Susan Brown

[Susan Brown]

Yes – I reset that quiz for you.

[Susan Brown]

Good! Let us know if you have any other questions.

[Susan Brown]

It looks like you figured it out…

[Susan Brown]

The easiest way to do this is in 2 steps.

1. Find the circuit current, then
2. Use the circuit current and the resistance of the load of interest to calculate heat (P).

We use these two steps in the video at the end of Unit 3 about the heat generated by a loose connection. It’s a similar scenario.

Watch that video and then let me know if you can follow those steps or if you have follow up questions.

[Susan Brown]

From what I said above, with emphasis added:

“Note – I = E/R, so it should have been 120/5005 for the current.”

Voltage drop is E = I x R

I’m sorry I wasn’t clear – that part of your answer was correct. You had not done the calculation of current (I) correctly.

But – you didn’t even need to do those calculations to answer the two questions. You just had to look at the relative amounts of the resistances.

[Susan Brown]

I just wanted you to tell me what Ohm’s Law formula you would use to find voltage drop.

I looks like you used E = I x R in your second step above.

Note – I = E/R, so it should have been 120/5005 for the current.

But the point of the questions you missed was
1. which resistor would drop more voltage, and
2. how much work is R1 doing compared to R2

You can answer both of those by just looking at and comparing the values of the two resistors, since you know that voltage drop and work are proportional to resistance.

[Susan Brown]

Yes, so R1 is 5 ohms and R2 is 5000 ohms.

Do you understand what we mean by voltage drop?

If you knew the circuit current, would you calculate voltage drop? (what is the formula?)

[Susan Brown]

No worries – you got it correct on your final attempt on the Midterm.

[Susan Brown]

Hi Jim – it’s in the Troubleshooting Forum, since that is what the question was related to.

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/troubleshoot-appliances-part-1/#post-22351

[Susan Brown]

Hi Jim,

First of all, do you know what k-ohms are?

It is written on the circuit diagram as a k in front of the omega symbol.

[Susan Brown]

Hi Jim,

For future reference, please use the Contact form at Appliantology.org for questions about Appliantology, since we have a different Admin for that site. Based on what I could see, I think your display name is jim ortiz

[Susan Brown]

Can you just tell me what you determine when you “become” the ignitor? It’s easy to see how you can reach out and get to L1. How would you get to N?

[Susan Brown]

Question 8: there’s an important thing going on in these circuits that when you see it, you will know the answers without having to do any calculations.

Do the “Zen trick” on the ignitor… how do you reach N? What about the booster?

Let me know, and we’ll go from there.

[Susan Brown]

Hi Jim,

First of all, did you get my email with feedback on the Midterm? You did get the answers correct to parts 2 and 3 on Question 7. Are you just trying to understand the answers better?

This is an example of where it’s good to cut through confusing-looking notations, etc., and try to see things more simply.

These are just two parallel circuits. The top one has a component that fails open.

We are checking to see if you understand the relationship parallel circuits have with each other. Does a failure in one affect the other?

Then, we want to see if you understand the effect that failure has from the point of view of the power supply (the overall current “draw”).

The videos at the end of Unit 4 go over this material.

Let me know if you have followup questions.

[Susan Brown]

Hi A.J,

We do the calculation in the video at the end of Unit 3, just with slightly different values for the resistances. Could you follow what we did?

You start with calculating the circuit current, I.

Then you use that and the resistance of the loose connection to calculate the heat, P, in watts.

A common mistake is to forget that the circuit current is determined by the source voltage and the total resistance in the series circuit.

Let me know what you get for I and then for P.

[Author]

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