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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.
You seem to still have some confusion about parallel circuits. It is crucial that you understand this!
Rewatch the parallel circuits portion of the first video and the entire second video in Unit 5. Then get back to me.
those definitions are correct. So, you can have voltage without there being current (if you’re measuring across an open circuit), but you can’t have current unless you have voltage (along with a closed circuit).
Yes, about the shunt. Which load will receive no current flow?
Once you know that, do the “Zen trick” (from unit 5) on the remaining loads to determine if they are in series with each other or parallel.
1.
you have to have current to have voltage
This isn’t correct! Please define voltage and current for me.
2.
and in order to have a voltage drop you have to have resistance correct?
When current flows through a load (which by definition has resistance) it will create a voltage drop across that load. So the first step in Question 8 is determining which of the loads will have current flowing through them. Are any of them shunted, for example.
Okay, I’ve got a couple of starter questions:
1. What creates voltage drop?
2. Why do you think the ignitor has zero?That’s correct. Current is the same throughout a series circuit, so you have to use the total resistance of all the loads in the calculation.
I’m going to hide the answer so we don’t just give it away to others.
Tell me in word how you got question 3. (That is the correct number – I just want to see how you got there.)
Start a separate topic for Question 6, showing your calculation. (That’s not the correct answer, so I have no idea where you are going wrong.)
Hi Terry,
You can look at other topics in the Forums and see how they are set up. You give the topic a title relating to what you’re having trouble with. For example, if you want to talk about using Ohm’s law calculations to figure out circuit current and voltage drop in series circuits (which is what questions 3 and 4 are about), you’d start a topic with a title like “Calculating Current and voltage in series circuits”, then write your question.
I don’t want you to “think” that you have question 3 down – it is such a fundamental, basic question that you need to KNOW you’ve got it.
Hi Pedro,
Questions 7 and 8 are about calculating the heat across each of two loads in series. We showed a similar scenario back at the end of unit 3, where we calculated the heat generated by a loose connection, which essentially became a load in series with the element in that circuit. Rewatch that video. You could use a similar technique on this question. Let me know what you get.
If you’re still stuck, I can step you through, but watch that video first.
This is the same general technique as the calculation of the heat generated by the loose connection that we showed in Unit 3, since that loose connection resulted in there being two “loads” in series with each other.
Do you remember that? The last video in that unit showed it, and you had a quiz question about it.
Go check that out, and let me know if you can understand how it’s done. If not, let me know and I’m glad to help you more here.
Yes – that is correct about current.
So that means there is an ____ in the circuit.
But based on the other info we give, that you measure 120 vac on one side of the circuit, that helps you to narrow the conclusion down even further.
The answer can actually be expressed in two words: there is an ____ _____ in the circuit.
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