Susan Brown

[Susan Brown]

We just want to think about the basics that these measurements are showing us. Voltage, current, open or closed circuit.

What do the measurements in Figure 1 tell you about current flow in the circuit?

[Susan Brown]

Correct! Note – I’ll need to hide your answer.

[Susan Brown]

You’re correct, although to be more precise, there is essentially zero resistance in the closed detector switch compared to the main coil, which makes the closed detector switch a shunt.

What does that mean for the main coil – does it have any voltage drop?

[Susan Brown]

I’m not quite following what you said.

In order to do the Zen trick, you “become” the load. In this case, you put yourself in the place of the ignitor. You then will reach out your arms to one side or the other, wanting to touch L1 with one hand, and N with the other. You will do so keeping in mind the behavior of electrons. If you are the ignitor, you’ll reach out one arm to L1 and just follow the path of the wire, down, over, and up, to L1. There are no other loads that you have to go through.

So, I want you to reach out with your other arm to touch N… what pathway will you take? It isn’t as straightforward as going to L1, but if you “think like an electron”, the path is clear. If you still don’t know, just say so and I’ll give more hints.

[Susan Brown]

In question #8 1.- Ignitor 120 vac , 2.- Booster 80 vac, 3.- 120 vac. 4.- 40 vac.

I’m going to step you through this. First question: When you do the ‘Zen Trick’ on the Ignitor, how do you reach N?

[Susan Brown]

In question #7 1.- the ignitor decreased, 2.- the booster coil No change, 3.- the safety coil decrease, 4.- the main coil No change.

You are mixing up #7 and #8. These are entirely different circuits.

[Susan Brown]

Let’s take these one at a time.

In question #5 my new answer will be: R(eq)= 1 divided by 10= .1, 1 divided by 20 = 0.05, 0.1+0.05 = 0.15 ohms. I = E/R = 10 + 15 + 20 = 45, 120/45 = 2.7 amps

When you start working on a question, pay close attention to what information you are given and what we are asking for. In this case, you have two loads that are in parallel, and are asked to calculate the “equivalent resistance” of those two loads. In other words, from the point of view of the power supply, what is the resistance? You started with the formula for R(eq) (which has a mistake, but I’ll get to that in a minute), but then did a calculation of current.

1. The R(eq) calculation. The correct formula (that we give you in Unit 5) is 1/(1/R1 + 1/R2 +…). So, you found the bottom part of that whole fraction (0.15) but didn’t do the final “1 over”. In other words, 1/0.15.

2. If you were asked to calculate the current in this circuit, you would just do I = E/R(eq)

[Susan Brown]

That’s correct. Now, what about the ignitor?

What do these two answers tell you about the function of the detector switch – how does it being closed effect the main coil?

[Susan Brown]

Hi Jordan,

Let’s start with Question 8.

Take a look at the diagram, and do the “Zen trick” on the booster coil. What path do you take to reach N?

[Susan Brown]

Hi Michael,
You assume correctly. Without voltage, there will be no current flow.

What two items are absolutely necessary for current flow to happen? (hint: you’ll need to be able to answer this on the final exam, so it’s good to make sure you know it now!)

[Susan Brown]

Hi Dylan,

See this video where we step through a similar calculation in more detail, and see if that helps. If not, you’ll need to show me your steps.

https://youtu.be/iHB3lxdc68E

[Susan Brown]

Hi Miguel,

Good question! I’ll give you a short answer here, because you’ll learn even more about AC power in the units ahead.

With DC, it’s easy to picture a battery with a positive and a negative pole or terminal, and how the electrons will flow in one direction if they have a complete circuit from one pole to the other. The terminals never change their charge.

In an AC circuit, the polarity of the power supply terminals switches constantly from positive to negative, causing the back and forth movement of the electrons.

Keep going through the next few units, and let us know if you need more details.

[Susan Brown]

Hi Richard,

Good question, and glad to see you are paying attention!

Part of the challenge in teaching Basic Electricity is that it is a pretty complex topic, and we’re only skimming the surface to equip appliance techs with enough knowledge to assist their troubleshooting, without diving so deep that it’s overwhelming.

When it comes to power transmission, remember that P = I x E. So, in that case, you can see that if you are given a fixed number for Power, increased voltage will result in decreased current.

But we also have the relationship of E = I x R, which shows voltage and current being directly proportional. In an appliance setting, if you are looking at an entire circuit, the value of E is usually fixed (either 120v or 240v). Thus, the variables tend to be I and R. However, as you will learn more about in Unit 8, we can also calculate the voltage drop across a particular load. The higher the current, the higher the voltage drop.

So, you not only have to understand the mathematical relationships between these electrical properties that Ohm’s Law is helping us to see, but the context that we’re using them in.

It takes some time and effort – but please keep working at it and asking us questions!

[Susan Brown]

Just to add to what Sam said…

Using a steamer to make sure you’ve removed all of the ice is one way to help make sure you won’t get a repeat of the problem.

Also, besides what Sam described, some models of fridges actually have a little cal rod heater that is used to prevent the freezing of the condensate drain.

I suspect that your resetting the timer or not was more of a coincidence, and didn’t really contribute to the fix.

[Susan Brown]

That’s correct!

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