Susan Brown

[Susan Brown]

1. A little simpler way to put that is current going through a load will create a voltage difference across that load. We call that voltage drop.

2. That’s not quite was we are looking for. We’ll get back to this in a minute.

3. Which load is in parallel? Please use its name so we can be sure we’re on the same page.

[Susan Brown]

Yes – what answer do you get?

[Susan Brown]

Remember to use a different title for each post!

There are 4 loads in this circuit: ignitor, booster, safety, main

Here are a few questions to get started:
1. What creates voltage drop across a load?
2. What effect does the closed detector switch have on the current flow in these circuits?
3. Can you describe to me which of these loads are in series, and which loads are in parallel?

Let me know what your answers are to those.

[Susan Brown]

No, remember I x R is the equation for voltage.

P is I^2 x R

(we use “^2” to designate I “squared”, or I x I, because I always forget how to type the little superscript 2)

• This reply was modified 3 years, 7 months ago by Susan Brown.

[Susan Brown]

great! good job.

[Susan Brown]

No – P (for Power)

A review of “work” would be helpful. Loads (motors, lightbulbs, elements, etc.) convert power (voltage and current) to some kind of desired output: motion, light, heat, etc.

So watts can both describe the power supplied to a load, or the output.

So “P” is used to represent wattage, whether we are talking about power supply or work output.

This means to find the heat generated when current flows through that loose connection, you are looking for a “P =” equation that uses the information you have so far on this circuit (current, I, and the resistance, R, of the loose connection)

[Susan Brown]

Yes! And what is the rule of thumb?

[Susan Brown]

And you read my post?

The most important thing is to understand what Req is, and what the rule of thumb is.

If you just need help calculating it, we break that down in more detail here:

https://youtu.be/iHB3lxdc68E

[Susan Brown]

Okay, good, that is correct.

But current times resistance (I x R) is how we calculate voltage (in this case, voltage drop), not heat/watts. That is why you came up with 240v, because that is the supply for this circuit. You were basically working backwards from how you calculated the current.

Remember to start by thinking “What are they asking me to calculate?”

What letter is used to represent heat/watts in the Ohm’s Law equations?

[Susan Brown]

We are not in control of this Google drive, and this is a new problem that has just recently occurred.

We are working right now on an alternative. Stay tuned…

[Susan Brown]

What is the current in the circuit?

[Susan Brown]

I changed the title for this post – it is confusing when they all have the same title. Please come up with unique titles for these questions.

How did you calculate that answer? It is not quite right… I’ll need to know how you did it.

[Susan Brown]

See this post I created.

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-vs-total-resistance/

Also, did you rewatch the video at the end of Unit 5?

[Susan Brown]

Hi Bryson – I’m working up a reply – hold on for a few more minutes.

[Susan Brown]

Hi Robert,

That is correct. It’s the section on Circuit Protection Devices.

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