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Hi Jim,
For future reference, please use the Contact form at Appliantology.org for questions about Appliantology, since we have a different Admin for that site. Based on what I could see, I think your display name is jim ortiz
Can you just tell me what you determine when you “become” the ignitor? It’s easy to see how you can reach out and get to L1. How would you get to N?
Question 8: there’s an important thing going on in these circuits that when you see it, you will know the answers without having to do any calculations.
Do the “Zen trick” on the ignitor… how do you reach N? What about the booster?
Let me know, and we’ll go from there.
Hi Jim,
First of all, did you get my email with feedback on the Midterm? You did get the answers correct to parts 2 and 3 on Question 7. Are you just trying to understand the answers better?
This is an example of where it’s good to cut through confusing-looking notations, etc., and try to see things more simply.
These are just two parallel circuits. The top one has a component that fails open.
We are checking to see if you understand the relationship parallel circuits have with each other. Does a failure in one affect the other?
Then, we want to see if you understand the effect that failure has from the point of view of the power supply (the overall current “draw”).
The videos at the end of Unit 4 go over this material.
Let me know if you have followup questions.
Hi A.J,
We do the calculation in the video at the end of Unit 3, just with slightly different values for the resistances. Could you follow what we did?
You start with calculating the circuit current, I.
Then you use that and the resistance of the loose connection to calculate the heat, P, in watts.
A common mistake is to forget that the circuit current is determined by the source voltage and the total resistance in the series circuit.
Let me know what you get for I and then for P.
Hi Ethan,
A shunt will result in all of the current taking that path rather than the alternate path with a resistance in it.
In the example you give, when the switch is closed the heater is completely bypassed – no current.
Correct – the supply voltage is what it is and won’t be affected by a failure of a load.
Did you try clicking the gear and improving the quality?
I took a screen shot – let’s see how that shows up here:
I am not looking at a diagram, but it sounds like you are describing a circuit where R1 and R4 are in series with each other and also with R2 and R3, which are in parallel with each other.
The total voltage drop across the entire circuit would always be 120vac. The voltage drops across each load would change, at least a little bit, with the failure of one of the parallel loads.
Would it be correct to assume that the total current would decrease because the resistance is actually increasing?
Yes – that’s exactly right. You can also think of it this way: you lose the current in one parallel circuit when it fails, but the others stay the same, so the overall current decreases.
Hi Michael,
Do you know how to make the video full screen? You click the little square with 4 arrows in the bottom far right corner of the player window. The images are a little blurry, but when I do that I can read them clearly.
Does that help?
Hi Kevin,
There are a lot of tests in the course, and many of them are not very difficult, particularly when you take into account that the tests are “open book” (meaning, you can look over the unit and your notes while you are taking the test) and you have more than one attempt at each test.
On the ones that are a lot more difficult, such as are found in the Basic Electricity module or the Midterm Exam, a student could have poor comprehension and receive low scores and yet still have an overall high average because of the large number of easier ones.
Since the information in the Basic Electricity module is arguably the most important for a student to master, we want to make sure they have the incentive to do so. That’s why we have the system we have, and it has worked very well.
Does that make sense?
Did that answer the question for you? I was expecting to step you through it more, starting with you answering the question I asked.
Hi Jim,
It’s always good to start by thinking about the basics: current, voltage, and open/closed circuit.
In order for loads, like a light bulb, to do work, they need Power, which is voltage and current. And current will only flow in a closed circuit.
In this scenario, we have two light bulbs in series. Do you know the physical effect on a circuit when a light bulb burns out?
Hi Jim,
When current flows through the two loads in this circuit, it will produce heat (and voltage drop). Heat is a form of power.
There are several different formulas for P in the Ohm’s Law chart that we gave you earlier. The easiest one for this application is to use P = I^2 x R, where I is the current in the circuit and R is the resistance of the load you are interested in.
(I^2 is another way of writing “I squared”)
So, this is a two-step calculation. First you need to find the current in the circuit, then when you have that you can calculate the heat generated by either load.
We show this type of calculation in a similar scenario in a video in Unit 3 on the “loose connection.”
See if that helps you to figure this out. If you have any followup questions, let me know.
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