Susan Brown

[Susan Brown]

If there is only one load in a circuit, do you know its voltage drop? (think about what Kirchhoff’s law teaches)

[Susan Brown]

Hi Troy,

No calculations are needed to answer Question 8, and you definitely don’t need to make something up. The key is seeing the impact that the closed detector switch has on the circuits.

Let’s back up a bit – if you look just at the Safety, do you know what its voltage drop is?

[Susan Brown]

Hi Brian,
I reset you. FYI – it’s best to use the Quiz & Exam Reset Request form when needed (in the “Campus Support” menu).

~ Susan

[Susan Brown]

That’s correct

[Susan Brown]

Also – go to the Core course, Mod 4, unit 4, and watch the second video. I start with a circuit with a single 60 ohm load, then add a 40 ohm load in series. The current in the circuit decreases, and the voltage drop across the 60-ohm load decreases with the addition of the second load. But obviously the 60-ohm load still has 60 ohms.

The main point of the question you are writing about is to make sure people understand that resistance is an inherent property of a load and in the types of Ohm’s Law calculations we are likely to do as appliance servicers, if you are given a value for resistance, that value is fixed. It won’t change in response to a change in current, for example.

(Note – one exception to this discussion is that some material’s resistance will change some based on temperature. Some types of sensors are an example of this.)

• This reply was modified 9 months ago by Susan Brown.

[Susan Brown]

Yes, if you add another load in series, that changes the total resistance within that series circuit.

Which unit is this question in? I want to see it in context to make sure I answer correctly.

[Susan Brown]

The voltage in these equations is the voltage dropped across the load, not the source voltage. So, if a load was originally by itself in a circuit, dropping the source voltage, then another load is brought into series with it (by a switch activating), then the voltage drop across the original load will decrease. The current will also change.

[Susan Brown]

One example I thought of where this could happen is based on Question 8 on the Midterm Exam in the Core course where the Main Coil was shunted by the closed detector switch, leaving the Ignitor and Booster in independent, parallel circuits (*each* having 120v drop). When that switch opens, it becomes a series-parallel circuit, where all 3 loads will now share the 120v drop.

So any scenario where a load is either bypassed or brought back into series by a shunt will affect current and voltage to loads.

[Susan Brown]

“k” stands for “kilo”, which is 1000 of something. (For example, a kilometer is 1000 meters). So 5k-ohms is 5000 ohms.

See Mod. 4, unit 7, near the end of the unit for a table of units like this.

[Susan Brown]

Hi Ronald,
I have not heard of this. But I know we have a number of techs from TX at Appliantology – I suggest you post this in the Dojo and see what you can find out.

A lot of these types of licenses have more to do with knowing Code or other regulations, rather than actually understanding technology and troubleshooting, in which case the info you need is likely in that book.

Can you send me more info on this test? (A link?)

[Susan Brown]

The answer we are looking for is “It’s a PTC start device”. Essentially no one uses true relays anymore in this application, so it’s safe to just assume “PTC”.

[Susan Brown]

Those are correct. Good job!

[Susan Brown]

I just replied to your email, Jonathan.

[Susan Brown]

Always check the schematic. Some include a very clear chart of which color is what. Some you have to figure out by using your own savvy. They don’t all stick to the usual black/white convention, unfortunately. We’d have to see the diagram to know what the white wire you saw was.

[Susan Brown]

Hi Raja,
First of all, note that the question asks for the two most likely causes.

Of the three that you gave, the one that is not correct is “A broken agitator”

Here’s what we said in the lesson:

“If the motor is running, but the machine is not agitating, then you are most likely facing one of two issues, depending on the method your model uses to drive the agitator. On belt-drive models, a worn or broken drive belt causes this issue. Alternatively, on direct-drive models where the motor directly drives the transmission, the most likely cause is a broken drive coupler. This is a small plastic component that connects the drive motor to the transmission.”

The agitator is just that plastic center post with the fins. It is the items that cause it to move (belt or coupler) that usually are the cause of the issue.

Does that make sense?

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