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Thanks for using the forums to ask a question!
We go through a very similar scenario in the last video of this unit, showing step-by-step how to calculate this (the numbers given in the scenario are different, but the procedure is the same).
Please re-watch that video and let me know if there’s a point at which you don’t understand something (you could tell me the time stamp). I can help you better if I know exactly where you start to have difficulty following it.
These are the kinds of questions we help techs with over at Appliantology.org. Are you familiar with that site? You can read about the professional membership here:
MRA students do get a discount – use our Contact form to ask about that if you are interested.
Student forums are for course-related questions. Appliantology is for service call support.
Yes – that’s correct! I see that you got the questions correct on your re-attempt. Good job!
The diaphragm moves up and down as the gas pressures in the intake and exit fluctuate. Seeing the cutaway of an actual valve helps us to visualize it.
It’s important to keep the terms clear.
The intake into the valve is “unregulated” or “upstream”
The gas leaving the valve is “regulated” or “downstream”
You do have to read the questions carefully, and review the info in the video in Unit 6 (the first 3 or so minutes) to answer these.
Here are the questions:
Which direction would the diaphragm in a gas pressure regulator move if the regulated pressure decreased?
Which direction would the diaphragm move if the unregulated pressure increased?
Do you know what we mean by regulated and unregulated pressure?
One thing you can do is go back and watch the video at the end of Unit 3 about the loose connection. It is also a 240vac circuit with two “loads” in series (even if one of the loads is accidental, from the loose connection.) It is similar to Question 7 in the Voltage Drop quiz.
The basic technique is to calculate the circuit current (and to remember that the current is the same throughout a series circuit), then knowing that you can calculate either voltage drop or power across each individual load.
Let me know if that helps, or if you have any more specific questions about what I wrote above.
That’s a little bit tricky because it’s close to being the right answer. Yet it’s not “heat” that you need, but an ignition source (a flame or spark).
Here’s what we say in Unit 1 of the Gas module: But all gas fuels work the same way because they all obey the Fundamental Law of Gas Appliances: If you have good fuel flow that meets a good ignition source, you will have flame.
Does that clarify it for you?
P.S. Glad that you are going over your quiz/exam results and thinking about the answers!
Yes it does! If electrons have a choice of going through a wire with no load and going through a wire with a load, they will choose the no-load path 100% of the time.
So, what do you think that does to the circuit configuration?
There is one thing you are missing about what’s going with these circuits. (Don’t feel bad – it’s the most commonly-missed part of the exam!)
Look at the branch with the detector switch in it. When that switch is closed, it creates a special circumstance that we talked about in Unit 5.
Do you see what I’m talking about? Let me know! (If not, I’ll try to give you another hint.)
We also teach you in Unit 3 the conversion between inches wc and psi:
1 psi = 27.7 inches wc
To figure out which choice is correct for Question 6, you have to do some converting of the answers to different units. It’s good practice for something you occasionally have to do in real life!
Yes, that is correct!
Hi David – the final answer is correct!
However, maybe you mistyped something. You wrote 240/6 = 6.66.
It should be 240/36 = 6.66
Good job. I’m going to hide your answer so we don’t just give it away to the next student 🙂
January 6, 2017 at 2:53 pm in reply to: Module 3 unit 5 question about paralell vs series curcuit #11085
- This reply was modified 4 years, 9 months ago by Susan Brown.
They can’t be in series with each other, or else as you say the other lights wouldn’t stay on. So the lights are loads that are somehow in parallel with each other. (I don’t know much about household wiring, but I’m sure you could do an internet search if you wanted to know more!)
Dotted lines often indicate an optional component or configuration, since diagrams/service manuals often cover several models of an appliance, and not all models will have all features. (See the diagram for Q. 7, for example).
For the purpose of question 8, however, it doesn’t influence the answer and you can ignore it. The thickness of the lines also have no meaning for this question.
The equation is
So, it is actually “1 divided by (1/20 + 1/10)” which is then “1 divided by .15”
That equation is a mathematical representation of what the equivalent resistance would be if all loads that are in parallel with each other in a particular configuration were somehow combined to be just a single load. In other words, it shows the net affect on the power source (in terms of resistance) of multiple loads that are in parallel.
When loads are in series with each other, you simply add the resistances to get the same kind of thing (equivalent, or “total”, resistance of the circuit).
Does that help?December 30, 2016 at 6:11 pm in reply to: Module 3 unit 5 question about light bulb brightness #11074
You might not have noticed it, but this explanation shows up at the quiz after you’ve submitted it:
“Remember that in a series circuit, the voltage drop across each load is proportional to the resistance of that load. In circuit A, the current must flow through the heating element and will drop some voltage across the element, leaving less voltage available to light up the light bulb. In circuit B, the light bulb gets all the voltage since the heating element is being bypassed.”
Does that make sense? You’ll get some more practice with voltage drop in Unit 8, too.