a 40 ohm load in series. The current in the circuit decreases, and the voltage drop across the 60-ohm load decreases with the addition of the second load. But obviously the 60-ohm load still has 60 ohms.

                              The main point of the question you are writing about is to make sure people understand that resistance is an inherent property of a load and in the types of Ohm’s Law calculations we are likely to do as appliance servicers, if you are given a value for resistance, that value is fixed. It won’t change in response to a change in current, for example.

                              (Note – one exception to this discussion is that some material’s resistance will change some based on temperature. Some types of sensors are an example of this.)

                              • This reply was modified 1 year, 2 months ago by Susan Brown.
                              in reply to: current and resistance #25835
                              Susan Brown
                              Keymaster

                                Yes, if you add another load in series, that changes the total resistance within that series circuit.

                                Which unit is this question in? I want to see it in context to make sure I answer correctly.

                                in reply to: current and resistance #25833
                                Susan Brown
                                Keymaster

                                  The voltage in these equations is the voltage dropped across the load, not the source voltage. So, if a load was originally by itself in a circuit, dropping the source voltage, then another load is brought into series with it (by a switch activating), then the voltage drop across the original load will decrease. The current will also change.

                                Viewing 15 posts - 181 through 195 (of 1,987 total)