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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?
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.
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?
That’s correct. Do you understand the whole scenario now for #8?
Correct! So, what does this mean for the main coil – does it have any voltage drop?
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!
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?
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?
That’s correct!
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.
Why would the electrons go through the Main coil, which has resistance, when they have the option to go through the closed detector switch?
That is absolutely correct!
Now, what happens when you “become” the Booster?
You are correct about the meter 1 reading. We know the element has continuity, yet there is no voltage drop across it, so we know there’s no current flow. We also know there is some voltage present, so this means we’ve got an open somewhere.
But, you aren’t quite right about Meter 2 and 3 readings. Remember, you can read voltage across an open switch, which is just indicating the potential for current to flow.
You know that the element is not open, so this means the open is on one side or the other. Your answer for Part 2 (on your second attempt) explains the measurements we’re getting.
Hi Darren,
In general, Appliantology questions are best asked using the Contact form at Appliantology.org, as that will go to the Appliantology Admin, Sam. Have you ever taken advantage of your Student or Alumni memberships?
Hi Dan,
I’m away from the office this weekend (where my copy of the book is), but I believe there are two or three example problems on that page that they show. We just want people to re-do those problems on their own and make sure they follow them.
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