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Glad it helped!
In a series circuit, the current is the same throughout the circuit. So the current is calculated from the *total* resistance and the voltage supply. In this case, 37 ohms and 240vac (using I = E/R)
Once you have that current, you can then calculate the heat generated by an individual load using P = I2R
Hi John,
I’ll give you a (big) hint: the water that results from defrosting the evaporator is called “condensate”.
I know it’s referred to in the last video in that unit.
You are correct!
Great attention to detail 🙂
Hello!
You have to make a couple of calculations to get the final answer. Let me help step you through this.
The problem gives you the voltage supply (L1 and L2, so 240vac) and the total resistance for the circuit. (Total resistance for a series circuit is just the sum of the individual resistances, so in this case, 5 + 32 = 37 ohms.)
Look at the Ohm’s Law chart. Do you see a way to calculate the circuit current, I, if you know the total resistance and voltage for the circuit?
Hi Steve,
There isn’t, but the Final exam is a bit more of a review than the midterm is. We’ve found that folks who did well on the quizzes and module exams generally do well on the final if they’ve taken the time to look back through those quizzes/exams to refresh their memories.
It’s pretty straightforward- you might be overthinking it.
If you know the circuit current, you can find the voltage drop across each resistance using a simple Ohm’s Law calculation.
Exactly! That’s the case for a series circuit. Just wanted to make sure you were clear on that.
E=IxR can be used in a few different ways, but as far as calculating voltage drop across an individual load, then the way you wrote it in detail (the second one) is correct.
One clarification: you said “I (current flow through load)”. If you had several loads in a series circuit, would the current change at all? Would it be different for different loads, or the same throughout the circuit?
Hi Steve,
3c) voltage drop is the product of current flowing through a resistance (load). Does that make you think of an Ohm’s Law equation that you could use?
5b) re-watch the video at the end of unit 6.
let me know if you have any other questions!
August 9, 2016 at 2:05 pm in reply to: Unit #5 Basic Electricity: Series and Parallel Circuits #10643Way to pay attention! 🙂
You are correct – that is a mistake. Current is indeed inversely proportional to resistance. Good eye! Thanks for letting us know so we can correct that.
Good! I’m glad that you asked the question. We get some students who hesitate to ask questions here, which is a shame.
Yep! And Figure 2 is shows the testing that can tell you which line is out of commission.
Does that all seem clearer to you now? Any more questions?
Exactly. It seems like you are pretty much there.
If L1 and L2 were both supplying 120vac each (out of phase), combined with a complete circuit and a good element, we’d have 240vac drop across that element and heat being produced.
But we’re reading zero across the element. This tells us there is no current flowing, so there’s an open on one side or the other.
All of this, combined with what I said earlier about the ends of the element being EEPs – can you see why we are measuring 120vac wrt N from either end of the element, even if we know there must be an open on one side of the circuit?
L1 and L2 being out of phase with each other is how a 240vac voltage can be made by combining L1 and L2. However, in this problem, we never measure 240 anywhere.
If the circuit were complete and operating as expected, what voltage drop would you measure across the element?
Sure – it’s in unit 6, at the end.
Note that this is a very similar, but not identical, scenario.
https://my.mastersamuraitech.com/module-3/basic-electricity-breaker-panel-basics/
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