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Hi John,
You’ve got the correct quantity (“about 1350”), but what are “heat units”? That’s not a normal way for us to quantify heat (and thus not the correct answer).
There was another choice for an answer that uses the correct units – do you know what that would be?
The concept of this problem is very similar to the “loose connection” problem that you successfully calculated on the Midterm Exam (Question 6). Two loads in series, and calculating the heat produced by one of the loads in particular. Look back over your answer for that and let me know if you still understand what you did there.
Hi Phillip,
Are you talking about this question?
You’re working on an appliance with the circuit configuration shown below. R1 = 5 ohms. R2 = 32 ohms. If everything is working correctly, what is the expected heat produced by R2?
April 15, 2018 at 11:42 am in reply to: EEPS Electric Dryer troubleshooting why not check out the other two devices? #14173Hi Phillip,
See if this video helps: https://youtu.be/cLkeBsDXjZc
For any more detailed advice on service calls, please use Appliantology.org.
Great!
In the problem you are talking about from the Midterm, you are given the resistance of each load, and you calculated the circuit current in the previous question. You should then be able to do a calculation for of voltage drop across each load. Each load has the same current going through it.
Hi Mariah,
Don’t you also know the circuit current?
Here are the critical things to remember about a circuit with loads in series:
1. The current will be the same at every point in the circuit.
2. The current flowing through each load will produce a voltage drop across that load.
3. The sum of those voltage drops through all of the loads will equal the source voltage.Look at the Ohm’s Law equations in Unit 3. If you know the circuit current and you know each load’s resistance, do you see how you could calculate the voltage drop across each load?
As far as resistance numbers go, practically speaking there’s no difference between 18.75 and 18.86. We don’t need to be that precise when it comes to resistance.
As far as a reset goes – please check your email.
The equation is Req = 1/(1/R1 + 1/R2).
In the example in the Unit 5 video, the resistances are 30 and 50.
You always calculate the stuff in parentheses first: 1/30 + 1/50 = 0.0333 + 0.02 = 0.0533
Then 1/0.053 = 18.75
Hi Robert,
Don’t feel dumb! This is a common question. Check out these two previous forum topics and see if they help you. Feel free to ask more questions after looking over these:
March 23, 2018 at 7:26 pm in reply to: gas pressure regulators question 7 explanation on the leak limter orfice #14128Yes, it’s purpose is to relieve gas pressure buildup above the diaphragm in case the diaphragm develops a leak.
March 15, 2018 at 9:44 am in reply to: Module 5: Troubleshooting- Unit 5 Using Schematics Part 2 Quiz #14122Hi Jason,
First of all, it would probably be helpful for you to rewatch the Half-splitting video, if you haven’t already, where we go over this.
But here’s some more info.
Question 16: we haven’t disconnected any wires, and we energize the circuit. Measuring voltage across a load is attempting to measure voltage drop. When current flows through a load, it will create a voltage drop. So if you measure zero voltage drop across a load, it means no current is flowing. This means the voltage we are measuring is potential voltage. Furthermore, the fact that we measure this 120vac wrt N at each terminal of the element also demonstrates that the element is not open – it is acting like an unbroken wire.
Does all of that make sense?
Question 17: the circuit is now half-split on the L2 side of the element. Yes, the measurements are wrt N. The side where we measure 0 vac wrt N is the side that has the open. In this case, it is the L1 side that has the various controls.
Let me know if this clears it up, or if you have follow-up questions.
Yes!
That’s correct.
That’s correct!
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