Susan Brown

[Susan Brown]

Hi Steve,

There isn’t, but the Final exam is a bit more of a review than the midterm is. We’ve found that folks who did well on the quizzes and module exams generally do well on the final if they’ve taken the time to look back through those quizzes/exams to refresh their memories.

[Susan Brown]

It’s pretty straightforward- you might be overthinking it.

If you know the circuit current, you can find the voltage drop across each resistance using a simple Ohm’s Law calculation.

[Susan Brown]

Exactly! That’s the case for a series circuit. Just wanted to make sure you were clear on that.

[Susan Brown]

E=IxR can be used in a few different ways, but as far as calculating voltage drop across an individual load, then the way you wrote it in detail (the second one) is correct.

One clarification: you said “I (current flow through load)”. If you had several loads in a series circuit, would the current change at all? Would it be different for different loads, or the same throughout the circuit?

[Susan Brown]

Hi Steve,

3c) voltage drop is the product of current flowing through a resistance (load). Does that make you think of an Ohm’s Law equation that you could use?

5b) re-watch the video at the end of unit 6.

let me know if you have any other questions!

[Susan Brown]

Way to pay attention! 🙂

You are correct – that is a mistake. Current is indeed inversely proportional to resistance. Good eye! Thanks for letting us know so we can correct that.

[Susan Brown]

Good! I’m glad that you asked the question. We get some students who hesitate to ask questions here, which is a shame.

[Susan Brown]

Yep! And Figure 2 is shows the testing that can tell you which line is out of commission.

Does that all seem clearer to you now? Any more questions?

[Susan Brown]

Exactly. It seems like you are pretty much there.

If L1 and L2 were both supplying 120vac each (out of phase), combined with a complete circuit and a good element, we’d have 240vac drop across that element and heat being produced.

But we’re reading zero across the element. This tells us there is no current flowing, so there’s an open on one side or the other.

All of this, combined with what I said earlier about the ends of the element being EEPs – can you see why we are measuring 120vac wrt N from either end of the element, even if we know there must be an open on one side of the circuit?

[Susan Brown]

L1 and L2 being out of phase with each other is how a 240vac voltage can be made by combining L1 and L2. However, in this problem, we never measure 240 anywhere.

If the circuit were complete and operating as expected, what voltage drop would you measure across the element?

[Susan Brown]

Sure – it’s in unit 6, at the end.

Note that this is a very similar, but not identical, scenario.

https://my.mastersamuraitech.com/module-3/basic-electricity-breaker-panel-basics/

[Susan Brown]

Load 1 is the ignitor.

Yes, everything is basically “connected”, like in any configuration of parallel and/or series circuits. But can the electrons get from the ignitor to L1 and N without having to go through another load? Yes or no.

Remember – if they have a direct path along wire, they will take it.

For now just analyze this with the detector switch being closed, essentially making that branch a wire. We’ll talk about if it were open later…

[Susan Brown]

Hi Matt,

The portion of the webinar referred to in this thread was not recorded (free-form Q&A parts usually aren’t).

All of the concepts you need to know to answer this question were covered in the module, mostly in unit 5. But this diagram is drawn a little oddly, which is one reason why we use it. You’ll encounter all kinds of variations when you look at schematics, so we want our students to have to stretch a little to help them gain experience and confidence!

You are actually making the circuits more complicated than they are. After all, if you were going to have to divide up the voltage drop across multiple loads, you would have to know something about their relative resistances, right? And we don’t give that in the problem statement. So step back and take another look at the diagram.

Become each load, one at a time. Any of the loads that can reach out and touch L1 and N without having to go through another load is by itself in a circuit, in parallel with the others. Don’t get fooled by how the lines are drawn – even if there is a bend or a junction, as long as the electrons can travel along a wire rather than through another load, they will.

Let me know for each load what you see when you do that Zen trick. (see the video in this unit for a refresher https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/)

[Susan Brown]

Hi Steve,

Read through this Forum thread on Req and see if that helps you figure it out.

Let me know if you still have questions after looking though this: Calculating Equivalent Resistance

[Susan Brown]

Hi Steve,

This is such a good question that I decided to add some material to Basic Electricity, unit 3, to help clarify using Ohm’s Law and specifically why using P = E²/R doesn’t work where E is the supplied voltage and R is the resistance of the loose connection. (you can look at that here: https://my.mastersamuraitech.com/module-3/basic-electricity-ohms-law/)

It has to do with the fact that in a series circuit, supply voltage will drop across the loads in proportion to the resistance of each load. The total of these voltage drops will equal the voltage supply. So, if you have two loads in this oven circuit, neither one has a voltage drop of 240 – the voltage drops will add up to 240.

We teach a lot more about voltage drop in the subsequent units, so don’t think you have to completely understand it at this moment. But it’s an important concept, so pay attention as we continue to talk about it!

I hope that helps. Let me know if you have any follow-up questions.

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