Susan Brown

[Susan Brown]

Hi Jose,

First question you had: The last division problem Scott shows is 1/0.12 (not 1/12) which gives you 8.33333… (but you can drop all the extra 3’s and just say 8.3).

Second: 0.1 and 0.10 are the same number. The 0 after the one doesn’t affect the calculation at all, so no need to add it.

Hope that helps!

[Susan Brown]

There’s a lot to your question – let me see if I can help.

1. Don’t get too caught up in the “nanosecond of current flow” bit – this was speaking theoretically, and really doesn’t matter because a nanosecond of tiny current flow will not affect the loads or light the bulbs.

2. Same thing with the shunt scenario – no appreciable or measurable current will flow through the shunted loads, and the circuit becomes a series circuit with the one load that is still energized.

In the last video in that unit, towards the end, Scott shows mathematically why a parallel circuit cannot have a branch with no loads (zero resistance) in it.

We’ll see if Scott has something different to say about this, but even if the teeny tiny amount of resistance in a wire results in a few stray electrons wandering through the shunted loads, that will have no practical effect on the operation of the circuit. The current flow will so overwhelmingly be through the shunt and the load after the shunt as to render anything happening in the shunted branch as completely insignificant.

Make sense?

[Susan Brown]

I encourage you to stay focused on that loose connection and play with the numbers.

For example, if you double the resistance of the loose connection from 5 to 10 ohms, what happens?

5 ohm calculation, circuit current is 6.5 amps: (6.5)(6.5)(5) = 211 watts.

If r=10 ohms, the circuit current goes down to 5.7 amps. But (5.7)(5.7)(10) = 327 watts. So the heat at that connection increases (and the element wattage continues to decrease).

Whether P goes up or down depends on the exact scenario and numbers. You have to do the math and see what happens!

[Susan Brown]

Hi Chris,

When I’m trying to understand this stuff, I always pull up that colorful Ohm’s law wheel and use the equations to show me the relationships between power, resistance, amps, and voltage. Math helps to keep our thinking straight!

One equation we use in the loose connection scenario is P = I² x R.

P in this case is the heat generated from the current going through something that has resistance. From this equation you can see that increased resistance can increase the heat generated, even if the current is a little lower.

Normally a wire has essentially zero resistance, which generates essentially no heat. But the loose wire connection creates a spot of resistance, which generates heat, which then causes the wire to deteriorate even further, which increases its resistance, and so on, until there is finally a failure at that point or somewhere else in the circuit.

This loose connection lowers the current in the entire circuit, and thus the power in the element. But that doesn’t make it easier for the circuit. The reason this puts stress on the circuitry is that the current has to flow for a longer period of time in order to get the oven up to temperature – beyond the design parameters.

Does that make sense?

[Susan Brown]

Hi John,

Sorry you are having a problem with one of the videos playing!

I just tried and it played fine for me. There may be something going on on your end.

If you haven’t already, please read the Video Playback section at the Student Resource page:
https://mastersamuraitech.com/appliance-repair-course-support/sta-resource-page/

Then please let me know if you get the video to play or not.

[Susan Brown]

You’re welcome!

Some students have difficulty with equations that have a “squared” number in them because they aren’t sure how to square a number on their calculators. The easiest way is to just multiply the number times itself.

For future reference, here’s info on how to start a topic the next time you have a question:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/please-read-how-to-post-a-topic-in-the-forum/

[Susan Brown]

Hi Jose,

First, note that I moved this question to a new topic. Your question should be a new topic with an appropriate title.

There is an equation you could have used to do the calculation in one step: R = E squared/P.

However, you can still get the correct answer by calculating current first, then calculate R from that and the voltage. Mathematically it works out the same either way.

[Susan Brown]

Try combining what you went over above on voltage drop with the information on supply voltage in parallel circuits in the first video in this unit, particularly starting at the 11 minute mark:
https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/

[Susan Brown]

Question 4 involves a series circuit, and question 8 involves circuits in parallel.

There’s an important fact about voltage in parallel circuits that you need to account for here! Watch the presentation again in Module 3 Unit 5 starting at about the 11 minute mark, and see if that enlightens you!
https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/

[Susan Brown]

The answer can vary depending on how you treat the 1/30, since in decimal form it is 0.03333….

But the slight difference caused by how many decimal places you use is trivial – around 19 ohms is certainly good enough!

The most important point is to see that the equivalent resistance is less than the smallest resistance value.

[Susan Brown]

Hi Alex,

You don’t need to do a calculation to get the correct answer – it’s pretty simple to eyeball it and get it right. However, I’ve updated the explanation on the quiz with math that is a little easier to explain. See if that helps.

[Susan Brown]

P.S. We’ve removed the statement about the power supply being verified good to simplify the scenario even further.

[Susan Brown]

We’re not asking questions in this exam that require you to know about the rest of the dryer circuit. The purpose of the midterm exam is to test your understanding of fundamental circuit concepts. As for troubleshooting problems using stock schematics on the entire appliance, you’ll be getting plenty of that in the troubleshooting module.

[Susan Brown]

It sounds like perhaps you are not clear on measuring voltage vs. voltage drop. (This is a concept that take a little while to grasp!)

When you take a voltage reading, that does not tell you that current is in fact flowing. It just tells you that there is a voltage difference between two points.

In the question it stated that there was zero volts across the heating element, which tells you that no work is being done, thus no current is flowing in that circuit (so, there isn’t a complete circuit).

It helped me to watch the video in Module 3 Unit 8 again when contemplating this scenario. Watch that and see if it helps. Let me know!

https://my.mastersamuraitech.com/module-3/basic-electricity-voltage-drop-and-load/

[Susan Brown]

It sounds like you are confusing measuring voltage potential vs. voltage drop.

That’s a hint… does that help clear it up? If not, we can discuss further!

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