Susan Brown

[Susan Brown]

Thank you!

[Susan Brown]

Hi Everardo,

There are two different things you are talking about.

1. Your business structure (that would be like an LLC)
2. State or local government licenses/permits to do a particular business activity

1. Forming an LLC or other business entity has to do with taxes and liability (for example, if your business gets sued but you are an LLC, your personal assets are usually protected). You can get advice on this from tax or small business advisors. You don’t have to have this set up before you start running service calls, but you do want to look into it and make a decision before too long.

2. You’ll have to check with your local/state government to see what, if anything, they require before you open an appliance repair business. Fortunately, licensing requirements for appliance repair tend to be fairly minimal compared to other trades.

[Susan Brown]

Hi Elizeu,

The Ask the Teacher forums are only for questions related to the coursework. Questions about jobs that you are doing should be posted at Appliantology.

Now that your membership is active at Appliantology, you can post this question in the Tech Forums over there.

[Susan Brown]

Hi Charlie,

Have you set up your MST Student membership at Appliantology yet? If not, you can do that by taking the free Appliantology 101 course that you have access to (go to “My Courses” page).

This is the type of question you can ask at Appliantology tech forums.

The Ask the Teacher Forums are for questions directly related to your coursework here at MST.

[Susan Brown]

Hi Aaron,

I hope that sounded like an awful lot of heat just for a loose connection!

We show a nearly identical scenario in the video at the end of Unit 3. The values of the resistances are a little different, but otherwise it’s the same. Go back through that video and recreate the calculation on your own. Do you understand what we did? If not, let me know and we can break it down more here.

[Susan Brown]

It sounds to me like you are thinking of this correctly. Here’s some more discussion of the concepts that will hopefully help to solidify your thinking.

Here’s one way to think about what happens when one parallel branch fails open. Let’s use a simple example of two parallel circuits. The total current from the power supply is the sum of the current flow in both circuits. If one fails open, that current goes to zero. The other current is not affected however. So, the total current decreases.

Thinking of it mathematically, with equivalent resistance, you can come to the same conclusion. Below is a clip I wrote to help someone else who was grappling with equivalent resistance and current flow in parallel circuits.
———–

This is all from the perspective of the power supply. Be Zen-like and become the power supply, which has constant voltage, and think about the affect that having multiple paths for current to flow (parallel circuits vs. just one series circuit) has on the number of electrons you can push out (current). You can push a lot more out when there are multiple branches, even though each branch has a load.

The best way to think this through is to draw out a couple of circuits on paper and play with the calculations.

The first one has two loads in parallel, R1 and R2. Assume a 120vac circuit, so L1 and N. Assign easy values to R1 and R2, say 10 ohms and 20 ohms. Then calculate the current flowing through each load. (I = E/R, and remember in parallel circuits each load gets the full 120 vac.) The total current draw from L1 would be the sum of those two different currents.

Now, let’s think about the “equivalent resistance” of this scenario. The equivalent resistance is taking the resistance of loads in parallel and theoretically combining them into a single load. So, do the math and come up with the equivalent resistance based on the loads above. Now use I = E/R to calculate the current draw from L1. Should be the same as what you came up with before.

Since current is inversely proportional to resistance, for the same amount of current to flow through our theoretical one load that was flowing through the two branches/loads in parallel, the single load, or equivalent load, would have to have a lower resistance.

———–

Adding on to that… if one of the two branches fails, then the only remaining branch results in a total circuit resistance that will be higher than the equivalent resistance of the two loads in parallel. So, as you correctly stated, the overall current flow will decrease.

[Susan Brown]

Hi Vince,

Thanks for the idea – I’ll pass it along to the Masters!

[Susan Brown]

Hi Wesley – great! There are lots of PDF reader apps out there that allow marking them up. Scott often uses an app called Vittle to do his presentations and markups. But he said to also check out Inkflow and Classroom PDF.

[Susan Brown]

Power is both the input and output of a load. The load transforms it from electrical energy to something else. For example, to thermal energy (heat). Does that help?

[Susan Brown]

Hi David,

No worries – we’ll discuss it here and I can hide certain portions afterwards. You answered “current”, which is not entirely wrong, but it is not the best one of the answer choices we gave you. “Power” is the best answer. Think about the Ohm’s Law equations. We use “P” (for Power) for work being done. P = E x I.

Does that make sense?

[Susan Brown]

At 70 deg. F it was 2000 ohms
At 32 deg. F it was 15,000 ohms

The temperature decreased, but the resistance increased.

[Susan Brown]

Hi Rigo,

The question is:

Question #6 – You’re testing an unknown sensor. You measure its resistance at 70F at 2 k-ohms. You then dunk the sensor into a glass of ice water and then measure its resistance at 15 k-ohms. From these test results, you conclude that the sensor’s type is:

Your answer on the second attempt was “PTC”

Do you think that is the correct answer?

[Susan Brown]

Good question!

The difference occurs because of the decrease in current once you have additional resistance in the circuit.

The current goes from 7.5 amps to ~6.5 amps when we add in the 5 ohm loose connection.

Knowing that the voltage dropped across the entire circuit will be a total of 240vac in either case (Kirchhoff’s Law), you can see that losing an amp of current will decrease the wattage by 240 watts.

P = I x E = 7.5 x 240 = 1800
P = I x E = 6.5 x 240 = 1560 (~1350 from the element and ~ 210 from the loose connection)

Does that make sense?

[Susan Brown]

Hi Jason,
Your initial one got lost in the mail. I put your name on the list for the next batch of Certificates that will be printed. I should have those next week and will put it in the mail. Sorry for the delay!

[Susan Brown]

Hi Samori,

I assume you are talking about the video at the end of Unit 6. The measurements “wrt N” would be taken with a multi-meter. One probe would be on a heating element terminal, the other would be put on a known-good neutral point that would be identified by looking at the schematic. You’ll be learning more about this as you go along!

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