Susan Brown

[Susan Brown]

Hi Michael,
Questions like that you can always email me about, FYI. But this is fine, too!

MST Courses are like parallel circuits – what happens in one does not affect the others 😀

The only thing you risk by taking a break is needing to spend more time refreshing for the Final Exam. However – that is actually a better method in terms of retaining the info. And hopefully you’ve been taking good notes to help you with this.

[Susan Brown]

Hi Dylan,

I’ll start by just telling you how I came to understand it, to see if it helps.

All voltage is a difference in charge between two points. For our purposes (dealing with circuits in appliances), there are two categories of voltage.

1. Voltage potential – the difference in charge between two points where there is no current flowing, but there is the potential for current to flow if there is a closed circuit between those two points. A common measurement of this type of voltage is putting one probe at one side of a load, and the other at a good neutral point. Or, measuring across an open switch.

2. Voltage drop – the difference in charge from one side of a load to the other, created by current flowing through the load. In other words, you will only measure voltage drop in an active circuit.

“Dissipation” sounds more like an idea from power transmission, where you can have some losses due to small amounts of resistance over very long stretches of power lines. Small amounts of the power are transformed into heat and lost to the environment. This is not something we deal with in appliance circuits, which are very small compared to transmission lines.

Here’s something else to keep in mind. In circuits, there are always two sides to every power supply. Either L and N, or L1 and L2. These two sides meet at loads.

Switches, on the other hand, are always on just one side of the power supply (otherwise they would create a short).

Let me know if this helps, or if you have any follow up questions.

• This reply was modified 10 months, 1 week ago by Susan Brown.

[Susan Brown]

Hi Douglas,

That is correct. The closed detector switch is a [answer hidden]
Good job!
Note – I will need to hide these answers.

• This reply was modified 10 months, 1 week ago by Susan Brown.

[Susan Brown]

Great! Keep up the good work.

[Susan Brown]

Hi Edwin,

I love to see students playing around with these calculations in order to understand!

Your mistake is in how you figured out the current in each circuit – trying to use proportions. Remember, current is inversely proportional to resistance. Did you notice that you were getting higher current flows in the circuits with higher resistances? That is not how it works. I = E/R

The voltage supply to each parallel circuit is 120v. This fact is true regardless of the state of any circuits in parallel to it. So the voltage drop across these single loads will always be 120v if they are active.

The resistance of each load doesn’t change – it is a fixed quantity. So, how can the current in each circuit change due to the failure of a circuit in parallel?

Let’s look at the calculations.

Given your scenario, with three parallel circuits with R1 = 100 ohms, R2 = 200 ohms, and R3 = 300 ohms, Req = 55 ohms, Itotal = 2.2 amps.

I1 = 120/100 = 1.2 amps
I2 = 120/200 = 0.6 amps
I3 = 120/300 = 0.4 amps

Totals up to 2.2 amps.
Notice that I1 is 3 times bigger than I3, which makes sense because R1 is three times *smaller* than R3. (Current is inversely proportional to Resistance.)

If you remove R2, you get Req = 75 ohms, Itotal = 1.6 amps
You can see from above that I1 + I3 = 1.2 + 0.4 = 1.6 amps

Same thing if you only have R2 and R3 active, It = 1 amp, which is what you get if you add I2 + I3.

Does that make sense?

[Susan Brown]

Hi Anthony,

The correct answer is “serial data transfer”, which is a type of digital data communication. You are not actually wrong, but we want the better, more specific terminology for the answer here.

We realize the distinction is pretty subtle, so we decided to take away “digital” as one of the answer choices.

[Susan Brown]

Hi Tiago,
All you need to know is that the detector switch is closed. Like any other closed switch, it acts just like a wire, with no resistance. So the question you ask yourself is: if you do the “Zen trick” on the Ignitor or the Booster, how will you reach N? Through the closed switch, the main coil, or both? Review the first video in Unit 5 for a refresher on the Zen trick and the effect of a pathway like this closed switch on circuits.

[Susan Brown]

Hi Travis,

The Advanced Courses assume that students have taken the Core course, so there may be questions that draw on fundamental topics like electricity and circuits.

But most of the time quiz questions are drawn from the material in the unit at hand.

For example, the question “What are two common operating voltages for door locks in computer-controlled washers? (Mark both correct answers)”
In the first video, we show different examples. At least one is using 120v, and the other 12vdc. (5vdc is just data, not operation).

For any of the questions you are stumped on, you can ask us for guidance. Just post it here.

[Susan Brown]

Question #9: The heater relay on the heater PCB switches _____ to the heater.

[Susan Brown]

Hi William,

The first sentence of the question has important technical terminology: You’re testing a micro switch in a live, 12 V DC circuit. That tells you all you need to know – the switch is in a circuit which will have at least one load.

[Susan Brown]

Okay – just did that for you.

[Susan Brown]

Hi Anthony,
I see that you completed the Module – would you still like me to reset you?

Right – the board would not be a suspect

[Susan Brown]

Hi Gustav,
If any of the lessons require a tech sheet to answer the quiz questions, we will provide it. Any other tech sheets you need can be gotten from Appliantology. I notice that you sent in your Student Membership request, but it doesn’t look like you set up your account first, which is necessary before we can upgrade it. Please see Unit 2 of the Appliantology 101 course on how to do that, then let us know.

[Susan Brown]

With the right messaging to your customers, a few days here and there would not be a problem – they wouldn’t even have to know. They could just assume that you are so busy you can’t schedule anything until the following week.

For the couple of months stretch, that’s a little tougher. You would need to let people who contact you know that you are temporarily unavailable for service calls until a certain date. This means that people will have to call someone else, which means they might become that person’s customer in the future, too. But not always. We’ve seen techs who needed to pause their business due to injury or something like that, and plenty of their customers still contacted them in the future, even if they had to use someone else in the interim. If people like you, they will call back!

[Susan Brown]

Hi Rudi,

That’s an interesting and great question. My first thought is that this would be great to post at the “Dojo” forum at Appliantology, so you can potentially get feedback from other techs.

How long are you typically unavailable? Are you talking days or weeks?

(We have a booth at several trade shows a year, so I have an idea of the type of work you are talking about.)

[Author]

Posts