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That is the explanation we give as a help for those having a hard time with that question. I’m not sure what your particular question is. These are the same steps that you took to get the answer for Question 6 on the Midterm Exam. We also show a similar calculation in the last video in unit 3 of Basic Electricity.
Is there a specific question you have about this?
Another important characteristic of a load is that when current flows through it, a voltage drop is created.
I assume you are talking about this question:
If the Defrost Heater and the Freezer Drain Heater are operating within specifications, what equivalent resistance would you measure from Pin 13 to Pin 7 in the diagram below:
Here’s your answer:
58 Ω because the equivalent resistance in a parallel circuit will always be equal to the smallest branch resistance.The heaters are in parallel, with resistances of 58ohms and 320ohms.
Is that the correct rule of thumb for equivalent resistance of parallel loads?
Correct! I’m glad to see that you are making sure that you understand all of the questions and answers.
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.
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