Susan Brown

[Susan Brown]

Hi Mike,
Your answer was, “58 Ω because the equivalent resistance in a parallel circuit will always be equal to the smallest branch resistance.”
That is close, but not quite right.

You also got the answer incorrect to this question:
Question #12 – In parallel circuits, the equivalent resistance will always be _____ the smallest branch resistance.

Look at the second video in Unit 5, around the 6 minute mark. Do you see what the “rule of thumb” is for equivalent resistance?

[Susan Brown]

Hi Clark,

#2 – You have a circuit with two lightbulbs in series. If one of the bulbs burns out, what change, if any, would you expect to see on an ammeter (current/amps) reading? [emphasis added]

A bulb burning out means that the filament (which is what the electrons go through) has opened. This means that current will stop flowing in the entire circuit, since it is a series circuit. (So the other bulb will have no current and therefore won’t light up.)

Make sense?

[Susan Brown]

Hi Lukas,

For Question 2, watch the video starting at about 1 min. 25 seconds.

For Question 4, he talks about this starting at about the 18 minute mark. The question is referring to where on the *diagram* is the refrigeration effect “happening”.

If you need more help, let me know.

[Susan Brown]

Hi David,
Whether you are doing the calculations yourself using the formulas or using an app, you need to understand the basics of these electrical properties and circuits in order to use them properly.

Things like:

• In a series circuit, the current is the same throughout the circuit, and is determined by the source voltage and total resistance in the circuit.
• When you have loads in series, they will have a voltage drop across them that is proportional to the resistance. In fact, the voltage drop is equal to the current times the resistance. The voltage drops will add up to equal the source voltage.
• Heat generated by a load (P) is also proportional to resistance and voltage drop across the load.

So, with the situation with the loose connection, what we are dealing with is two resistances in series. This results in a higher total resistance, thus a lower circuit current, than when the loose connection was not there.

In order to find the heat generated by just the loose connection, if you use a formula for P that involves voltage, it must the be voltage dropped across the loose connection. If you use the source voltage (e.g., 240v), then you will end up finding the heat generated by the entire circuit.

Does that help?

(I am not sure what you mean by a cheat sheet.)

[Susan Brown]

Hi Allan – please check your email – you are all set!

[Susan Brown]

I reset you

[Susan Brown]

Sure! That will be a good review for you. These units are very common ones to need a little more help on (and often resets). Be sure to ask questions if you need to.

One other thing about your original question. On my dashboard, I can only see the score from the most recent attempt. If you ever have a situation where your first score was higher (and qualified for Certification), you would have to send me the email you received with that score. Otherwise, I’d have no way to know.

[Susan Brown]

Hi Ethan,
First of all – you indicated in your student info that you want to earn Certification for the course. Do you recall the requirements for that?

You must earn 80% or higher on EACH unit quiz, and
90% or higher on EACH exam in the course.

So, not only would you need to retake the exam anyway, but there are two unit quizzes that are below 80%. In order to retake those, you would need to get set back to Mod 4, unit 5.

Let me know!

[Susan Brown]

Hi Michael – I reset it so you can start fresh and answer all the questions.

[Susan Brown]

You got it!!!

(FYI, I’ll hide this answer so we don’t just give it away to other students. They need to experience their own “a-ha!” moment 🙂

[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.

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