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Hi James,
You state it correctly. It’s very frustrating when faced with a situation like this, when you aren’t given enough specs to narrow the troubleshooting down to a specific board. You know the customer probably isn’t getting the best deal on a possible repair, through no fault of your own. But you can’t afford to buy multiple boards to then experiment with the one that will repair the machine. (And, it is conceivable that both boards could have been damaged by whatever event caused the failure.)
Fortunately, not all manufacturers do this, and we are hopeful that the ones who tend to (currently, Whirlpool comes to mind) will up their game. It’s not something to be overly concerned about, but to be on the lookout for.
Hi Jay,
Good question!
We do mention safety at various places in the course, but I’ll summarize some safety tips for you here.
The number one way to increase safety is to understand electricity! So, learning the material in the Basic Electricity module as thoroughly as possible is your first step towards being safe on the job.
The second way is to apply common sense and be observant.
Pay attention to safety information in the service manual for the appliance you’ll be working on.
Disconnect power to the appliance before doing any disassembly or replacing parts.
Some measurements must be taken when power is connected. Reapply power for the test, and disconnect when you are done. During the test, make sure you are just touching your meter probes, and nothing else, to the test points.
We caution against wearing metal jewelry on your hands or wrists, and any metal neck chains should be secured or removed.
We show you how to use a voltage stick (or “sniffer”) to check for the presence of voltage in many of our videos. That’s a smart thing to do regularly.
For a more thorough description of safety procedures, see Part 1, Chapter 2 of the Kleinert reference book.
March 14, 2019 at 7:36 pm in reply to: How to find voltage drop if I don't know resistance of loads #15561Hi Chase,
First of all, I moved your reply to its own topic since it has to do with a different question.
You are correct in general. If you have two loads in series, for example, you would have to know something about their resistances in order to calculate the voltage drop across them. Either the resistance values themselves, or how their resistances were related. Such as: if you knew they had equal resistances, then you’d know that they split the voltage drop evenly (each one would drop half of the source voltage). Of course if you have just one load in a circuit by itself, it drops the whole source voltage and no calculations are needed.
Since we don’t give you any info on the resistances of the loads in Question 8, that is a big hint that all 4 of these loads cannot be receiving current. If they were, then there would be a slightly complicated series-parallel calculation that you’d have to do, and you’d definitely have to know resistances.
What we are hoping is that when a student faces this conundrum, they will take a closer look and realize that there is something going on with the circuit configuration that makes the answer easy – no calculations needed. We suggest doing the “Zen Trick” on the booster and igniter to see if that helps illuminate things.
Take another look and let me know what you think.
If your answer is close, it may just be due to differences in how you round the decimal numbers. What number did you get?
Here’s a Forum topic where we show a calculation in more detail – see if this helps:
Hi Phillip – don’t you have a membership at Appliantology? That is the place to get repair advice. The Forums here at the MST Academy are for asking questions related to the training courses.
But, doing the Ten Step Tango is always the approach. As you know, just because a load isn’t working isn’t proof that it is “bad”. You have to do electrical testing to find out if the load is faulty or if it’s not getting the power it needs.
Review the last video in Unit 6. Let me know if you have any questions about what we’re showing in that video.
Hi Robert,
You’ll need to give me more details on this. Is this related to a Question on the Midterm? (Question 9?) If so, please rewatch the last video in Unit 6 for a similar, but not identical, scenario. Let me know!
Great! Thanks for using the Forums 🙂
Hi Kenneth,
These loads are in series in one circuit. Will the current change throughout the circuit? (if you think “yes”, then review the first video in Mod. 3, unit 5).
You do need to use the equation E = I x R to calculate each voltage drop, but you have to use the correct value for the current (I) to get the correct answers.
Hi Rees,
Thanks for posting in the Forums!
A great video to rewatch that shows these types of calculations is the last video in Mod. 3, unit 3 on the heat generated by a loose connection. Watch that, and re-create the calculations in your notebook. Then look at the questions in the Unit 8 quiz and see if they make more sense.
If you don’t follow something we’re showing in that video, please let me know and I’m happy to help you further.
No problem! Sam (who is a mister) is one of the folks helping you. One of us will get back to you ASAP.
Hi Samuel – could you please be more specific about which question you’re referring to? Thanks!
Hi Michael,
The fact that the terminal is labelled NO, and it’s in the “normal” state, is what tells you that the switch is open. And that is why you would measure 12vdc.
Does that answer your question?
Hi Ryan,
Thanks for posting in the forums!
One of the keys when looking at schematics is to “see” it from the electron’s point of view. Electrons don’t “see” bends in a wire, for example, like we do when looking at a diagram.
That’s one reason we suggest doing the “Zen trick” on the loads, when you aren’t sure if they are in series or parallel. Try doing that on the light bulb – “become” the light bulb – both with the switch open and then the switch closed. Does that help you figure it out? Let me know!
That’s correct! (I’m going to hide those answers so future students won’t see them.)
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