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Susan Brown

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Viewing 15 posts - 1,366 through 1,380 (of 1,889 total)
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  • in reply to: Voltage drop #16892
    Susan Brown
    Keymaster

      Hi Josh,

      Thanks for posting a question!

      A great video to review is the one at the end of Unit 3 in Basic Electricity. The one about the heat produced by a loose connection. Watch that and see if you can duplicate the calculation on your own. If there are any steps that you don’t understand, let me know here and we’ll discuss it further.

      Let me know if that helps.

      in reply to: Service Manual Wiring Diagram #16867
      Susan Brown
      Keymaster

        Hi Mike,

        Fundamentals has taught you the basics of reading schematics. But that’s like learning to read music. You will need to practice in order to get more “fluent”.

        Oven & Range Repair and Advanced Troubleshooting have the most practice on schematics compared to the Refrigerator course.

        Then, once you’re done with the courses, many of our webinar recordings over at Appliantology give you more exposure.

        Susan Brown
        Keymaster

          Hi Josiah,

          Did you see the explanation that showed up after you took the quiz?

          Although it is true that a heating element is composed of high-resistance wire, this is not what makes it a load. For example, a heating element in a box sitting on the shelf is also made of high resistance wire but it is not a load because it isn’t in a circuit with current flowing through it.

          A load in the context of appliance repair is:
          – technically speaking, a component that does work – produces motion, heat, light, etc. – when it gets power
          – electrically speaking, a component that produces a voltage drop when current flows through it.

          Susan Brown
          Keymaster

            It’s just a humorous one-question quiz. 🙂

            in reply to: Schematic Reading and Hand-on practice #16798
            Susan Brown
            Keymaster

              Hi Mike,

              It’s not so much of what type of learner you are, it’s more that you just need to practice. That’s true for everyone who takes our courses. You have to practice to really get everything to click and to gain proficiency.

              We suggest starting with your own appliances. Get the tech sheets and compare them with what you see. Look for free or cheap appliances that are banged up or even broken (Craigslist, town dump, etc.). Use Appliantology to get the tech documents and start playing around.

              The only way to post an image directly in these Forums is if the image is on a website and you can give the link. You can also just email the photo to us and we can upload it here.

              in reply to: Relationship between Voltage and Current #16750
              Susan Brown
              Keymaster

                Just did!

                in reply to: Relationship between Voltage and Current #16734
                Susan Brown
                Keymaster

                  Hi Mike,

                  Good question! If you re-listen to that part of the video, Samurai mentions that the power company is trying to deliver a certain amount of power along the lines. And we know that Power = Current x Voltage (P=IxE). So if you have a certain goal for the amount of power you want to deliver, the higher your voltage, the lower the current needs to be.

                  But you are correct that if we are looking at a circuit, and it’s got a fixed resistance, if you changed the voltage supply from a 120vac to a 240vac, then the current would increase according to I = E/R.

                  But, say you need to generate 1000 watts of power. Play around with the numbers, using P=IxE. What would the current need to be if you used a 120 vac supply and then a 240vac supply? You’d need half as much current if you use 240vac.

                  Make sense? There are various “moving parts” to creating the electrical situation that an engineer wants in a circuit: power, voltage, current, and resistance. And they all act according to Ohm’s Law equations!

                  in reply to: Common math expressions used #16727
                  Susan Brown
                  Keymaster

                    That will work ONLY if you have 2 resistances in parallel.

                    If you take 1/(1/R1 + 1/R2) you can rearrange it to equal (R1xR2)/(R1+R2)

                    This does NOT work for 3 or more loads in parallel.

                    in reply to: Common math expressions used #16724
                    Susan Brown
                    Keymaster

                      Thanks, Gaspare! I was just getting on here to link you guys to the previous threads on Equivalent Resistance.

                      Jonas – be sure to click and read that thread above. And in that thread, you’ll see a link to another one where we break down the calculation for you.

                      Let me know if you have any other questions.

                      in reply to: Module 4, Unit 3, Quiz Quest #6 – Clarification Required #16710
                      Susan Brown
                      Keymaster

                        Hi Raymond,

                        Those two statements are actually saying the same things. We just changed the wording slightly for the sake of being a little challenging.

                        The main point is the direction of electron movement. Both cases indicate moving from the cathode end to the anode end and not vice versa. This makes sense, given that the voltage is more positive at the anode end than the cathode end, and electrons move towards more positive charge.

                        Hope that helps!

                        in reply to: Short or Shunt? #16709
                        Susan Brown
                        Keymaster

                          “Shunt” is correct in that scenario.

                          Or does shunt lead to another load, and that’s what makes a shunt (vs. a short)?

                          That’s more like it. A shunt will be deliberately used to bypass one or more loads, but there will still be a load in the circuit somewhere.

                          A short creates a condition where there is no longer a load in a path from line to neutral/ground, so there’s nothing to hinder the flow of electrons, leading to a failure of some type (blowing a fuse, burnt wire, etc.).

                          in reply to: Common math expressions used #16706
                          Susan Brown
                          Keymaster

                            Hi Jonas,

                            Thanks for posting a question!

                            Yes, the * symbol is the same thing as x . Both are “times”, or multiplication.

                            You might have just made this up as an example, but I want to make sure you know that V=W*R is not a valid Ohm’s Law equation! It would be V= I * R (voltage equals current times resistance)

                            in reply to: Resistance in series circuit #16608
                            Susan Brown
                            Keymaster

                              Hi Abbey,

                              Thanks for posting a question! We show this in several examples in the Basic Electricity module, but just to be clear: when you have loads in series in a circuit, the total circuit resistance is simply the sum of the resistances of the loads.

                              Again, this is a series circuit only. Loads in parallel are different. We can calculate an equivalent resistance with those, which we describe more in Unit 5, and also in this Forum thread:
                              https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-in-parallel-circuits/

                              in reply to: Basic Electricity: Voltage, Current, Resistance, and Power #16590
                              Susan Brown
                              Keymaster

                                Hi Abbey,

                                It all depends on what information you are given.

                                If you look on the Ohm’s Law “Pie chart”, there are 3 equations that can be used to calculate current (I). You choose the one that best fits the scenario you have.

                                What info about the circuit do you have?

                                in reply to: Midterm question 8 #16584
                                Susan Brown
                                Keymaster

                                  Hi Desert,

                                  The first step on this question is to figure out how the loads are laid out in relation to the others. Are they in parallel? Series? Are all loads receiving current? (that’s an important one!)

                                  The “Zen trick” is a great help in figuring this out.

                                  let me know what you think.

                                Viewing 15 posts - 1,366 through 1,380 (of 1,889 total)