Susan Brown

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  • in reply to: Question 4. #15328
    Susan Brown
    Keymaster

      You’re welcome. And glad to hear it! That’s what we’re trying to do with the Midterm 🙂

      in reply to: Question 4. #15322
      Susan Brown
      Keymaster

        Exactly. The circuit current is the same throughout a series circuit, and will be determined by the total resistance in the circuit.

        The current flowing through each load will produce a voltage drop in proportion to the resistance of the load, with all of the voltage drops adding up to equal the source voltage.

        (Note: I will hide part of your answer so we aren’t just giving it away to other students!)

        in reply to: Calculating heat produced across a load #15320
        Susan Brown
        Keymaster

          Hi Kenneth,

          Good question – I moved it so that it could be its own topic.

          First of all, we do have a video showing the calculation of something very similar to this – the heat produced by a loose connection – the last video in Unit 3 of Basic Electricity.

          If that doesn’t clear it up for you, tell me what Ohm’s Law equation you are trying to use to calculate the heat produced by that load.

          in reply to: midterm exam question #15317
          Susan Brown
          Keymaster

            Correct – The fan motor circuit is parallel to the element circuit, thus unaffected by its failure.

            Also basically correct about the element branch – current will stop when it fails open, since there’s no longer a closed circuit.

            Do you know what the final answer is – what happens to the total current draw from L1?

            in reply to: Midterm Question 8 #15316
            Susan Brown
            Keymaster

              Not quite. Read what Michael said about the main above. And do the “Zen trick” on the booster or ignitor – what do you find?

              in reply to: Question 4. #15315
              Susan Brown
              Keymaster

                Sort of! But you’ve got a mistake. The source voltage is 120vac. All of the loads combined must drop 120 – as you knew in your first response. So, calculating 120 as the drop across just one of the loads can’t be right.

                The correct formula for voltage drop is E = I x R. We give you each resistance, so you’ve got to figure out what to plug in for the circuit current. (And what you just did isn’t correct.)

                Does current vary throughout the series circuit or is it the same?

                (And – for another hint – look at what you did for Question 3 on the Midterm.)

                in reply to: Module 3 unit 5 question #18 #15258
                Susan Brown
                Keymaster

                  Hi Gilbert,

                  No, 0.013 is not correct. You would expect to get something less than 103, obviously, but not *that* much less!

                  However, 1/.013 does get you the correct answer, which is around 78 ohms.

                  Check out this previous forum topic where we show a similar calculation in more detail, and see if this helps:
                  https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

                  Let me know!

                  in reply to: Unit conversion #15242
                  Susan Brown
                  Keymaster

                    “Mega-” is a million of something. So, 1 Mega-watt is 1 million watts (1,000,000 watts). But you can change it to any other unit of measure, too. A “Giga-” is 1 billion (1,000,000,000) of something. That’s why converting a mega-watt to a giga-watt results in a smaller number. 1 mega-watt is 0.001 giga-watt. Is that what you wanted to know?

                    in reply to: What do loads need to operate? #15238
                    Susan Brown
                    Keymaster

                      Hi Terry,

                      This can seem like a fine point, but we do emphasize (and want to encourage) techs to think in terms of “power” when it comes to loads doing work. The specs for loads are often given in watts (which is voltage times current). For a load to do its work properly, there must be the correct amount of both voltage and current to create the necessary wattage. (So, the correct answer was “power”.)

                      in reply to: Unit conversion #15234
                      Susan Brown
                      Keymaster

                        Hi Kenneth,

                        What is your question about 4 megawatts? Are you trying to convert it to something different?

                        in reply to: midterm exam question #15229
                        Susan Brown
                        Keymaster

                          Hi Joshua,

                          No point in googling other resources when you’ve got the ultimate resource right here – Team Samurai! A lot of what your tuition pays for is the ability to ask us questions here in the Forums.

                          We need to address these questions one at a time. Let’s start here with question 7.

                          (By the way – I’m going to hide the question in your post above, so that other students won’t see it before they take the Midterm themselves.)

                          Question 7 has two parallel circuits, and at first we assume everything is behaving normally – current is flowing through both branches.

                          Then, the top circuit (with the element in it) fails open. You gave us the correct answer for what happens to current in that branch. But then we ask what will happen to the voltage drop and current in the fan motor circuit.

                          Do you recall what we taught about parallel circuits, in terms of how a failure in one branch affects another?

                          in reply to: Question 4. #15228
                          Susan Brown
                          Keymaster

                            Hi Joshua,

                            Happy to help! First of all, these loads have different resistances. Will that affect the voltage drop across them? Isn’t there a relationship between resistance and voltage that we taught?

                            in reply to: Unit # 3 question #15224
                            Susan Brown
                            Keymaster

                              Hi Mohamed,

                              We’re talking about the size of the letters or numbers used on a schematic. They tend to use large letters for the various safety warnings (which professional techs are already aware of). But the key information needed for troubleshooting – specifications for the loads, for example – are often in small letters/numbers. In other words, for a professional tech, the most important information is usually the smallest.

                              in reply to: Voltage types #15215
                              Susan Brown
                              Keymaster

                                Hi Robert,

                                Think of their function in the circuit, and that will help you know the answer to this. Also, we teach this in Mod. 3, unit 7 (skim through the text of that unit and you’ll see).

                                in reply to: Basic electricity #15212
                                Susan Brown
                                Keymaster
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