Susan Brown

[Susan Brown]

240v appliances experience all the same types of problems as a 120v appliance (failed loads, switches, or boards), but a unique issue that can happen is that one of the breakers will trip and the appliance will only receive 120 V AC. A classic example is when an electric dryer will tumble but have no heat.

[Susan Brown]

My first suggestion is that you rewatch the AC Voltage measurement video in Mod. 4, unit 7.
https://my.mastersamuraitech.com/module-4/basic-electricity-electrical-measurements-in-appliance-repair/

And also read the section further down in that unit about Loading meters.

The point of a loading meter is to allow more current to flow in the circuit to eliminate the possibility that you are reading “ghost” voltage.

Do you recall where that video is that you are referring to?

[Susan Brown]

Correct.

[Susan Brown]

Hi Tracy, you are on the right track. Main coil is shunted by the closed switch, so 0v.
The Safety is easy to see – it is an independent circuit with one load, in *parallel* with the others, so 120v.

The Ignitor and the Booster are in parallel – why would they split up the voltage?

[Susan Brown]

What a coincidence – I was just asking the Samurai this question myself yesterday!

It is definitely not neutral, neither is it line voltage. But it is “hot”. In other words, you would measure some voltage in between the two loads (with respect to neutral).

As an example, if the two loads had the same resistance, then you would measure 60v wrt N in between the loads, because each one would drop 60v.

Make sense?

[Susan Brown]

Hi Anthony,
Just to make sure we understand your question – are you asking if the explanation (“Both the main and sub PCBs would be replaced because 1) the Fast Track says to replace the PCB *assembly* which means BOTH boards, if they meant just the main PCB, they would have said that”) comes from somewhere on the Fast Track, or just from our general knowledge of these things?

[Susan Brown]

Hi Mitchell,

Most of our videos are hosted on Vimeo, but a few, like these, are YouTube videos. They work fine for me right now, so I assume you were having some temporary issue with those.

See the Video Playback section on this page for more suggestions to try, if they still aren’t playing for you.

https://my.mastersamuraitech.com/appliance-repair-course-support/student-resource-page/#video-playback

Let me know if you continue to have this issue, and I’ll make sure you get the video links another way.

[Susan Brown]

Yes, 2 is the correct answer. It is great to see you working on understanding the questions you missed, even though you “passed.” Not all students do that.

[Susan Brown]

Hi Michael,
Questions like that you can always email me about, FYI. But this is fine, too!

MST Courses are like parallel circuits – what happens in one does not affect the others 😀

The only thing you risk by taking a break is needing to spend more time refreshing for the Final Exam. However – that is actually a better method in terms of retaining the info. And hopefully you’ve been taking good notes to help you with this.

[Susan Brown]

Hi Dylan,

I’ll start by just telling you how I came to understand it, to see if it helps.

All voltage is a difference in charge between two points. For our purposes (dealing with circuits in appliances), there are two categories of voltage.

1. Voltage potential – the difference in charge between two points where there is no current flowing, but there is the potential for current to flow if there is a closed circuit between those two points. A common measurement of this type of voltage is putting one probe at one side of a load, and the other at a good neutral point. Or, measuring across an open switch.

2. Voltage drop – the difference in charge from one side of a load to the other, created by current flowing through the load. In other words, you will only measure voltage drop in an active circuit.

“Dissipation” sounds more like an idea from power transmission, where you can have some losses due to small amounts of resistance over very long stretches of power lines. Small amounts of the power are transformed into heat and lost to the environment. This is not something we deal with in appliance circuits, which are very small compared to transmission lines.

Here’s something else to keep in mind. In circuits, there are always two sides to every power supply. Either L and N, or L1 and L2. These two sides meet at loads.

Switches, on the other hand, are always on just one side of the power supply (otherwise they would create a short).

Let me know if this helps, or if you have any follow up questions.

• This reply was modified 1 year, 3 months ago by Susan Brown.

[Susan Brown]

Hi Douglas,

That is correct. The closed detector switch is a [answer hidden]
Good job!
Note – I will need to hide these answers.

• This reply was modified 1 year, 3 months ago by Susan Brown.

[Susan Brown]

Great! Keep up the good work.

[Susan Brown]

Hi Edwin,

I love to see students playing around with these calculations in order to understand!

Your mistake is in how you figured out the current in each circuit – trying to use proportions. Remember, current is inversely proportional to resistance. Did you notice that you were getting higher current flows in the circuits with higher resistances? That is not how it works. I = E/R

The voltage supply to each parallel circuit is 120v. This fact is true regardless of the state of any circuits in parallel to it. So the voltage drop across these single loads will always be 120v if they are active.

The resistance of each load doesn’t change – it is a fixed quantity. So, how can the current in each circuit change due to the failure of a circuit in parallel?

Let’s look at the calculations.

Given your scenario, with three parallel circuits with R1 = 100 ohms, R2 = 200 ohms, and R3 = 300 ohms, Req = 55 ohms, Itotal = 2.2 amps.

I1 = 120/100 = 1.2 amps
I2 = 120/200 = 0.6 amps
I3 = 120/300 = 0.4 amps

Totals up to 2.2 amps.
Notice that I1 is 3 times bigger than I3, which makes sense because R1 is three times *smaller* than R3. (Current is inversely proportional to Resistance.)

If you remove R2, you get Req = 75 ohms, Itotal = 1.6 amps
You can see from above that I1 + I3 = 1.2 + 0.4 = 1.6 amps

Same thing if you only have R2 and R3 active, It = 1 amp, which is what you get if you add I2 + I3.

Does that make sense?

[Susan Brown]

Hi Anthony,

The correct answer is “serial data transfer”, which is a type of digital data communication. You are not actually wrong, but we want the better, more specific terminology for the answer here.

We realize the distinction is pretty subtle, so we decided to take away “digital” as one of the answer choices.

[Susan Brown]

Hi Tiago,
All you need to know is that the detector switch is closed. Like any other closed switch, it acts just like a wire, with no resistance. So the question you ask yourself is: if you do the “Zen trick” on the Ignitor or the Booster, how will you reach N? Through the closed switch, the main coil, or both? Review the first video in Unit 5 for a refresher on the Zen trick and the effect of a pathway like this closed switch on circuits.

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