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Susan Brown

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Viewing 15 posts - 1,921 through 1,935 (of 1,941 total)
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  • in reply to: Final Exam time frame. #7392
    Susan Brown
    Keymaster

      As you discovered, your exam was graded shortly after you posted this! (Congratulations!)

      It can take anywhere from 1 to 3 days, usually, for an exam to get graded. It happens to be our busy season right now, so sometimes it might take a little longer.

      Thanks, and great job on Fundamentals!

      We’ll be sending out your certificate soon.

      in reply to: Module 3: unit 7 #7295
      Susan Brown
      Keymaster

        It may not be stated outright, but hopefully you can figure it out based on what you know so far. DC has polarity (+ or -), because it moves in one direction. AC is constantly switching directions. Can you see why measuring DC voltage will result in a positive or negative reading, depending on where you put your leads, whereas measuring AC will not? (It will just be a number with no + or – sign.)

        Does that help you figure out the answer?

        in reply to: Module 3: unit 7 #7286
        Susan Brown
        Keymaster

          Hi Dennis,

          Thanks for asking questions!

          First question – Sometimes it will be as easy as reading the label on a motor, but the main way that you’ll know what type of power a particular component is getting is when you do a load analysis using the schematic diagram for the appliance. You will be learning how to do this as you continue to go through Fundamentals, so be on the lookout for it!

          As for your second question, appliance techs never need to measure DC current, that’s why we had this message above the video: “Here’s a quick video showing the use of a digital multi-meter, to help clarify the reading. The amp reading he does at the end is not something you’ll ever need to do so you can stop watching at about 3:30 into the video – I’ll show you in another video how to use a clamp-on ammeter (a picture of which is on page 95 of the Kleinert text).”

          However, if you are asking just to try to understand the difference, it is because DC current moves in one direction only, so in order to read it, the meter has to physically be in the circuit. AC current, on the other hand, is constantly changing direction which creates a moving magnetic field that the clamp-on meter can read. Hopefully, you’ll be able to have a better feel for this after you have learned about transformers and motors, later in the course.

          in reply to: Module 5, Unit 2, Questions 5 & 6 #7166
          Susan Brown
          Keymaster

            Great! Glad to help.

            in reply to: Module 5, Unit 2, Questions 5 & 6 #7163
            Susan Brown
            Keymaster

              I just went to look at your quiz, and it was questions 6 and 7 that you missed. 6 is above, here is 7:

              Question #7 – In order to calculate the voltage drop across a load in series with another load, we first need to calculate the ______ .

              I’ll deal with this one first. The choices are total circuit resistance, current, or voltage. We can rule out total circuit voltage, because that is not calculated, it’s given.

              To calculate voltage drop across a load, we need to know both the current in the circuit (which is constant throughout the circuit) and the resistance of the load, right? E=I x R.

              The resistance of the load is what it is. In other words, it’s an intrinsic value of the load and must be specified on the tech sheet.

              Current, however, has to be calculated knowing the circuit supply voltage and the total resistance of the circuit. I = E/R. Does that make sense? So, the first step in eventually being able to calculate the voltage drop across a particular load is to calculate the total circuit resistance so that you can then calculate the circuit current.

              For Question 6, the possible answers are:
              The current flow through each load will be proportional to their resistance.
              The current flow through each load will be proportional to their voltage drop.
              The voltage drop across each load will be proportional to their intrinsic resistance.
              (Note – we clarified the wording just a bit, changing “them” to “each load”)

              We are talking about each load in the series. The current is the same throughout the circuit, right? So do the first two answers make sense? The current flow in the entire circuit is a function of the TOTAL resistance and supply voltage, but does not vary through each load.

              E = IxR. Current is constant, but each load has it’s own resistance, so the voltage drop for a load will be proportional to the resistance of that load.

              Hope that helps! Let me know if you have any follow-up questions.

              in reply to: Module 5, Unit 2, Questions 5 & 6 #7162
              Susan Brown
              Keymaster

                Hi Tyler,

                Don’t feel bad – this is supposed to be a challenging course! The struggle is just part of the learning process. And we’re here to help ya.

                First, let me just verify that these are the two questions you are asking about:

                Question #5: What are loads in series?
                Question #6: What is the true statement about loads in series?

                Are those them?

                in reply to: equivalent resistance #7159
                Susan Brown
                Keymaster

                  It’s hard to write out equations here, but I’ll try to show it.

                  It’s 1/(1/58 + 1/320)

                  1/58= 0.017
                  1/320= 0.003
                  Add those together = 0.02
                  Then 1/0.02 = 50

                  I rounded the decimals a bit, which is fine. To do the fractions on your calculator, you just enter 1 and then divide by the denominator.

                  Is this what you needed?

                  in reply to: equivalent resistance #7157
                  Susan Brown
                  Keymaster

                    Hi Robert,

                    That particular question was set up so that you just needed to know the rule of thumb about equivalent resistance: that it will be something less than the smallest resistance in the parallel circuit.

                    You can calculate the exact value of the equiv. resistance by using the formula that is shown in the last presentation in that unit (starting at about 2 1/2 minutes in), but you didn’t need to for this question.

                    Does that answer your question?

                    in reply to: "using Ohms Law" ? #7118
                    Susan Brown
                    Keymaster

                      Hi John,

                      Since March, we’ve added 9 new presentations to Fundamentals, including the one you are asking about. We almost immediately decided that the Ohm’s Law presentation went beyond the scope of Fundamentals, which focuses on basics, and was advanced enough that we should include it among the presentations in a new course on Advanced Schematic Analysis and Troubleshooting. We left the presentation in the Fundamentals course for the time being while we built the ASAT course, but moved it over to the ASAT course last week when that course was launched.

                      So you were one of the few students who happened to catch it during that brief window that it appeared in Fundamentals and got a bonus lesson (and a preview of the ASAT). Lucky you! 🙂

                      We’ve noted this and all other updates in the Course Updates log at the site.

                      in reply to: Calculating ohms law equations. #6651
                      Susan Brown
                      Keymaster

                        Hi Glenn,

                        I’m not sure what the “ugh of ohms law” is that you mentioned – perhaps that was a typo? 🙂

                        To cover all the bases, I’ll go over the basic math functions that are in the Ohm’s law pie chart that first makes its colorful appearance in Unit 3 of Basic Electricity.

                        1. Multiplication: as in, E = I x R
                        2. Division: as in, R = E ÷ I (written in the pie chart as E / I, which is just another way to show a division problem)

                        Those two are easy to do on any calculator.

                        3. Squared: as in, P = I2 x R. A number that is “squared” is simply that number times itself. For example, 2 squared is the same as 2 x 2 which is 4. So, if your calculator doesn’t have a squared function button (usually labelled as x2) just enter the number and multiply it by the same number.

                        4. Square root: as in, E = √ P x R 
                        A square root is the opposite of a square. You are basically saying, what number was squared to get this number? If you want to know the square root of 4, you ask “what number times itself will equal 4?” The answer is 2. For all but the simplest numbers, you’ll need a calculator that has a square root button to calculate this. If you are using the calculator on an iPhone, turn the phone sideways to get a lot of extra buttons, including the square root. It looks something like this: 2 x 

                        I hope that helps! Lemme know if you have any other questions.

                        in reply to: "Using Ohms Law" module question #6576
                        Susan Brown
                        Keymaster

                          That’s some wise advice, Smashy!

                          You can retake the quiz when you are ready.

                          in reply to: "Using Ohms Law" module question #6573
                          Susan Brown
                          Keymaster

                            Listen to the video starting at about the 6:20 mark 🙂

                            It’s great that you’re going back and studying new material, even though you finished the course a while back. Have you seen all of the new screencasts that were added to the Basic Electricity module over the last month? You should check them out if you haven’t – it’s killer stuff!

                            in reply to: Would this be possible? #6101
                            Susan Brown
                            Keymaster

                              Hi Smashy!

                              Thanks for the suggestion. That would be a good thing to be able to show in the results. I played around with the quiz settings in the software we use, and unfortunately it doesn’t work quite the way we need it to for our particular style of quizzes. I’m putting in a ticket with the software company to see if they can make this possible in the next update. If we can do this, I’ll let you know!

                              By the way – props to you for going back and studying the quizzes!

                              in reply to: upgraded bussines part #5375
                              Susan Brown
                              Keymaster

                                Hi – sorry for the delay, there was a glitch in the system that notifies me of posts!

                                No, when new content is added to a course you will have to take any new quizzes. If you have already done that and you still have problems accessing the units, please let us know. There have been a couple of students who took the business class early on who have had problems getting back in, but we have a way to reset you if that happens. If you do end up needing to retake any quizzes, just think of it as a good review! Fortunately, they are short.

                                in reply to: Possible website error- no business modules #5374
                                Susan Brown
                                Keymaster

                                  Thanks for the heads-up! Yep, we made that change because we realized how long that sidebar was going to be as we kept adding courses. Now you only see the modules/units for the course you are in.

                                Viewing 15 posts - 1,921 through 1,935 (of 1,941 total)