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Hi Sean,
That’s awesome that you are taking the time to understand this material better. Unit 8 is challenging! The quizzes are an integral part of the learning experience. Some of the questions deliberately push you to grapple with the material, and not just be a quick check on what you learned. These questions pull together what you’ve learned in several units in this module.
Questions 6, 7, and 8 are all based on the same circuit. There are two loads in series, and it’s an L1-L2 circuit.
Question 6 asks for the current you will measure at a certain point. So, you have to think about the characteristics of current in a series circuit, what the voltage supply is, and which Ohm’s Law equation will be appropriate.
Questions 7 and 8 ask for the heat produced by each load. If you’ve answered Question 6 correctly, you’ve got one of the things you need for this calculation. Then it’s a matter of selecting the correct Ohm’s Law equation.
One of the helpful videos to review, because it’s showing a similar scenario, is the “loose connection” video in Unit 3. Rewatch that, and see if you follow the calculation, and if it helps you to understand these questions better. If not, let me know and I’ll help you further.
Hi Terry,
You have a good point. We were obviously thinking of a load when we wrote the question, but the word “something” could be interpreted the way you have. I’m going to run it by the rest of the team, but we’ll probably update the wording to be more clear. But for your reference, we are talking about a load (which, naturally, has resistance as an inherent property).
The main point we are trying to drive home is that you must have current flowing through a load to get voltage drop. Many techs forget that it takes voltage AND current to power a load, and the concept of understanding voltage vs. voltage drop can help clarify troubleshooting.
Thanks for putting so much thought into this and for posting your comment!
Hi Michael,
You are correct about the safety and also the main.
Electrons will flow through any valid, non-shunted path between Line and Neutral if they have voltage “pushing” them.
So now you just need to figure out what the relationship of the booster and ignitor are. Are they in series or in parallel? Do the Zen Trick on those two loads. What do you find?
Hi Phillip,
Yes – that is correct! You also correctly stated the rule of thumb. So, for Question 18, the correct answer choice is “something less than 58 ohms…”Do you know the rule of thumb for equivalent resistance of loads in parallel? That’s what you need in order to find the correct answer out of the choices we give you.
You are correct about the safety. It’s all by itself in a circuit, so will drop the entire source voltage. (No need to know its resistance, since you didn’t have to actually do a calculation.)
But, there are a couple of things we still need to straighten out.
First of all, the detector is a switch, not a load. (Is there any voltage drop across a closed switch?)
Second, remember that electrons don’t “see” any turns or bends in a wire. Unless there is a shunt, they will take any valid path. So, if the main coil were getting current, then it would be in series with both the igniter and the booster.
But – do the “Zen trick” on the booster. How do you reach out and touch Neutral?
Hi William,
Each load does have a resistance, but we don’t tell you what it is because you actually don’t need to know it in this particular scenario.
The first step is to decide if each load is in series with any other loads. The “Zen trick” on the booster, igniter, and safety is a helpful tool.
Then, if you have that determined correctly, you just use what you know in general about voltage drop and voltage supply to answer the question – no calculations necessary!
What do you think? How are the loads in relation to each other? Do they all have current flowing through them?
The reason I asked about the supply voltage is that the answer you gave – 40 vac and 80 vac – only adds up to 120vac. As we taught you, the sum of the voltage drops across loads in series will equal the source voltage.
But we indeed were asking you to calculate the voltage drop across R1 AND the voltage drop across R2.
And the correct answer is one of the choices given.
Hi Jim,
Thanks for posting a question!What is the source voltage for this circuit?
Hi Jim – when I look at your results for that quiz, there is no answer given for Question 8. It appears to have been left blank. If you’d like to retake the quiz so that you can answer all of the questions again, let me know. Otherwise, your score is fine for Certification.
You’re welcome!
Yes, there are only those two answers correct.
The “Protector” bimetal is not in the circuit with the “Hot Ind Lights”
Hi Samuel,
Here’s the problem statement: “You’re called in to troubleshoot a refrigerator for a warm temperature complaint. You use your IR temperature gun and measure the fresh food compartment at 37 degree F and the freezer temperature at 1 degrees F.”
The customer claims it is warm, but are those temperatures that you measure out of range?
The most common mistake folks make on this one is to look at the wrong indicator light on the schematic. There is one “surf ind light” and 4 “hot ind lights”. Which one are you looking at? Re-read the problem statement to make sure you are looking at the correct one.
HI Darron,
What is the power source for the light – L1-N, or L1-L2?
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