Susan Brown

[Susan Brown]

Hi Bill,

I moved your question to a new topic, since it didn’t really have to do with that other thread.

See the video in this forum topic – it should help you!
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

[Susan Brown]

There’s a big difference between a path with resistance (a load) in it and a path with none. A shunt is not like a parallel circuit, because ALL of the electrons will choose that path over one with a load in it.

Keeping that in mind, take another look at the circuits and see what you think.

[Susan Brown]

When electrons have two paths available to them, and one has a load and the other does not, what happens? (This is a particular circumstance that we address in Unit 5.)

[Susan Brown]

Forget about main for the moment.

If you are the igniter or the booster, don’t you have more than one option to get to N? Remember, electrons don’t “see” bends or connections in the wires.

[Susan Brown]

Hi Duane,

Thanks for posting a question! I did move it to the Basic Electricity forum, and gave it a new title.

Short answer to your question is no – there will never be a partial voltage drop across a load if it is the only load in the circuit.

It’s an observable law – “Kirchoff’s Law” – the sum of the voltage drops in a series circuit shall always, everywhere, and forever add up to the supply voltage. It’s related to the conservation of energy, but there’s no need to try to understand it deeper than that. We just have to accept it!

In the video you referred to, if we changed the lightbulb to something with a different resistance, that would change the current flow in the circuit.

I = E/R

The voltage is determined by the power supply, and the resistance in the circuit is determined by the loads that are in the circuit. The current will be whatever it takes to drop all the voltage across the load(s) in the circuit.

Does all of that help?

[Susan Brown]

By the way, here’s a great webinar recording at Appliantology that would be helpful, if you haven’t seen it yet:

https://appliantology.org/topic/72423-voltage-voltage-drop-loads-switches-jumpers-cheaters/

Also, we’ve got some recent videos at our YouTube channel that show voltage measurements on various jobs.

https://www.youtube.com/user/SamuraiRepairman/videos

[Susan Brown]

Hi Nate,

They don’t give us the strip chart for this washer or enough information to know exactly what that “full – empty” area is doing, with the two switches and the wire between them. All we can do is go by what they say in the text, which is when the tub is empty, the water level switch is as drawn, which energizes the circuits in the bottom half of the schematic that have closed contacts. When the tub is full, the switch flips up and energizes the top circuits.

When the motors and agitate solenoid are energized, switch 8 is open. No electrons will flow through that circuit.

Does that help?

[Susan Brown]

Hi Nate,

A Watt is by definition 1 Joule per second. The time increment is part of what a Watt is.

You could say that 1 BTUH (or, BTU/hr) is equal to 0.29307107 Joules/second if that helps you picture it.

There’s no need to make the time increment equivalent, since we are only ever going to use the units BTUH or Watts. We don’t need to know Joules/hour or BTU/second, for example.

When you say a car is going 55 mph that doesn’t imply that the car is going to drive for an hour. It just says that IF the car drove at that rate for an hour, it would traverse 55 miles.

Joules and BTUs are different from each other, just like kilometers are different from miles. We could convert 55 miles/hour to whatever number km/minute, and that car would be still driving at the same rate of speed.

Does that help?

[Susan Brown]

Might have been! Of course, there are sharp edges on other brands as well. It’s definitely something to watch out for.

[Susan Brown]

Yes, that’s a joke 🙂
Some of Frigidaire’s models over the years have been frustrating to work on. I remember Samurai once saying he thought that Frigidaire engineers hated appliance techs, since they made their machines so hard to work on, with lots of sharp metal edges, etc.

[Susan Brown]

There is an awkwardness to the course software’s messaging on quizzes. Once you begin a quiz, it immediately shows the message that will be accurate once you have finished the quiz.

For example, when you begin your second attempt, it will give the “maximum limit” message that will be accurate once you have submitted that attempt.

Sorry for the confusion! We have complained to the software company that it confuses some students, but it must not be an easy thing for them to change.

I hope that helps!

[Susan Brown]

Try recreating the scenario and calculations yourself on a piece of paper. That can help you think it through a little better. If at any point you feel there’s still something you’re not getting, feel free to follow up with more questions!

[Susan Brown]

Voltage drop is simply this: when current flows through a load, it will create a difference in voltage. In other words, if you put your meter probes on either side of a load while it is doing work, you will get a non-zero voltage measurement. We call this voltage drop, to distinguish it from generic voltage, which is simply the difference in electrical potential, or charge, and isn’t related to whether or not a load is doing work.

If there is just one load in a circuit, and current is flowing, the one load will drop all of the source voltage. (e.g, 120vac in an L1-N circuit). If there are two or more loads in series, then the sum of the voltage drops will equal the source voltage. The individual voltage dropped across each load would be proportional to the resistance of the load (E = IxR).

If you encounter a series circuit in an appliance, it was designed to be that way, so thinking in terms of “peak efficiency” of the load(s) doesn’t really make sense. Now, if you got an unintentional additional resistance in the circuit, due to a loose connection for example, then you could see that the loads wouldn’t perform as designed.

That’s your little intro to voltage drop. You’ll see more in Unit 8!

[Susan Brown]

Hi Shawn,

I took a look at the questions you missed. I’m glad you posted here so we can make sure you get a better understanding.

A couple of questions had to do with two light bulbs in series with each other, and asking what happens when one of the bulbs burns out.

I would recommend that you rewatch the beginning portion of the first video in this unit, and make sure you understand the behavior of current in a series circuit. Particularly – what happens when a load fails open? (like the burned out lightbulb)

As for voltage drop that is mentioned in a few questions and explanations, for now just remember this answer that you gave correctly: “The sum of the voltage drops across each load will equal the supply voltage.” You’ll learn a lot more about voltage drop in Unit 8. Once you’ve done that unit, come back to these questions and explanations and see if they make more sense.

[Susan Brown]

That’s correct!

There’s a longer explanation of why that is in this Forum thread, if you’d like to check it out:

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

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