Susan Brown

[Susan Brown]

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.

[Susan Brown]

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.

[Susan Brown]

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.

[Susan Brown]

Okay, I’ve got a couple of starter questions:

1. What creates voltage drop?
2. Why do you think the ignitor has zero?

[Susan Brown]

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.

[Susan Brown]

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.)

[Susan Brown]

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.

[Susan Brown]

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.

[Susan Brown]

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.

[Susan Brown]

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.

[Susan Brown]

Hi Colin,

This is a very commonly-missed question on the Final exam, and I’ve been surprised I don’t get more questions about it, so I’m glad you posted this.

The answer is actually simple once you see it. I’ll try to give you some hints and see if that helps you get it, instead of just telling you the answer.

In the problem statement we rule out the bulb or the receptacle as being the culprit.

In other words, we have a good load, and we know we have voltage.

When we measure across the load (across the two terminals), we measure zero volts. This is the first thing to focus on. What does zero voltage drop across a known-good load tell us?

[Susan Brown]

Hi Kyle,

We’re glad to help. First of all, what type of element are we talking about – oven or dryer? Also, is this a 3-wire or 4-wire power cord configuration?

~ Susan

[Susan Brown]

HI David – check your email. I sent you detailed info on your results this morning.

[Susan Brown]

Hi Jared,

My biggest tip for taking the Module Exams is to remind you that these questions are taken from the Unit quizzes. So, you have seen all of the questions before. The questions are chosen at random, and the answers are rearranged, to make sure that you aren’t just answering based on pattern recognition which helps to make sure you’re learning the material.

To study, you should look back over each unit quiz in the module and make sure you really understand the questions and answers, using these forums for any clarification you might need from us.

[Susan Brown]

The light’s bimetal switch is the switch shown just to the left of the light. The protector bimetal is labelled “protector” on the schematic.

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