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Hi Kenneth,
Good question – I moved it so that it could be its own topic.
First of all, we do have a video showing the calculation of something very similar to this – the heat produced by a loose connection – the last video in Unit 3 of Basic Electricity.
If that doesn’t clear it up for you, tell me what Ohm’s Law equation you are trying to use to calculate the heat produced by that load.
Correct – The fan motor circuit is parallel to the element circuit, thus unaffected by its failure.
Also basically correct about the element branch – current will stop when it fails open, since there’s no longer a closed circuit.
Do you know what the final answer is – what happens to the total current draw from L1?
Not quite. Read what Michael said about the main above. And do the “Zen trick” on the booster or ignitor – what do you find?
Sort of! But you’ve got a mistake. The source voltage is 120vac. All of the loads combined must drop 120 – as you knew in your first response. So, calculating 120 as the drop across just one of the loads can’t be right.
The correct formula for voltage drop is E = I x R. We give you each resistance, so you’ve got to figure out what to plug in for the circuit current. (And what you just did isn’t correct.)
Does current vary throughout the series circuit or is it the same?
(And – for another hint – look at what you did for Question 3 on the Midterm.)
Hi Gilbert,
No, 0.013 is not correct. You would expect to get something less than 103, obviously, but not *that* much less!
However, 1/.013 does get you the correct answer, which is around 78 ohms.
Check out this previous forum topic where we show a similar calculation in more detail, and see if this helps:
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/Let me know!
“Mega-” is a million of something. So, 1 Mega-watt is 1 million watts (1,000,000 watts). But you can change it to any other unit of measure, too. A “Giga-” is 1 billion (1,000,000,000) of something. That’s why converting a mega-watt to a giga-watt results in a smaller number. 1 mega-watt is 0.001 giga-watt. Is that what you wanted to know?
Hi Terry,
This can seem like a fine point, but we do emphasize (and want to encourage) techs to think in terms of “power” when it comes to loads doing work. The specs for loads are often given in watts (which is voltage times current). For a load to do its work properly, there must be the correct amount of both voltage and current to create the necessary wattage. (So, the correct answer was “power”.)
Hi Kenneth,
What is your question about 4 megawatts? Are you trying to convert it to something different?
Hi Joshua,
No point in googling other resources when you’ve got the ultimate resource right here – Team Samurai! A lot of what your tuition pays for is the ability to ask us questions here in the Forums.
We need to address these questions one at a time. Let’s start here with question 7.
(By the way – I’m going to hide the question in your post above, so that other students won’t see it before they take the Midterm themselves.)
Question 7 has two parallel circuits, and at first we assume everything is behaving normally – current is flowing through both branches.
Then, the top circuit (with the element in it) fails open. You gave us the correct answer for what happens to current in that branch. But then we ask what will happen to the voltage drop and current in the fan motor circuit.
Do you recall what we taught about parallel circuits, in terms of how a failure in one branch affects another?
Hi Joshua,
Happy to help! First of all, these loads have different resistances. Will that affect the voltage drop across them? Isn’t there a relationship between resistance and voltage that we taught?
Hi Mohamed,
We’re talking about the size of the letters or numbers used on a schematic. They tend to use large letters for the various safety warnings (which professional techs are already aware of). But the key information needed for troubleshooting – specifications for the loads, for example – are often in small letters/numbers. In other words, for a professional tech, the most important information is usually the smallest.
Hi Robert,
Think of their function in the circuit, and that will help you know the answer to this. Also, we teach this in Mod. 3, unit 7 (skim through the text of that unit and you’ll see).
Hi Kenneth,
This is the Ask the Teacher Forum! If you want help with something, you can start a new topic. Did you watch this video on how to do that?
Do you check the little box that says “Notify me of follow-up replies via email” ? That’s a very helpful thing to do. As for looking at this page, it doesn’t auto-refresh. You have to reload the page to see new replies.
I’ve reset you for that quiz. For future reference, there’s a Quiz & Exam Reset Request form that we want you to use. Just look under the “Contact Us” item in the main menu.
Also, keep in mind that Module 3 is very challenging. Please keep asking questions here in the Forums so we can help you.
We didn’t discuss Question 11. We have a video in Unit 3 that goes over this calculation (with slightly different numbers for the resistances). Watch that again, and try to do the same calculations we show you in your own notebook. If there are steps that you don’t quite follow, just let us know and we can help you.
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