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Which measurement makes you conclude that L2 is missing?
Also, back to what I was trying to lead you to. You already correctly said no work was being done by the element (zero voltage drop measurement across it).
It takes current flowing through a good load to produce work.
We’ve asked you to assume that the load (the element) is perfectly good.
We’re measuring some voltage present, even if it’s not exactly what we know we should have.
So what can we conclude at this point that we absolutely do not have happening in this circuit?
We’ve already told you that the element is good. We want you to assume that it is a perfect element. So if no work is being done, but we do know at least some voltage is present, what is the only other possible conclusion?
In this question, we don’t need to think about the phase relationship of L1 and L2. It is true that 240vac is created by L1 and L2 being out of phase, but it isn’t necessary to talk about that for this question.
in order for heating element to do any work current must flow
That is correct. So, there’s a measurement of zero voltage drop across the element, and we’ve already told you the element is known to be good. So, what can we conclude about current flow in this circuit?
If you are the booster, you can obviously reach out and touch L1 without having to go through a load. But can you reach out to N without having to go through any other loads? (electrons will always take a path with no load if they have the option)
Yes, that’s correct. For loads in a series, the current is the same throughout the circuit. So when you know the current and the resistance of a load, the voltage drop is an easy V = I x R calculation.
Hi James,
Did you receive the email I had written back to you with comments on your Midterm?
I had explained which parts you got correct/incorrect on Questions 7 and 8.
And, yes, you must have timed out with Question 9.
Here’s part of what I said about 8:
Your answers for parts 1 and 2 are not correct (the ignitor and the booster). Are these in series or parallel with each other? Knowing that is the primary step towards getting the answer correct.Exactly! Does that all seem clear to you now, or do you have any further questions?
If loads are in series, then they will split up the voltage proportional to their resistance.
AC power goes back and forth (rapidly) between L1 and N in a 120vac circuit like this, so a load needs a connection to both or else you don’t have a circuit.
You’re close, but not quite seeing where I’m going. Electrons will always seek a path with no load in it over one with a load in it. There is a special name for this situation – when there is an alternate path for current that has no load. Do you remember what that is called? Can you see how when the detector switch is closed, it creates this situation?
The booster goes through the detector so its 120 in VD, Right?
Yes. So then why would you think that the main will drop any voltage? Will any of the current that goes through the booster and/or ignitor also go through the main?
(We’re almost there!)
Zen trick: Become the Booster. It’s easy to see that you can touch L1 directly, without having to go through another load. What happens when you reach out to touch N? (Remember, electrons don’t see bends or turns in the circuit – they are just finding a path to the power source.)
A good first step for this problem is to examine the layout of the circuits and determine if current is flowing through each load. What do you think?
Hi David,
#6: You didn’t carry the calculation out correctly.
[answer hidden]
# 7: correct. This is actually very straightforward, but I think a lot of students think it’s more complicated. You have a branch with current flowing then suddenly a component fails open.
#9: The measurements in Figure 1 tell us for sure that no current is flowing (since the element is known to be good, it would produce a voltage drop if current were flowing through it). But we do know there is some voltage present from somewhere. When you have voltage but no current, that mean there is an open in the circuit.
Does that help?
P.S. I am going to come back later and hide parts of these answers just so we don’t give it away too readily to other students.
Hi Phillip – here’s another hint.
This is the answer you chose: “58 Ω because the equivalent resistance in a parallel circuit will always be equal to the smallest branch resistance.”
That is close, but not quite correct. As Sam advised, we tell you that “rule of thumb” in the video.
Posting here is exactly what you need to do! Lots of folks find Unit 5 challenging, so I’m glad you’ve asked for help.
We can try to step through some of the concepts you need to understand these questions here.
The first few questions you had trouble with involve a series circuit. A series circuit is where you have the ends of the circuit connected to line and neutral (or, L1 and L2, if it’s a 240vac circuit) with one or more loads in the circuit. The current flowing between L1 and N has to go through all of the loads in the circuit – there are no alternate (or parallel) paths.
Does this make sense? Can you envision this?
What happens in a series circuit if there is an open – whether it is a switch that is designed to open or one of the loads has a failure and fails open.
What does that mean – will current flow through the circuit?
Hello!
We get a Belkin that can handle a max of around 1000 joules.
This is the one: http://amzn.to/2iL5tKR
As for marking up parts – we use the Blue Book to price out our jobs, and they mark up parts for you. I know that for inexpensive parts, the rule of thumb is usually to double your cost. As the parts get more expensive, the markup percentage goes down somewhat.
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