Susan Brown

[Susan Brown]

Hi Victor – I reset that quiz for you.

[Susan Brown]

Hi Kenneth,
Right, because if you only have one load in a circuit, the voltage drop will equal the source voltage.

The key here is the impact of the closed detector switch on the circuits with the Ignitor, Booster, and Main.

If you do the “Zen trick” on the Ignitor or the Booster, how do you “reach” N? Through the closed switch, through the Main, or both?

[Susan Brown]

Hi Chris,

“1 over” is the same as “1 divided by”. So, for example, 1/10 is “1 divided by 10”. When you do that on your calculator, you should get 0.1 (“one tenth”)

1/2 should result in 0.5 (“five tenths”, which is the same as one half)

So, let’s say you have two 10-ohm resistances in parallel.

The calculation would be
1/(1/10 + 1/10) = 1/(0.1 + 0.1) = 1/(0.2) = 1 divided by 0.2 on your calculator = 5 ohms.

Do you get that when you do it on your calculator?

What do you get if you try it again but with two 20-ohm resistances in parallel? Let me know!

[Susan Brown]

It would be similar to a refrigerator – the point is that it’s a cold, damp environment. So the choice of connector and other precaution are important. You had a quiz question about this in that unit.

[Susan Brown]

Hi Raja,

This is a little Basic Electricity practice.

In order to get current to flow through a load, you need a voltage and a complete circuit.

And by voltage, we mean a voltage difference. All voltage is expressed as the difference in charge between two points. A classic reading of voltage potential is measuring from L1 with respect to Neutral (where you use a known-good neutral point as reference). Reading voltage drop is when you are measuring across a load (a component with resistance that does work – element, pump, bulb, etc.). Current flowing through a load produces voltage drop.

So – if you have L1 on both sides of a load, there is no difference in voltage that will drive current through that circuit. So, no current.

[Susan Brown]

If there is only one load in a circuit, do you know its voltage drop? (think about what Kirchhoff’s law teaches)

[Susan Brown]

Hi Troy,

No calculations are needed to answer Question 8, and you definitely don’t need to make something up. The key is seeing the impact that the closed detector switch has on the circuits.

Let’s back up a bit – if you look just at the Safety, do you know what its voltage drop is?

[Susan Brown]

Hi Brian,
I reset you. FYI – it’s best to use the Quiz & Exam Reset Request form when needed (in the “Campus Support” menu).

~ Susan

[Susan Brown]

That’s correct

[Susan Brown]

Also – go to the Core course, Mod 4, unit 4, and watch the second video. I start with a circuit with a single 60 ohm load, then add a 40 ohm load in series. The current in the circuit decreases, and the voltage drop across the 60-ohm load decreases with the addition of the second load. But obviously the 60-ohm load still has 60 ohms.

The main point of the question you are writing about is to make sure people understand that resistance is an inherent property of a load and in the types of Ohm’s Law calculations we are likely to do as appliance servicers, if you are given a value for resistance, that value is fixed. It won’t change in response to a change in current, for example.

(Note – one exception to this discussion is that some material’s resistance will change some based on temperature. Some types of sensors are an example of this.)

• This reply was modified 11 months ago by Susan Brown.

[Susan Brown]

Yes, if you add another load in series, that changes the total resistance within that series circuit.

Which unit is this question in? I want to see it in context to make sure I answer correctly.

[Susan Brown]

The voltage in these equations is the voltage dropped across the load, not the source voltage. So, if a load was originally by itself in a circuit, dropping the source voltage, then another load is brought into series with it (by a switch activating), then the voltage drop across the original load will decrease. The current will also change.

[Susan Brown]

One example I thought of where this could happen is based on Question 8 on the Midterm Exam in the Core course where the Main Coil was shunted by the closed detector switch, leaving the Ignitor and Booster in independent, parallel circuits (*each* having 120v drop). When that switch opens, it becomes a series-parallel circuit, where all 3 loads will now share the 120v drop.

So any scenario where a load is either bypassed or brought back into series by a shunt will affect current and voltage to loads.

[Susan Brown]

“k” stands for “kilo”, which is 1000 of something. (For example, a kilometer is 1000 meters). So 5k-ohms is 5000 ohms.

See Mod. 4, unit 7, near the end of the unit for a table of units like this.

[Susan Brown]

Hi Ronald,
I have not heard of this. But I know we have a number of techs from TX at Appliantology – I suggest you post this in the Dojo and see what you can find out.

A lot of these types of licenses have more to do with knowing Code or other regulations, rather than actually understanding technology and troubleshooting, in which case the info you need is likely in that book.

Can you send me more info on this test? (A link?)

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