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Question 1 asks: In a gas burner spark ignition system, how does the spark current return to the spark module after it leaves the electrode?
The Hint says: In a non-reignition system, the spark current doesn’t have a return path to the spark module — it simply grounds out in the burner head.
I answered that it returns through chassis ground. Compared to the other answers I don’t see how I got that wrong.
Grounding out is not the same as returning to the spark module. So, the answer is basically, “it doesn’t”. In fact, we had a humorous option that was supposed to be the correct answer. Most people probably don’t get the old pop culture reference, so I just changed it.
Question 2 asks: In a gas burner spark reignition system, how does the spark current return to the spark module after it leaves the electrode?
The Hint says: Since we’re talking about a reignition system, the spark current must return to the spark module so that it can be sensed.
This is the one where the answer is that is returns via the chassis.
Question 10 asks: What is a reasonable component to select as our LOI?
The Hint says :Since we hear a spark, we know the spark module is at least trying to do its job. Also, since flame is established, we know the spark is being delivered to the correct location on the burner head and that the burner head is delivering the fuel-air mixture to the spark location. Therefore, there’s no reason to suspect the spark module, board, or electrode. The problem must be in the return current.
Now that you know more about where the return current comes from, you know that your LOI would be the part of the chassis. The burner head is a good place to start as an LOI.
Hi Rees,
The premise of that statement is that there is usually only one thing broken at a time, even when there are multiple loads not doing their thing. It is that one fault that is causing multiple symptoms.
It is very rare, but if there are actually two different points of failure in one machine, then starting at one malfunctioning LOI may or may not lead you to both failures.
This would become apparent in the troubleshooting process.
Make sense?
The Final has questions from Modules 3 through 8 (so, not from Mod. 9)
Hi Joe,
From the unit:
A dryer’s heating system must have proper air flow in order to operate correctly. If it does not, it may trip a hi-limit thermostat or a thermal fuse.
So – no heat is also a symptom of an air flow problem.
That’s correct – good job!
I am a little nervous I haven’t attempted to retake this since you reset it. what are my options if i bomb again? I hope that’s not the case but i’m asking what if?
You didn’t exactly bomb it – you were one question away from passing!
The questions don’t change, so just make sure you’ve gone through and figured out correct answers to as many of the ones you missed as possible, and make notes for yourself. If there are any particular questions/answers that you don’t understand, as us and we’ll help you.
But don’t worry – we’ll keep working with you until you pass everything.
Glad you figured it out! And don’t ever feel bad about posting a question… that’s what we’re here for!
Since you paid for your own course, then it is completely up to you if you want to earn Certification or not. There is no requirement that you earn a certain score in order to move on in the course or take other courses.
The only exception to this, as you’ve seen, is Part 1 of the Final exam. We do have that set to block someone from taking Part 2 if they didn’t get 90% on Part 1. That’s just because most students are aiming for Certification, and we want to make sure they understand the questions and answers on Part 1 before doing the open answer final.
As a reminder, the requirements for Certification are 80% or higher on each unit quiz, and 90% or higher on each exam. I think your scores (except for the Final) meet that criteria, unless I missed something when I glanced over your results.
-voltage itself does nothing
Be a little careful with putting it that way. Voltage causes current to flow, if there is a complete circuit. Without voltage, nothing will happen in a circuit.
But, just because you measure voltage, this doesn’t mean that current is flowing. For example, if you have an open switch in a circuit, you will measure voltage across that switch, but it is “potential”. Since the switch is open, no work is being done in that circuit. But we know if we close the switch, current will flow and work will be done.
As for the quiz question you are talking about, “select the true statement about voltage in a series circuit”, your answer was: “The first load always drops the most current.”
As you can probably see now, there are several problems with that statement. There is no such thing as “dropping current”, since current is constant in a series circuit. Also, if we were talking about dropping voltage, then it is the resistance of a load that determines the amount of voltage drop, not where the load is in the circuit relative to the other loads. One more thing – in an AC circuit, current moves back and forth, so there is no “first” load.
The correct answer, of the choices we gave, is, “The sum of the voltage drops across each load will equal the supply voltage.”
Otherwise, your statements look fine.
One thing I find helpful is to remember that parallel circuits are just series circuits that are tied to the same power supply. So, each “branch” is just a series circuit.
Hi Desert,
This is good stuff to get straight. I’ll go through the concepts:
A load is a component that will do work when current flows through it. Work is something we can detect with our senses, such as heat, light, motion. (So, yes, a light bulb is a load.)
For a load to do work it needs power: voltage AND current.
Voltage alone creates the potential for work to be done. Voltage and a complete circuit will cause electrons to move (current), which will result in work being done by any loads in that circuit.
In a series circuit, the current is the same at every point in the circuit. That current is determined by the voltage supply and the total resistance in the circuit.
A circuit will have a given power supply voltage (120vac, 240vac, 12vdc, etc.). As current flows through a load, it will create a voltage difference (we call it “drop”) across each load. The amount of each individual voltage drop is proportional to the resistance of each particular load. The sum of the voltage drops will equal the source voltage.
In the circuit with several bulbs in a row, remember: the current will be the same through the whole circuit. If those bulbs are identical, then each would have the same amount of voltage drop, and the sum of those voltage drops would equal the source voltage.
Does that help?
Don’t be too hard on yourself! The Midterm is a challenge for most students. It’s designed to really help some of the material we’ve been teaching click in. It takes many students 2 or 3 attempts to pass, but the effort pays off.
Hi Michael,
It’s just the way you are rounding the 1/30 result. It is actually 0.0333333….
So, if you use a calculator in a way that preserves those decimal points, you end up with 18.75. If you round down to just the 0.03 then you end up with 20.
It’s not a very significant different when we’re talking about resistance.
okay – I just did that for you!
The GE example (around 5-6 minute mark) had a 12vdc voltage, so yes – that was the other answer.
Hi Ruslan,
The reason you know you are missing half the voltage is that you have zero voltage drop across the heater. No current is flowing through it, no work is being done.
This means that one of the lines is open, and the 120vac you are measuring from each side of the heater wrt N is coming from the same source – either L1 or L2.
This is like the last question we asked you on the Midterm exam.
So, the correct answer would be #2 above.
Does that make sense?
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