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