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Hi Danny – I will send you an email with your options, so keep an eye out for that.
1. That’s correct – the answer is “It does not have a supply voltage, it switches 120 VAC at its working contacts”
2. We have never seen an NTC thermistor that didn’t have a 5vdc supplyOkay, I had stated that the Resistances were R1 = 100, RA = 80 and RB = 480.
You are trying to do a similar calculation using different resistances?
If R1 is 84, RA is 83, and RB is 423:
The Equivalent resistance of RA and RB would be 69.4 ohms [1/(1/83 + 1/423)]
If you put that in series with R1 you have a total resistance of 69.4 + 83 = 152.4 ohms.
So I = 240/152.4 = 1.6 amps
Note – you won’t have to do series-parallel calculations for the Midterm Exam. You just have to understand how voltage and current work in series, parallel, and series-parallel circuits.
Hi John,
What is the scenario you are talking about? Is this from one of the videos or a quiz question?
A couple of volts plus or minus is fine.
The reason you got 119 is just because 120v/85 is 1.4118, not just 1.4. If you preserved all the decimal points, you would get exactly 120v
Hi John,
Yes, I = E/Rtotal, so in a 120v circuit with a total of 85 ohms, the current would be 1.4 amps.
Then, if you want to calculate voltage drop across a load, you would use E = I x R, where R is the resistance of the load in question.
If there is just a single load in the circuit, then you know that it will drop the full source voltage without having to do any calculations. You actually showed that in your example by ending back at 119v for the voltage drop calculation.
Amps is the one property/measurement where it is usually best to preserve the digit after the decimal point rather than rounding up. Volts, Ohms, and Watts *usually* do not require this level of precision.
For example, on gas valve specifications, you will often see ranges that look like 2.5-3.0 amps or 3.2-3.6 amps, etc. So, several 1/10s of an amp can be significant.
You will encounter this on the Midterm Exam. You’ll be asked to calculate the circuit current for a series circuit, then later use that as part of your calculation of voltage drop across each load. If you round the current up to the nearest whole number, you would end up with voltage drops that add up to exceed the source voltage, which is not possible.
Hi Droo – you can use our Contact form (look under “Campus Support” in main menu), or just email me directly: susan@mastersamuraitech.com
My apologies – I had made a mistake earlier in this thread. The correct answer is
To relieve gas pressure buildup above the diaphragm in case the diaphragm develops a leak
Hi Ray,
You already did the first step of creating your “Grasshopper” account at Appliantology last month. Now you need to go to Unit 3 of the Appliantology 101 course (from your “My Courses” page) and submit the form requesting your upgrade to the 6-month Student Membership. Once your account is upgraded, you will have full access to everything.The main point of that video is to teach about using EEPs for measurements. But, you are correct that we caution techs about Ohm measurements. However, they have their use. For example, if you determine that a circuit or element is open using an ohm measurement, then that is reliable information. If you get a “good” ohm reading, then you often have to push that investigation further, but you have at least ruled out an open fault.
Does that make sense?
See the Orientation Module, unit 1: Pre-flight checklist item 4.
Hi Andrea,
#16: this is basically Question 9, Part 1 from the Midterm Exam, if you remember that one. Review that answer and see if it helps you.
The Element (“heater”) is not connected to a neutral line in any way, so part of your thinking here is not valid.
In order to measure voltage across the element, either current needs to be running through it so that you have voltage drop, OR the element would have failed open and you would be measuring voltage potential. In either of those situations, in an L1-L2 circuit, you would measure 240v.
So measuring 0v across the element tells us one side of the circuit is open.
17. Again, this is similar to the Midterm #9, part 2. Half-splitting occurs when we disconnect the L2 side from the element and re-measure.
You measure 120 V AC at the disconnected wire and close to 0 V AC on the wire still connected.
In other words, we measure 120v wrt N on the L2 wire, and 0v wrt N on the L1 wire (referred to as the “control circuit” in the correct answer choice). So the L1 side is the one that is open somewhere.
21. “Missing leg” is referring to one of the lines (L1 or L2) being open. In the video, it turned out that both lines were fine and providing voltage, so the answer is “neither.”
The answer is “Load”.
See the definition we give for a load in Mod. 4, unit 1. And in the section on Power in unit 3 we mention loads.
Let me know if you have any other questions.
Hi Effy,
180 degrees out of phase means completely opposite. A circle is 360 degrees. 180 degrees is halfway. So if a person turns 180 degrees, it means they are opposite of the starting point.
So, when we are talking about L1 and L2, it just means that when L1 is positive, L2 is negative, and vice versa.
If you haven’t already, rewatch the following videos:
Mod. 4, unit 6, second video: “Household 120-240 VAC…”
Mod. 4, unit 9 videoOne other video you can watch, to see a different way of presenting it, is this one:
https://youtu.be/vOh2OSJ44eE
I hope these help. Let me know! -
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