Susan Brown

[Susan Brown]

If an appliance is plugged into (or in the circuit with) a GFCI outlet, and the outlet trips for some reason, that will cut off power to the appliance. Resetting the outlet should restore power.

Non-GFCI outlets don’t have that feature – if they are not providing power it is because the circuit breaker tripped or a wiring issue with the outlet/circuit.

[Susan Brown]

Hi Douglas,

From the unit: “Depending on the model, you can either have a single spark module that sends sparks to all the burners (called multi-point ignition) or you can have a separate spark module for each burner (called single-point ignition).”

I know it seems counterintuitive at first, but if each burner has its own spark module, then it only ignites at that one point. If there is only one spark module for the whole cooktop, that one module needs to ignite at multiple points.

Does that make sense?

[Susan Brown]

Hi Andrea,

The key here is the weird 78v reading. This is the kind of thing you get with “ghost voltage” which is often related to an open neutral.

So the two answers we are looking for are:
That you are probably seeing ghost voltage and should have set your meter on LoZ or used a loading meter to make this measurement
That you are probably dealing with an open Neutral situation

[Susan Brown]

Hi Raja,

#9: we are using “switch” as a verb there. In other words, what does the PCB control: line or neutral?

For #10, see this section of the schematic:

[Susan Brown]

Done!

[Susan Brown]

Hi Denis,

I mentioned this in the Video in Unit 3, around 12 minutes or so in.

It depends on what relationship you are focusing on.

E = I x R is talking about voltage drop (because we’re dealing with a resistance).

Here are some examples that hopefully will help.

Scenario 1
Let’s say we’re dealing with a 120v circuit with two loads in series, R1 with a 40 ohm resistance and R2 with an 80 ohms resistance. So, total of 120 ohms. The current flowing through the whole circuit would be I = 120v/120ohms = 1 amp.

So the voltage drop across R1 would be I x R = 1 x 40 = 40 volts.

Scenario 2
Now let’s say we swap out a different load for R2, which only has 20 ohms of resistance. This will change the current in the circuit.

Our total resistance in the circuit is now 60 ohms, so the current is I = 120v/60 ohms = 2 amps.

The voltage drop for R1 is now 2 amps x 40 ohms = 80 volts. The current going through R1 doubled, as did the voltage drop. So – directly proportional.

Okay, now to talk about power. A typical way we talk about power in appliance circuits is the wattage specification for a load. Let’s say we have a load that needs to generate 500 watts.

If this is a 120v circuit, and this load is the only one in the circuit, then we can figure out what current needs to flow through the load to get 500 watts. P = I x E, so I = P/E = 500/120 = 4.2 amps.

But what if we have a 240v circuit instead? then I = P/E = 500/240 = 2.1 amps. So, when voltage supply went up, current went down.

Does that help?

[Susan Brown]

Hi Quentin,

The short answer is, it depends on the load you’re dealing with and the problem the machine is having. And as you gain experience reading schematics and more familiarity with the various loads in appliances, you’ll get stronger at knowing which tests you want to do when.

Also, which measurement(s) you make will be clearer if you are following the Ten Step Tango. If you have a good hypothesis (Step 5), you’ll know what you need to test. Review Mod. 6, unit 3 of the Core course.

It might also be good to review Mod. 4, unit 7 (Electrical Measurements) as well.

But, in general, the go-to tests are voltage and amps. We have many videos showing us using these to troubleshoot.

Voltage, to see if voltage is getting to a load. If not, you then know you’re looking for an open somewhere between the load and the power supply. If you get voltage at the load, then you’d want to check for current (amps) for your next piece of info. Resistance is the measurement we use the least, as the others are more instructive and reliable.

Let me know if you have any specific follow up questions.

[Susan Brown]

You’re troubleshooting a problem with the oven where neither bake or broil are working. You use your voltage light stick at the ignitors and see that neither is getting voltage. What’s your next move?

Look at the schematic on the left hand side and you’ll see the two ignitors (bake and broil) and the test points that correspond to each ignitor. What you will want to do is energize either the bake or broil, and then test at the appropriate points for that ignitor.

Do you see what the test points are for the
1. Bake ignitor
2. Broil ignitor
?

Let me know what you think they are.

[Susan Brown]

Are you clear on this answer now?

[Susan Brown]

You are correct!

[Susan Brown]

Hi Raja,
I can help step you through these. Please answer the questions in bold.

Question 5:
If you look at the J1 connector, what is the voltage supply to it? That is what will be available to the optional evaporator fan.

Question 14:

The hint for this question is:
Use the watts specification for the condenser fan motor to calculate its resistance.

Were you able to do this? What did you get?

The explanation that showed up after you took the quiz for the first time is:
There are 3 resistances in parallel here. The resistance of the condenser fan motor is *much* higher than the other two, and thus has little impact. The answer will be, basically, the equivalent resistance of the two compressor windings.

What are the specifications for the compressor windings?

[Susan Brown]

#3: yes, triacs and relays
#7, 12, 16, correct
#20: the answer is “Set the washer to fill and see if water comes into the drum.”
#21: the other answer is “Wiggy or solenoid voltage tester”

As for your question connected to #16 – remember that the drain pump and water valve are in series, so they will have the same current flowing through both of them (determined by the total resistance in the circuit). So, if the R of the water valve is much higher than the pump, very little voltage will be dropped across the pump, so it won’t have enough power to operate. Does that help?

[Susan Brown]

Hi Denis,

There are three answer choices on this question:
Yes
No
Probably, but it depends on the model

The text in the unit is:

The water supply from the home connects to a water inlet valve. When this valve receives voltage from the control, it allows water to enter the dishwasher tub. How long this valve is kept energized depends on the model, but it is usually under 2 minutes to get the specified amount of water in the tub.

So the answer we are looking for is that third choice.

Let me know if you have any other questions!

[Susan Brown]

Hi Joshua,

The reading across the element is looking for voltage *drop*, since we are measuring across a load. If current flows through a load, then it will create a voltage drop. Since we measure 0v, that confirms that no current is flowing in the circuit. In Figure 2, since we have deliberately disconnected the circuit (and current will not flow in an open circuit), we would not expect to measure any voltage across the load.

Does that help?

[Susan Brown]

My iPad is pretty old, so I don’t know much about the newer ones. Generally, we always get as much memory as we can afford. However, unless you are going to store a lot of video, you don’t need to max it out.

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