Susan Brown

[Susan Brown]

Hi Chase,

First of all, I moved your reply to its own topic since it has to do with a different question.

You are correct in general. If you have two loads in series, for example, you would have to know something about their resistances in order to calculate the voltage drop across them. Either the resistance values themselves, or how their resistances were related. Such as: if you knew they had equal resistances, then you’d know that they split the voltage drop evenly (each one would drop half of the source voltage). Of course if you have just one load in a circuit by itself, it drops the whole source voltage and no calculations are needed.

Since we don’t give you any info on the resistances of the loads in Question 8, that is a big hint that all 4 of these loads cannot be receiving current. If they were, then there would be a slightly complicated series-parallel calculation that you’d have to do, and you’d definitely have to know resistances.

What we are hoping is that when a student faces this conundrum, they will take a closer look and realize that there is something going on with the circuit configuration that makes the answer easy – no calculations needed. We suggest doing the “Zen Trick” on the booster and igniter to see if that helps illuminate things.

Take another look and let me know what you think.

[Susan Brown]

If your answer is close, it may just be due to differences in how you round the decimal numbers. What number did you get?

Here’s a Forum topic where we show a calculation in more detail – see if this helps:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

[Susan Brown]

Hi Phillip – don’t you have a membership at Appliantology? That is the place to get repair advice. The Forums here at the MST Academy are for asking questions related to the training courses.

But, doing the Ten Step Tango is always the approach. As you know, just because a load isn’t working isn’t proof that it is “bad”. You have to do electrical testing to find out if the load is faulty or if it’s not getting the power it needs.

[Susan Brown]

Review the last video in Unit 6. Let me know if you have any questions about what we’re showing in that video.

[Susan Brown]

Hi Robert,

You’ll need to give me more details on this. Is this related to a Question on the Midterm? (Question 9?) If so, please rewatch the last video in Unit 6 for a similar, but not identical, scenario. Let me know!

[Susan Brown]

Great! Thanks for using the Forums 🙂

[Susan Brown]

Hi Kenneth,

These loads are in series in one circuit. Will the current change throughout the circuit? (if you think “yes”, then review the first video in Mod. 3, unit 5).

You do need to use the equation E = I x R to calculate each voltage drop, but you have to use the correct value for the current (I) to get the correct answers.

[Susan Brown]

Hi Rees,

Thanks for posting in the Forums!

A great video to rewatch that shows these types of calculations is the last video in Mod. 3, unit 3 on the heat generated by a loose connection. Watch that, and re-create the calculations in your notebook. Then look at the questions in the Unit 8 quiz and see if they make more sense.

If you don’t follow something we’re showing in that video, please let me know and I’m happy to help you further.

[Susan Brown]

No problem! Sam (who is a mister) is one of the folks helping you. One of us will get back to you ASAP.

[Susan Brown]

Hi Samuel – could you please be more specific about which question you’re referring to? Thanks!

[Susan Brown]

Hi Michael,

The fact that the terminal is labelled NO, and it’s in the “normal” state, is what tells you that the switch is open. And that is why you would measure 12vdc.

Does that answer your question?

[Susan Brown]

Hi Ryan,

Thanks for posting in the forums!

One of the keys when looking at schematics is to “see” it from the electron’s point of view. Electrons don’t “see” bends in a wire, for example, like we do when looking at a diagram.

That’s one reason we suggest doing the “Zen trick” on the loads, when you aren’t sure if they are in series or parallel. Try doing that on the light bulb – “become” the light bulb – both with the switch open and then the switch closed. Does that help you figure it out? Let me know!

[Susan Brown]

That’s correct! (I’m going to hide those answers so future students won’t see them.)

[Susan Brown]

Correct!

There’s a lot to learn in the Basic Electricity module, and it can be quite a challenge. Please feel free to ask more questions here as needed – we’re glad to help!

[Susan Brown]

You are essentially correct!

A. Don’t think in terms of “least resistance” – that can lead you astray. Current will take any valid, complete path unless there is a shunt. When it has an option to take a path with (essentially) zero resistance, then all the current will flow through that path instead of alternate paths that have loads (resistance). So, that’s the case here. 100% of the current will flow through the branch with the closed switch instead of the path with the element.

B. Since the switch is open, all current will flow through the branch with the element.

C. Current will flow through both loads, in inverse proportion to the resistance of those loads.

Remember the formula for current? I = E/R

In other words, if R2 is twice as large as R1, then it will have half the current flow that R1 does.

And here’s where saying current follows the path of least resistance is flawed. R1 has the least resistance in the scenario I just proposed, but it does not get all of the current. It just gets more.

Let me know if you have any follow-up questions.

BONUS QUESTION: In Circuit A, if we did not have the light bulb in the circuit, what would this circuit configuration be depicting?

[Author]

Posts