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.)

[Susan Brown]

They do need to total 120v, but they will only have equal voltage drops if they have equal resistances. Remember, the voltage drop of a load is proportional to its resistance. (Thus the equation E = I x R)

[Susan Brown]

Yes, [answer hidden] What effect does the detector switch have on the circuits? In other words, what is its function?

• This reply was modified 1 year ago by Susan Brown.

[Susan Brown]

Hi Ian,
Some of this should become more clear as you continue through the Module. Units 8 and 9 especially.

Here’s a quick summary.

L1 and L2 each provide 120v to the element. Because they are out of phase with each other (when one is negative, the other is positive), there will be a 240v voltage difference driving current. If the circuit is working (current is flowing), then you would measure 240v (voltage drop) across the element – from one side to the other, or “L1 wrt L2”.

If you measure 0v across the element, that means one side is open and no current is flowing.

When you have a 240v circuit where one side is faulty (open somewhere), it is hard to know which side has the fault because you’ll still measure the 120v (with respect to N) present at either end of the element, coming from the side that is “good”. (You don’t have this issue with a 120v circuit, since only one side is “hot”)

That is where “half-splitting” the problem – disconnecting one side – can show you where the fault is. The open side will go to 0v wrt N.

Does that help?

[Author]

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