Forum Replies Created
-
AuthorPosts
-
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!
Hi Ronny,
The Samurai just misspoke at that point (~ 18:58). Just go by what is written on the slide! He does go on to say that resistance increases as temperature increases, and that it is a mathematically “direct” relationship.
Sorry for the confusion!
July 10, 2020 at 7:40 am in reply to: Refrigeration….Just when you thought it was safe to be here; I’m Baaaaaack..! #19372Ah – he’s talking about the cabinet being “out of square” which means the right angles of the cabinet aren’t exactly right angles. In other words, slightly warped in some way so the doors don’t seal properly.
Hi Everardo,
Sure – I just set you back to that quiz.
We’ve got a Quiz and Exam reset request page that you should use if this happens again – you’ll find it in the sub-menu under “Contact Us” in the main menu. Here’s the direct link:
https://my.mastersamuraitech.com/quiz-exam-reset-request/
Basic Electricity is a challenging module – we’re happy to help you here in the Forums if you need it – just start a topic.
That’s what we’re here for!
Figure 1 is the schematic – it won’t show what the error is. You have to use the measurements to determine what is going on – just like in real life! You aren’t able to see the entire circuit in the appliance – you are using measurements to deduce them.
In Figure 2, we deliberately disconnect a wire in order to do the half-splitting technique. That’s not the same thing as showing you where the fault was in Figure 1.
When we have an L1-L2 circuit, and there is an open on one side or the other, you cannot tell from measurements which side is open because both sides have a hot line.
In an L1-N circuit, you would be able to tell without half-splitting. If you measured 120v on one side of the load wrt N, you would know that the open was NOT on the L1 side, otherwise the voltage wouldn’t be felt at the load.
If you measured 0 volts from both sides of the load wrt N, you would know the open IS on the hot side.
But with both sides being hot, you can’t tell from the scenario in Figure 1 which side is at fault without disconnecting the circuit and re-measuring.
Do you see how that works?
Yep! So, that’s the important bit of info that Figure 1 is giving us. Do we know – in Figure 1 – where the open is?
Good!!
What is the general cause of current not flowing, even though there is some voltage present?
Correct again!
However, the measurement from L1 to L2 in Figure 1 is zero. So, we have no heat and no voltage drop. Does this change your mind about current flowing in the circuit?
Remember, you can measure voltage (potential) when current isn’t flowing.
-
AuthorPosts
