Susan Brown

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  • in reply to: Module 5 Unit 1 Quiz question on calculating voltage drop #13397
    Susan BrownSusan Brown
    Keymaster

    Hi David,

    In order to calculate the circuit current, you first need to know the total resistance in the load so you can do I = E/R.

    Remember the Midterm exam question about the heat generated by the loose connection? You mentally added 30 + 6 to get 36 ohms, then went on to calculate the current.

    in reply to: How do I solve this problem. #13393
    Susan BrownSusan Brown
    Keymaster

    Yes, that’s correct for Part 2 – L1 is the one with the fault. (Note: your original answer to Part 1 on the exam was correct. It could use a little more detail, but you’ve got the basics correct.)

    in reply to: How do I solve this problem. #13390
    Susan BrownSusan Brown
    Keymaster

    Reminder: the Help Page I linked you to in your Midterm Feedback email has suggested units and videos for each question.

    For #7, rewatch the videos at the end of Unit 4.

    For #9, Look at Figure 2 on the Midterm and let me know why you think it is L2 that is at fault?

    #9 midterm

    in reply to: How do I solve this problem. #13386
    Susan BrownSusan Brown
    Keymaster

    That’s correct. Note – I’ll have to hide your answer so other students won’t just be able to copy it.

    in reply to: How do I solve this problem. #13384
    Susan BrownSusan Brown
    Keymaster

    Yes – did you see my explanation above about Total resistance in series circuits vs. Equivalent resistance in parallel circuits?

    in reply to: How do I solve this problem. #13381
    Susan BrownSusan Brown
    Keymaster

    Yes, that is why its Total Resistance is 70 ohms.

    in reply to: How do I solve this problem. #13379
    Susan BrownSusan Brown
    Keymaster

    70 ohms is the answer.

    I think you are mixing up Total Resistance of loads in series with the Equivalent Resistance of loads in parallel.

    Series circuit: total resistance is the sum of all resistances (loads) in the circuit.

    Parallel circuits: equivalent resistance will always be less than the smallest of the resistances, and can be calculated with the formula
    Req = 1/(1/R1 + 1/R2 +…)

    Equivalent resistance is a way of taking two or more parallel loads/circuits and combining them into one theoretical resistance. It lets you know what the resistance of those parallel circuits are from the perspective of the power supply.

    in reply to: How do I solve this problem. #13377
    Susan BrownSusan Brown
    Keymaster

    When loads are in series, the total resistance in the circuit is just the sum of the resistances.

    You did this on Question 3 of the Midterm as part of finding the circuit current:

    I=E/R=120vac / 10+20+40=1.71 amps

    (that is copied from your answer)

    in reply to: How do I solve this problem. #13375
    Susan BrownSusan Brown
    Keymaster

    What is the circuit? I need to know how the resistances are laid out.

    in reply to: How do I solve this problem. #13364
    Susan BrownSusan Brown
    Keymaster

    It sounds like you are looking over the Midterm study sheet. (Good!)

    That’s referring to the information in Unit 6, particularly the video at the end.

    in reply to: How do I solve this problem. #13362
    Susan BrownSusan Brown
    Keymaster

    Hi Randy,

    We were in the middle of working on this together in the thread above.

    First step: rewatch the video at the end of Unit 3 about calculating the heat generated by a loose connection. (That loose connection is acting like a “load” in that it has resistance and generates heat when current is going through it.)

    Note that we first calculate the current going through the circuit, I.

    Then we use a formula for P to find the heat generated by that particular “load” in the circuit. Write down the steps in your own notebook and make sure you understand what we did there.

    Let me know!

    in reply to: unit 3 quiz reset. also question #7 #13360
    Susan BrownSusan Brown
    Keymaster

    I assume you mean P=I*E

    If you do that equation, when you are just trying to find the heat (P) generated by one of the two loads, then E has to be the voltage dropped just by that load, not the entire E of the circuit.

    It’s easier, in my opinion, to use a different formula for P. First find the circuit current, I, which is straightforward because you know the source voltage and the total resistance in the circuit. Then you can use a formula for P that just uses current and R of the load of interest.

    We used this technique to calculate the heat generated by a loose connection that was in series with an element at the end of Unit 3, if you want to review that.

    I’ll reset you, but for future reference, please use the Quiz and Exam Reset Request form under “Contact” in the main menu – thanks!

    in reply to: Various Questions on the final exam part 2 #13357
    Susan BrownSusan Brown
    Keymaster

    Adam, you need to do some of this footwork yourself. These questions are very similar to ones you had throughout the Fundamentals course, so you should know where to go look for the answers.

    The Forums are when you’ve looked up the information in the course but you still can’t figure it out.

    in reply to: Various Questions on the final exam part 2 #13355
    Susan BrownSusan Brown
    Keymaster

    Troubleshooting Module: Troubleshooting Basics and Strategies (unit 2)

    in reply to: Various Questions on the final exam. #13352
    Susan BrownSusan Brown
    Keymaster

    Yes. You need your reference point for the voltage measurement to be a known value. More often than not that means a known-good neutral point. Then you know that the numerical reading on your meter is with respect to 0vac. In this case, we’d be looking for 120v at P6 wrt N.

Viewing 15 posts - 16 through 30 (of 367 total)