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Hi Ethan,
A shunt will result in all of the current taking that path rather than the alternate path with a resistance in it.
In the example you give, when the switch is closed the heater is completely bypassed – no current.
Correct – the supply voltage is what it is and won’t be affected by a failure of a load.
Did you try clicking the gear and improving the quality?
I took a screen shot – let’s see how that shows up here:
I am not looking at a diagram, but it sounds like you are describing a circuit where R1 and R4 are in series with each other and also with R2 and R3, which are in parallel with each other.
The total voltage drop across the entire circuit would always be 120vac. The voltage drops across each load would change, at least a little bit, with the failure of one of the parallel loads.
Would it be correct to assume that the total current would decrease because the resistance is actually increasing?
Yes – that’s exactly right. You can also think of it this way: you lose the current in one parallel circuit when it fails, but the others stay the same, so the overall current decreases.
Hi Michael,
Do you know how to make the video full screen? You click the little square with 4 arrows in the bottom far right corner of the player window. The images are a little blurry, but when I do that I can read them clearly.
Does that help?
Hi Kevin,
There are a lot of tests in the course, and many of them are not very difficult, particularly when you take into account that the tests are “open book” (meaning, you can look over the unit and your notes while you are taking the test) and you have more than one attempt at each test.
On the ones that are a lot more difficult, such as are found in the Basic Electricity module or the Midterm Exam, a student could have poor comprehension and receive low scores and yet still have an overall high average because of the large number of easier ones.
Since the information in the Basic Electricity module is arguably the most important for a student to master, we want to make sure they have the incentive to do so. That’s why we have the system we have, and it has worked very well.
Does that make sense?
Did that answer the question for you? I was expecting to step you through it more, starting with you answering the question I asked.
Hi Jim,
It’s always good to start by thinking about the basics: current, voltage, and open/closed circuit.
In order for loads, like a light bulb, to do work, they need Power, which is voltage and current. And current will only flow in a closed circuit.
In this scenario, we have two light bulbs in series. Do you know the physical effect on a circuit when a light bulb burns out?
Hi Jim,
When current flows through the two loads in this circuit, it will produce heat (and voltage drop). Heat is a form of power.
There are several different formulas for P in the Ohm’s Law chart that we gave you earlier. The easiest one for this application is to use P = I^2 x R, where I is the current in the circuit and R is the resistance of the load you are interested in.
(I^2 is another way of writing “I squared”)
So, this is a two-step calculation. First you need to find the current in the circuit, then when you have that you can calculate the heat generated by either load.
We show this type of calculation in a similar scenario in a video in Unit 3 on the “loose connection.”
See if that helps you to figure this out. If you have any followup questions, let me know.
Check your email – we need to work on these questions privately.
Hi Jim,
Question #11 is “In which circuit will the heating element be getting power and heating?”
Question 12 is “In which circuit will the light bulb glow brighter?”
Which one are you wanting help on?
The thing to notice is the effect that the closed switch has on the circuit. Do you know what it’s function is?
Got it. I’ll be able to go over these this weekend and will email you back – so keep an eye out
Hi Everardo,
No, I haven’t gotten an email from you since Sunday. Please send them again. The best way is to reply to the email thread that we had going – the one where I gave you the questions I wanted new answers to.You’ve raised a more interesting question than you probably realize!
First of all, pure water is not a good conductor. It is actually the impurities (ions) in water that carry the current. A biggish drop of dirty water might create a low-resistance connection that might carry enough current to cause the breaker to pop before it just boiled away, but it’s far from a given. Also, note that it would not be a short, as the drop of water would have some resistance. But it could be a low-resistance connection.
I think the point of your question was to imagine something creating an accidental, highly conductive connection between the various spades/posts of the plug. So, let’s instead imagine a bit of copper wire that somehow contacted the prongs in the 3 ways you described.
In that case, your imagined outcomes would most likely occur, with a big IF.
And that is IF the ground is valid – that it is bonded to neutral at the breaker box. That’s how it’s supposed to be, but you don’t want to take that for granted.
Lemme know if you have any followup questions.
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