Susan Brown

[Susan Brown]

Yes, each machine will have its own “key dance” to enter diagnostic mode, so the correct answer is, “Look it up on the tech sheet”

Let me know if you have any other questions.

[Susan Brown]

Okay!

[Susan Brown]

4 Megawatts is 4,000,000 watts (You can type 4 megawatts in your search engine to see how many zeros it has, if you forget that Mega means 6 zeros. Mega = million)

The possible answers are:
A. 400,000 watts
B. 400 kilo-watts
C. 0.004 Giga-watts

So A is obviously not right.
A kilo is 1000, so 400 kilowatts is 400,000 watts (add 3 zeros to the end of the 400). So also not right.
C. Must be it. Let’s see: a Gigawatt is 9 zeros, or 1 billion. 0.004 x 1 billion = 4,000,000. yes!

Occasionally you have to know these things – but you can use your search engine to help you do the conversions. It’s good to know how to do!

[Susan Brown]

Hi Yehoshua,

Read this topic and see if it helps. You haven’t gotten into parallel circuits yet, so don’t worry about that part.

If there is a particular quiz question you want me to step you through, let me know and I’m happy to.

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/voltage-and-voltage-drop-summary/

[Susan Brown]

We don’t currently have access to a scale, but our best guess is about 30 pounds.

[Susan Brown]

Hi Jesse,
No, we don’t have something like that. The examples we use in the course are on just a small percentage of the thousands of models out there.

Are you wanting the manuals for when you are working on jobs? That is what Appliantology is for!

Be sure to activate your free 6-month Student Membership there. See the “Appliantology 101” short course at MST for how to do that.

[Susan Brown]

Do you recognize these? You had the same scenario on the Midterm Exam in the Core course (the diagram just looks a little different).

Remember that when you have a single load in a circuit, that it will drop the full source voltage. Only loads in series will split the voltage drop. Loads in parallel will behave independently of the others.

The Safety is the clearest example of this – it is parallel to the other circuits.

The Main coil is wired in series with the Ignitor and Booster, which are in parallel to each other. However, the closed detector switch has a big impact on these circuits – it is what allows us to answer the questions without doing any calculations.

Is this ringing any bells?

[Susan Brown]

“open” is a better term than “missing”, but yes, L1 is the side with the fault.

[Susan Brown]

The fact that I’m reading 0 across the heating element means that there is an issue with one side of the 120s

That is correct, although you can be more specific about what you mean by “issue”.

240 being direct current only reading 120

I’m not sure what you mean here. This is an AC circuit, not DC. Is any current flowing in this circuit anywhere?

When we measure L1 wrt N or L2 wrt N, we would expect to read 120v each, because we’re just measuring one phase of power.

Where would we normally expect to read 240v in an L1-L2 circuit?

L2 is not present. When you disconnect it it’s just reading L1 threw out the circuit . l2 is open

Are you looking at the diagram on the Midterm? (It is a little different than the similar one in unit 6.) Which measurement changes after we disconnect one side?

[Susan Brown]

Correct! I moved your topic on Question 9 to a new topic

[Susan Brown]

Okay, good!

The Safety is 120v.
The Main Coil is [answer hidden]

How does this leave the Ignitor and the Booster?

• This reply was modified 4 months ago by Susan Brown.

[Susan Brown]

Main coil igniter and booster 40 vd each due to being in series

There are several reasons this is not correct. For one thing, voltage drop is proportional to resistance, and we haven’t given you any information on the resistance of these loads. Second, if all 3 of these loads are getting current, this is a series-parallel circuit scenario, which would require some complicated calculations.

However – none of that is necessary. Big hint: One of these loads is not getting current.

The key is to recognize the function of the closed detector switch.

Do you have any ideas about that?

[Susan Brown]

Okay. If you haven’t, see the Appliantology 101 short course, specifically units 2 and 3.

[Susan Brown]

Hi Jacob,

Question #4 – What does an oven’s gas valve need in order to open and release gas?

From the unit:

The gas valve has a specification for how much current must flow through it in order for it to release gas into the burner. The ignitor’s resistance lowers the hotter it gets, letting more and more current through. Once it is hot enough to allow the current to reach the valve’s spec, the valve opens,

I bolded the pertinent parts – heat is involved, but not the key component.

• This reply was modified 4 months ago by Susan Brown.

[Susan Brown]

Correct!

Now the next step is to figure out what is going on with the other 3 loads, if they are in series or parallel with each other.

The Main coil is wired in series with the Ignitor and Booster, which are in parallel with each other. Normally, when we have a series-parallel configuration, the voltage drops are impossible to determine unless we give you the resistances so you can do some calculations (like I show in one of the videos for Unit 5).

***But that closed detector switch changes things significantly, and allows us to know the voltage drops without needing to do any calculations.

Do you have any idea what that might be, or do you need another hint?

[Author]

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