Susan Brown

[Susan Brown]

Hi Darren,

Things have been very chilly up here, but we’re doing great, thanks!

If you have a DMM with LoZ, then you don’t need a Wiggy.

We have considered selling merch, but haven’t implemented that yet. We love orange, too! (Partly from being Clemson alum.)

Our graphic designer had started working on some ideas a while back – we’ll get him going again and let you know.

[Susan Brown]

Why are you trying to calculate amps there at the end?

[Susan Brown]

Hi Tyler,

Rewatch the first video in Module 3 Unit 5 starting at about 11 minutes.

https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/

[Susan Brown]

To successfully answer Question 9 you just need to know the basics of voltage, voltage drop, current, and L1-L2 power supply.

Look at Figure 1 and think about what measurements you would expect to get if the circuit were functioning properly. Remember that we tell you that the element has continuity.

IF the element was heating, what voltage drop would you expect to measure across the element?

[Susan Brown]

Hi Mark,

You are all set – I see the 100% score.

The way our course software words that message is a little awkward. As soon as you start the second attempt, it changes it to say you’ve already used two attempts. It’s just jumping the gun a little bit with that message, for some reason. Sorry for the confusion!

[Susan Brown]

Hi Cole,

I know that can be a little tricky, but just do your best. I will write out fractions like this: 1/2. If there’s something more complicated, use parentheses, like this: 1/(2 + 3) (which would mean that it is 1 over 2+3, in other words, 1/5. Note that you can substitute the words “divided by” for “over”.)

Does that help?

[Susan Brown]

Make sure you are looking at the page numbers on the pages of the book itself, and not the numbering according to the pdf reader. Page 98 in the text has Figure 6-31 at the top of it.

[Susan Brown]

No, you don’t really have it correct.

This is an L1-L2 circuit. There is no neutral line, but we do use a neutral point to do some of the voltage measurements. That’s important to keep in mind.

Have you rewatched the video at the end of Unit 6 to help prepare for this? And then re-read the Midterm Help Page?

[Susan Brown]

That’s correct. Do you understand the whole scenario now for #8?

[Susan Brown]

Correct! So, what does this mean for the main coil – does it have any voltage drop?

[Susan Brown]

is the detector the shunt or the thing being shunted? (That’s what I had asked: “which one is the shunt, and which one is being shunted?”)

I just want to make sure we’re clear!

[Susan Brown]

A shunt is a path that has no resistance. Shunts are used to bypass one or more loads at certain times during an appliance’s operation.

So, between the closed detector switch and the main coil (which are arranged in parallel to each other, so electrons have a “choice” of which path to take), which one is the shunt, and which one is being shunted?

[Susan Brown]

The “Zen trick” is very helpful in determining how a load is laid out in relation to the power supply and other loads. However, you have to take the whole circuit into account.

You’ve already determined that current that flows through the ignitor and the booster will go through the closed detector switch on its journey to and from N.

Do you recall what a shunt is and how it functions in circuits?

[Susan Brown]

That’s correct!

[Susan Brown]

You are correct for the Ignitor and Booster.

However, is the main coil receiving any current? Remember, a load has to have current flowing through it in order to have voltage drop.

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