Forum Replies Created
-
AuthorPosts
-
Hi Jason,
The answers to this question usually click into place once you figure out what is going on with the main coil.
If you do the “Zen trick” on the booster and the ignitor, how do you reach N?
Hi Jordan,
If we thought there was a better or easier way to explain Ohm’s Law, we would have done it that way! 🙂
To some extent, it just takes wrestling with it and some time. The quizzes should help you with that, as well as just rewatching the videos, taking notes, and practicing the calculations.
What I found, when I was learning it, is that as I went along suddenly a piece of the puzzle would really click in for me, and then I could go back and understand even more in the earlier videos. It’s a process.
I created this blog post to help people who struggle with the math aspect of it. Have you gone through this? It’s designed for you not just to read, but to use pencil, paper, and calculator to “play around” with it.
Lastly, asking more questions of us here in the Ask the Teacher Forums is a key component to learning. Just take anything you don’t understand – a video, or a quiz question – and use that as a starting point, and we’re happy to go back and forth and help you.
July 22, 2020 at 11:30 am in reply to: Refrigeration….Just when you thought it was safe to be here; I’m Baaaaaack..! #19420There are occasional mentions in the course, but there’s not that much that’s different about these units compared to the usual stand-alone ones other than configuration. The machine compartment is often located on top. And, as you would expect, they are designed to be worked on from the front, since you can’t just roll them out to access the back.
Question #1 – What characteristic of a heating element does NOT make it an electrical load?
So, we’re looking for the characteristic in the list that is not the thing that makes a heating element a load.
You answered:
It has a voltage drop across it when current flows through it.
This is one of the key characteristics of a load.
Thanks for asking!
We do not have a hard copy available. One reason is that we update the material over time. Another is to keep our proprietary teaching secure. Unfortunately, there are people out there who try to copy and sell our information.
What we strongly encourage people to do is take thorough notes as they go through the course, especially from the videos. And, as long as you have access to the courses, to review the course material and add to your notes.
July 20, 2020 at 11:35 am in reply to: Refrigeration….Just when you thought it was safe to be here; I’m Baaaaaack..! #19413Most residential refrigerators are “stand alone” – meaning, they are not installed or attached to cabinetry in any way. They are rolled into place (and perhaps connected to a water line.)
“Built in” would be like most Sub-Zeroes and other high end refrigerators that are actually installed into the cabinetry in some way.
Yes, the answer is power. For a load to do work, it needs both voltage and current, which is “power” (P = I x E)
You are on the right track.
Resistance is not the right answer because a load has a resistance by definition. Whatever material it is made from has a certain resistance.
According to that understanding a correct answer would the load would require voltage and current to do work.
That is correct. And the term we use for the combination of voltage and current is power.
Think about the common Ohm’s Law formula for power: P = I x E
Power is voltage times current. So, that is the correct answer here.
Hi Shane,
This forum is for us in Team Samurai to help you out.
Understanding the material in Module 3 often involves watching videos more than once – and pausing frequently to take notes and try to duplicate any examples we show of calculations. It also involves asking us for specific help when you still feel stuck.
Please give me an example of a question that you are having a difficult time understanding.
Also – did you do the assignments from the Kleinert book?
Did you need any other help from us? Do you have any other questions about the material in that unit?
The Quiz and Exam Reset Request Form is at this link:
https://my.mastersamuraitech.com/quiz-exam-reset-request/
You can also find it in the Sub-menu under “Contact Us” in the main menu.
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?
Although someone comfortable with a scientific calculator can do this calculation more quickly, you can also just use a regular calculator – just follow the technique in the video above.
For example,
1/10 is 1 divided by 10 which is 0.1
and so on.
The second part of Ronny’s comment above has to do with the later part of the video back in Unit 5 where the equivalent resistance of the two loads in parallel is then used with the other series loads in that series-parallel circuit to find circuit current.
Hi Everardo,
Check out this video and see if it helps you understand how to do the calculation:
https://www.youtube.com/embed/iHB3lxdc68E
Another note – depending on how many decimal places you preserve when you do the fractions, you may end up with a slightly different answer – it will be close, though.
Let me know if you need further help.
hello!
On Question 7 of the midterm, you do not need to calculate anything. You just need to know if the things we’re asking about (current or voltage drop) are increasing, decreasing, or staying the same. We don’t give you enough information to actually do the calculation. The videos at the end of Mod. 3, unit 4, are particularly helpful with this question!
I didn’t take it that way! We’re always glad when people ask us for clarification or help.
Glad you are enjoying the course!
-
AuthorPosts
