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Hi Steven,
I moved your questions to their own topic, just to keep things a little tidier here in the Forums 🙂
For your first question:
There are several different equations for P. How you use them depends on the scenario and what you are trying to find.The “E” in the equations is the voltage drop of the resistance that you are trying to find the heat for.
Our scenario is two resistances in series with each other. If you want to just find the heat generated by the entire circuit, then you could do
P = E squared over R
Where E is the source voltage (because the total voltage drop will = source voltage) and R is the total resistance in the circuit.But if you only want the heat generated by one of the resistances, then E has to be the voltage drop across just that resistance, and of course R = the ohms of that load.
That’s doable for question 11 in the Module 3 quiz, but I think the technique we show in the video at the end of the unit is easier. Find the circuit current first (using the total resistance in the circuit, because that’s what determines current), then P = I squared * R (where R is the resistance of the load you are interested in)
For your second question:
We are assuming that the element is the only load in this circuit. We give you E (voltage) and P (watts). We want you to find R (what the resistance of that element must be). Which equation will you use?
Hi Darren,
Always nice to hear from you!
No, we don’t have any plans for a dishwasher course. There just really isn’t enough unique technology beyond what we already teach in the other courses. Same thing with water heaters (very little demand for that – I think that’s mostly the domain of plumbers).We’re good, thanks! Hope your side effects were minimal.
that’s correct – there is no current flowing in the circuit. We know that there’s some voltage present, so what is wrong with the circuit?
(Think about what is necessary for current flow – voltage and a ____ circuit)
It’s a little easier to help you with this if I know what your answer is for Part 1. Based on Figure 1, what can you say about the general nature of the fault in the circuit?
Hi Jeremy,
Read the section near the end of Unit 7 about Loading vs. Non-loading Meters
Short answer: you want to use a loading meter or a DMM that has the LoZ function to measure AC voltage.
Hi Jad – I’m going to email an answer to you!
yes it is
The USPS has been terrible since December. We’ll send another one out to you next week and hope it makes it to you!
Best wishes on the new job!
Voltage is correct.
Not a load, though. A load is necessary in a circuit to prevent it from being a short circuit, but a load is not necessary for current. In fact, a load *resists* electron flow.
We’ve emphasized this in the Module – besides voltage, you need a ____ _____ for current to flow.
You are correct that there is no current flowing in the circuit.
But the problem is NOT that there isn’t “enough” voltage.
What two basic things do we need in order to get current flow?
those are correct
Hi Scott – thanks for letting us know – there is definitely something wrong with that link! I’ll get my IT guy to figure that out and fix it.
Sorry for the inconvenience!
the answers to Part 3 and Part 4 are not correct.
Did you rewatch the two videos at the end of Unit 4? They explain the scenario nicely (two circuits in parallel – what happens when one of them fails open).
Yes, that’s the answer. Do you understand it now?
First of all, did you rewatch the video at the end of Unit 5 where we talk about this?
Equivalent resistance is mathematically reducing two or more loads into just one.
It’s a little easier when we’re talking about loads in series, because all you have to do is add them (we call that “total resistance”). For example, if you have a 10 ohm load and a 20 ohm load in series, from the point of view of the power supply, it “feels” 30 ohms of resistance in the circuit. It doesn’t know that the resistance is divided up into two different loads.
Similarly, if you had a 10 ohm load and a 20 ohm load in parallel, the power supply would “feel” a resistance 6.67 ohms (that’s the equivalent resistance of those two loads). It’s just the way electricity works.
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