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Hi James,
In this question, R1 is 5 ohms and R2 is 5000 ohms.
We know that voltage drop is proportional to resistance (E = I x R)
We also know that work/power is proportional to voltage drop (P = I x E)
So, you can figure out logically that if R1 is 1000 times less than R2, it will also do 1000 times less work than R2.
The reason your calculation didn’t come up with the same conclusion is that when you are using P=E^2/R in a scenario with multiple loads in series, the “E” has to be the voltage drop across that particular load, not the source voltage.
You could try calculations again to confirm this, but first you would need to calculate the circuit current, I, then you could use P = I^2 x R. You could also calculate the voltage drops across each load and then P=I x E.
Does that make sense?
That’s correct!
The element has continuity.
If current were flowing through it, we would measure a voltage drop from L1-L2. Since we don’t, we know that there’s no current, and we have an open on one side of the circuit. The element is basically going to act as a wire (conductor).
Sketch that out on a piece of paper and think about how the voltage will be present and what you will measure, and see if that helps.
Hi Darryl – did you get my email? I was trying to step you through it there.
Hi Melissa,
No worries! You’ve got a Student Membership over at Appliantology, which is where you can get help on jobs that you are currently doing. Post a topic in the Appliance Repair Tech Forum about this.
Also, you will learn about this technology in more detail in the Oven & Range course.
It takes some effort to learn this stuff – we are glad to help anytime!
That’s a good and important question!
The answer is because of Kirchhoff’s Law. The sum of the voltage drops will equal the voltage supply for a circuit. It is a 120vac circuit, so we know the two voltage drops will need to add up to 120.
60+30 is only 90
80 + 40 = 120vac AND 80 is two times 40.
Make sense?
Hello!
The best place to start is with the Help Page that we linked you to when we emailed you after the midterm.
I’ll resend that email to you – look for it!
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!
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