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Yes – that is correct about current.
So that means there is an ____ in the circuit.
But based on the other info we give, that you measure 120 vac on one side of the circuit, that helps you to narrow the conclusion down even further.
The answer can actually be expressed in two words: there is an ____ _____ in the circuit.
Hi Colin,
This is a very commonly-missed question on the Final exam, and I’ve been surprised I don’t get more questions about it, so I’m glad you posted this.
The answer is actually simple once you see it. I’ll try to give you some hints and see if that helps you get it, instead of just telling you the answer.
In the problem statement we rule out the bulb or the receptacle as being the culprit.
In other words, we have a good load, and we know we have voltage.
When we measure across the load (across the two terminals), we measure zero volts. This is the first thing to focus on. What does zero voltage drop across a known-good load tell us?
Hi Kyle,
We’re glad to help. First of all, what type of element are we talking about – oven or dryer? Also, is this a 3-wire or 4-wire power cord configuration?
~ Susan
HI David – check your email. I sent you detailed info on your results this morning.
Hi Jared,
My biggest tip for taking the Module Exams is to remind you that these questions are taken from the Unit quizzes. So, you have seen all of the questions before. The questions are chosen at random, and the answers are rearranged, to make sure that you aren’t just answering based on pattern recognition which helps to make sure you’re learning the material.
To study, you should look back over each unit quiz in the module and make sure you really understand the questions and answers, using these forums for any clarification you might need from us.
The light’s bimetal switch is the switch shown just to the left of the light. The protector bimetal is labelled “protector” on the schematic.
Okay, great.
So for one of those lights, do the “Zen Trick”. Become that light. In order to work, you need to be able to reach L1 with one hand and L2 with the other. This question asks about L1.
What are the ways that you can reach L1?
You listed 4 of them above. When you do the Zen Trick, do you still think all 4 of those are ways to reach L1?
Hi John,
Happy to help!
Our load of interest is one of the hot surface indicator lights.
First I want to make sure we’re looking at the same thing on the schematic, since there are a couple of similarly-labeled lights.
The lights we are interested in are labeled “Hot Ind Lights” and we are wanting to see how those lights get L1.
Are we looking at the same thing? Let me know!
Hi Chip,
Happy to help!
First of all, do you remember how to calculate (or what the rule of thumb is) for the equivalent resistance of loads in parallel? You covered this in Fundamentals.
Yay! Exactly correct on all accounts. (Note: I’m going to hide your answers)
Just to summarize:
The general term “voltage” is the difference in electrical charge between two points.
When current is flowing through a load, there will be a difference in voltage between one side of the load and the other. This is “voltage drop”. And yes, loads in series will behave according to Kirchoff’s law.
Measuring a voltage across an open switch (or a load that has failed open) is NOT voltage drop, because there is no current flowing.
I’m glad that you are working on understanding this!
The short answer is “voltage drop is created by current flowing through a load”
Let’s unpack that just a bit.
How would you define “voltage” in your own words? (Just the word “voltage” – not necessarily voltage drop.)
Hi Josh,
Let’s talk about Question 9 first. I hid the last part of your answer, so we wouldn’t give it away to other students who haven’t taken it yet. But that is correct.
Another thing to think about with the measurements shown in Figure 1 is – what does it mean if you know that the element is good, but there is zero voltage drop across it? What creates a voltage drop? We are measuring some voltage, but what is missing?
Question 8: Please read through this topic and see if it helps. Be sure to read the whole thread, as several students chime in with questions. If not, ask another followup question.
Awesome! That’s what I like to hear 🙂
If you are the booster, is there not a way to get to N without going through the main coil? (twists and turns don’t matter to an electron: wire is wire)
Hi Sam,
This can be a tricky one! But you’ll get it. Let’s try this:
For each load, try the Zen trick that we describe in Unit 5. If you can become that load and reach L1 and N without going through another load, then you will drop 120vac, correct?
Are there any loads that you think you can’t reach L1 and N without going through another load? If so, which one(s)?
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