Susan Brown

[Susan Brown]

It’s not an inconvenience at all! It’s all part of the training to help you out here 🙂

[Susan Brown]

Correct, so the black jumper wire branches off of the wire that goes to the hot indicator light.

The question you were not sure of asked for the color of wires that supply power to the LF hot indicator light. You seem to be looking at the correct load – that little light symbol on the wire labeled LF-Y. Do you still have some difficulty in answering the question?

[Susan Brown]

Yes!

Here’s the explanation that we gave on the quiz results – had you seen this?

“When the surface element is first switched on, the light gets L1 through the L1 to H1 contacts of the infinite switch and the black jumper wire. This can be easy to miss because the two ends of the black jumper aren’t explicitly shown as being connected.

During the element’s operation, once it gets hot enough, the light’s bimetal switch closes, providing another path for L1. This means that even after the infinite switch is opened, turning the element off, the light stays lit until the element cools off enough for the light’s bimetal to open again.”

[Susan Brown]

Here’s what you wrote in email:

I am looking at the schematic diagram. The light appears to be labeled LF-Y. So following L1 it seems like it would go through an infinite switch (H1) and through a jumper.

I’m glad to see you are on the right track! Correct.

But further down that wire H1 is on there is the Surface Element and then a switch labeled protector, next to that i see BR or Y making me think the protector bi-metal (if that’s what it is and I am not mistaken) making that another way of travel.

The Surface Element is a load in that circuit. Will the wire to the right of it be L1 or L2 (if the switches are closed and current is flowing)?

I was also wondering though that the wire that goes straight down from L1 and then third wire down goes right to the hot light switches, what is it for or what does it indicate? At first i thought that was an indicator that L1 was hot wired into L1 but did not end up being the case.

(I’m assuming you meant to say that the light was hot wired into L1.)

If something is “hot wired” to L1 or L2, that means there’s a direct line to the power supply with no switches or loads in the path. That’s obviously not the case here.

Do you know what this symbol is?

[Susan Brown]

Yes it does! It’s amazing how our eyes and brains conspire against us sometimes.

[Susan Brown]

Hi Paul,

American circuits will be L1-N or L1-L2.

What do electric ovens and dryers need?

And, what is the notation for the circuit shown in Question 6?

(Review Unit 6 re: household power supply)

[Susan Brown]

You’re welcome. And glad to hear it! That’s what we’re trying to do with the Midterm 🙂

[Susan Brown]

Exactly. The circuit current is the same throughout a series circuit, and will be determined by the total resistance in the circuit.

The current flowing through each load will produce a voltage drop in proportion to the resistance of the load, with all of the voltage drops adding up to equal the source voltage.

(Note: I will hide part of your answer so we aren’t just giving it away to other students!)

[Susan Brown]

Hi Kenneth,

Good question – I moved it so that it could be its own topic.

First of all, we do have a video showing the calculation of something very similar to this – the heat produced by a loose connection – the last video in Unit 3 of Basic Electricity.

If that doesn’t clear it up for you, tell me what Ohm’s Law equation you are trying to use to calculate the heat produced by that load.

[Susan Brown]

Correct – The fan motor circuit is parallel to the element circuit, thus unaffected by its failure.

Also basically correct about the element branch – current will stop when it fails open, since there’s no longer a closed circuit.

Do you know what the final answer is – what happens to the total current draw from L1?

[Susan Brown]

Not quite. Read what Michael said about the main above. And do the “Zen trick” on the booster or ignitor – what do you find?

[Susan Brown]

Sort of! But you’ve got a mistake. The source voltage is 120vac. All of the loads combined must drop 120 – as you knew in your first response. So, calculating 120 as the drop across just one of the loads can’t be right.

The correct formula for voltage drop is E = I x R. We give you each resistance, so you’ve got to figure out what to plug in for the circuit current. (And what you just did isn’t correct.)

Does current vary throughout the series circuit or is it the same?

(And – for another hint – look at what you did for Question 3 on the Midterm.)

[Susan Brown]

Hi Gilbert,

No, 0.013 is not correct. You would expect to get something less than 103, obviously, but not *that* much less!

However, 1/.013 does get you the correct answer, which is around 78 ohms.

Check out this previous forum topic where we show a similar calculation in more detail, and see if this helps:
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

Let me know!

[Susan Brown]

“Mega-” is a million of something. So, 1 Mega-watt is 1 million watts (1,000,000 watts). But you can change it to any other unit of measure, too. A “Giga-” is 1 billion (1,000,000,000) of something. That’s why converting a mega-watt to a giga-watt results in a smaller number. 1 mega-watt is 0.001 giga-watt. Is that what you wanted to know?

[Susan Brown]

Hi Terry,

This can seem like a fine point, but we do emphasize (and want to encourage) techs to think in terms of “power” when it comes to loads doing work. The specs for loads are often given in watts (which is voltage times current). For a load to do its work properly, there must be the correct amount of both voltage and current to create the necessary wattage. (So, the correct answer was “power”.)

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