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First of all, what is the rule of thumb for equivalent resistance of parallel circuits? (It is explicitly stated in the video on equivalent resistance in Mod. 3, unit 5.)
Hi Josiah,
Yes – you have to score 90% or higher on EACH exam in the course. The Midterm is perhaps the most important one of all! It really pushes students to help them understand important electrical concepts. I’ll be grading yours soon and will email you.
Correct. The distance between each pulse (the “square waves”) will vary to do things like control the speed of a motor.
Be careful about that expression “path of least resistance”. If there is a path of NO resistance, such as a wire with a closed switch, then all of the current will flow through that instead of through a load. This is a “short”.
However, if there are two or more paths for current to flow, which all have loads, then some current will flow through all of them, not just the path with the least resistance.
I don’t know what the “120 volts” means at the end of your sentence – would you please explain?
Hi Angel,
“Constant duty cycle” means that the frequency of those square wave peaks are always the same. Does that make sense?
The big question is: DOES current flow through the main coil with the detector switch closed.
Hi Kelly,
We’re glad to help! Can you tell me which video (there are 3 in that unit) this is, and the timestamp for that section?
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?
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