Susan Brown

[Susan Brown]

Hi Chris,

1. ANY circuit must have a load for it to be a circuit, so having loads doesn’t determine if it is parallel or series.

Have you seen the new videos in Module 3, Unit 4? They may have been added after you did that unit. Go check ’em out! This question comes from the first video, where he talks about parallel circuits in the second half.

https://my.mastersamuraitech.com/module-3/basic-electricity-circuit-components/

2. This question also comes from those videos, although the info is in other parts of the module as well. When you have L1 and L2 being supplied to a circuit, what’s your total supply voltage?

[Susan Brown]

Hi Chris,

First of all, thank you for taking the time to explain your thoughts on this! It’s very valuable for us to be able to see where any points of confusion are so we can get better and better at communicating these things.

As to the example you discussed of a series circuit with two resistors, one is twice the resistance of the other. (And you are correct – a tech sheet would never say that, but often times when learning concepts it’s helpful to use streamlined scenarios like this, before adding the various complications and details that come with using actual tech sheet info.)

You can answer this question without doing “E=IxR”, but you need to *know* E=IxR. In other words, you need to know that voltage drop is directly proportional to current and resistance.

Then there are some fundamental truths about circuits that you need to keep in mind. One is that the supply voltage will completely drop across the circuit, with the drop being divided up across the loads based on their resistance. The second is that in a series circuit, the current will be the same at every point in the circuit.

So, knowing those two things and that E=IxR, you can think through this particular problem without having to actually plug any numbers into E=IxR.

The loads in the circuit need to drop 120vac total. Since the current is constant, the voltage drop across each load will be completely determined by the resistance of each load. If you know the percentage of the total resistance that each load represents, then the voltage drop will divide up the same way. If there was only one load, it would drop 120v. If two equal loads, they would each drop 60v, etc.

You could be given numbers for R, so you could then calculate the total resistance in the circuit, calculate the current, then calculate the voltage drops. But this was more about getting the foundational concepts of what goes on in a series circuit.

Also my perception of voltage drop is a bit fuzzy, I am getting the math (I think) to figure it out. But what it actually IS I dont quite get. I thought of it has a measurement of power(volts) being absorbed by a load. Which would leave me to believe that if a single bulb in a circuit absorbs 120v, then adding a second bulb further down on the neutral side would do nothing(which I know is wrong)

Be careful not to confuse power and voltage! Power is work being done. Voltage is basically like “pressure”. The equations for power show its relationship to current, resistance, and voltage.

As to the lightbulb question – keep reading…

I am having difficulty reconciling these two perceptions I have. I thought of voltage drop as a sponge. for example: If a sponge could soak up 2 cups of water, and if the quantity was 2 cups to start with then a second sponge added to the line up would have nothing to absorb.
But from your lesson and Kirchhoffs law, I am finding out that this is not true, that voltage drop is not a set amount for any given load but is completely dependent upon resistance and current.

The thing that is set for any load (light bulb, motor, element, etc.) is its resistance. It is what it is. The voltage that is dropped across that load is dependent on the current that flows through it. The power that it can produce is also dependent on the current that flows through it. And the current that flows through it is dependent on the total circuit configuration (supply voltage, total resistance of all loads).

Equations are essential to understanding electricity, but you have to use the equations in the context of understanding some fundamental physical concepts of electricity and circuits.

Does this all help you get any closer to understanding?

[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.

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