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Susan Brown

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Viewing 15 posts - 1,171 through 1,185 (of 1,968 total)
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  • in reply to: resisters #19314
    Susan Brown
    Keymaster

      Hi Ronny,

      That is all correct except for the conversion at the end.

      A milli ohm is a smaller unit of measure than a regular ohm. Dividing the 250,000 by 1,000 would give you kilo ohms, or k-ohms. You would multiply by 1,000 to get milli-ohms.

      in reply to: Why was question 10 wrong? #19308
      Susan Brown
      Keymaster

        Hi Ted,

        According to my notes voltage is known as potential energy and the Electromotive Force. The force that causes electrons to move from negative to positive. Measured in volts and referred to as V or E.

        That’s correct – voltage is the force that causes the movement. It is not the movement itself.

        The “flow of electrons” is referring to that movement of electrons. That’s what we’re asking for in Question 10. Do you know what we’re asking for now?

        Susan Brown
        Keymaster

          Haha – always good to hear from you!

          Yeah, it looks like our steamer is no longer available. Search Amazon for “handheld steamer” and look at reviews.

          These are steamers used for cleaning, not for clothes.

          Here’s a similar one that I found

          https://amzn.to/2ZeNscU

          in reply to: Midterm #19299
          Susan Brown
          Keymaster

            Did you watch the videos at the end of Unit 4? We cover this exact scenario.

            Remember, you got part 1 correct on your exam. In our question, the diode fails open. So we want you to tell us what happens to the current when that failure happens. In other words, the circuit was operating correctly then had that failure. So, it goes from having current to having no current.

            You also got part 2 correct.

            3. The fan motor is the only load in that particular circuit. And the resistance of the motor is what it is – it can’t change. Rule of thumb – what happens in one parallel circuit doesn’t affect others that are in parallel. (The overall current will change.)

            Part 4 is correct.

            in reply to: Midterm #19298
            Susan Brown
            Keymaster

              You’ve got this correct now! The 3 answers add up to 119, which is close enough. It’s just off a little because of rounding of numbers along the way.

              in reply to: Midterm #19297
              Susan Brown
              Keymaster

                Hi Ronny,

                Glad to help! Loads will only split up the voltage drop if they are in series with each other, but the ignitor and booster are in parallel with each other. Plus, we didn’t give you any idea of what the resistances are of the loads to know how to divide them up.

                Hmmmmm….

                This question trips up a lot of folks, but once you see what’s going on, you’ll have an “a-ha!” moment. I’ll help step you through.

                First, try doing the “Zen trick” on the booster. How do you reach N?

                in reply to: Module 1 Unit 9 Defrost Systems #19280
                Susan Brown
                Keymaster

                  Hi Zachary,

                  If you go back to the Operational Overview video in unit 2, starting around 1 minute or so, you’ll hear that the refrigerant enters the evaporator as a liquid. It absorbs heat from the air and then boils and turns into a gas.

                  Does that help?

                  in reply to: module 3 unit 1 #19275
                  Susan Brown
                  Keymaster

                    Look at the video again, for example around the 10 minute mark, and again later in the video, his discussion about Cam 6. When those contacts are closed, what is being supplied to the motor windings?

                    in reply to: voltage drop #19270
                    Susan Brown
                    Keymaster

                      Yes, that works.

                      in reply to: module 3 unit 1 #19266
                      Susan Brown
                      Keymaster

                        Hi Sajjad,

                        Do you know how timer contacts work? They open or close a circuit for one or more loads. So, they are either supplying L1, L2, or Neutral to a load, depending on the configuration.

                        in reply to: voltage drop #19262
                        Susan Brown
                        Keymaster

                          Hi Ronny,

                          I think you’re just mixing up the math

                          He’s telling you that the resistance of R1 is two times that of R2.

                          So, whatever R2’s resistance is, R1 will be two times that. If R2 is 10, R1 would be 20. If R2 is 50, R1 would be 100.

                          Since we know that R1 + R2 has to equal the voltage supply (120), then there is only one pair of values that will meet both criteria:

                          R1 = 2 x R2

                          AND

                          R1 + R2 = 120.

                          R1 = 80
                          R2 = 40

                          [R1 + (2 x R2) = 120 is not a correct way to rewrite the equation. You could substitute (2 x R2) for R1 and write (2 x R2) + R2 = 120. This then equals 3 x R2 = 120, which gives us R2 = 40.]

                          Does that help?

                          in reply to: loading down in DC power supplies webinar #19253
                          Susan Brown
                          Keymaster

                            Thanks for the head’s up!

                            Would you please tell me which course, module and unit that appeared in so I can fix that link? Thank you!

                            in reply to: module 7 unit 4 #19233
                            Susan Brown
                            Keymaster

                              Hi Rodney,

                              1. you are close! remember that an electrical circuit has to end up where it started. Your need to select the answer that adds “module” at the end.

                              2. “What current carriers are present in a gas flame with a voltage difference across it?”

                              When you have a flame, you’ll have both electrons and ions. We show that in a couple of places in the video.

                              in reply to: 3.4 #19223
                              Susan Brown
                              Keymaster

                                It does take some thinking and working in order to figure this stuff out!

                                Think of parallel circuits as individual series circuits that are tied to the same power supply.

                                When we are talking about parallel circuits, we often refer to different “branches”. Each branch is actually a little series circuit.

                                So, each individual series circuit will follow the rules of series circuits. However, when there are two or more circuits (or branches) in parallel, the total current to/from the power supply will be the combination of the current flow in each branch.

                                Look at the figure below. Each branch (B, C, D) is an individual series circuit. The resistance of the load in each circuit will determine the current in that circuit.

                                If you had more than one load in any of the branches, then the total resistance in that branch would determine the current flow in that branch.

                                In this figure, if each load is identical, then the current in B, C, and D would be the same.

                                The sum of the currents in B, C, and D would equal the current flow found at points A. So, A = B+C+D

                                This is only true for current.
                                Does that help?

                                Parallel circuits

                                in reply to: Module 3 unit 3 #19220
                                Susan Brown
                                Keymaster

                                  Hi Michael,

                                  There were 3 correct answers (we asked you to select all that apply)
                                  LoZ setting on your DMM
                                  Wiggy or solenoid voltage tester
                                  Any low input impedance meter that measures AC voltage

                                  You had only answered wiggy and solenoid tester

                                Viewing 15 posts - 1,171 through 1,185 (of 1,968 total)