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All that depends on where you, the technician, are relative to the board. Which is why it’s best to avoid trying to think about whose right is whose left and so on. If you imagine that your face is the clip, regardless of whether it’s attached to the connector or the board, you should be able to find pin 1 of the connector by looking to your left.
Hi Grasshoppah,
The trick is to become the harness connector clip and face away from the harness connector.
So, depending on which side of the connector the clip is located, you may or may not end up facing away from the entire board. It doesn’t really matter. The important thing is that you “become” the clip and face away from the connector, then find the pin that’s farthest to your left.
Power companies are sending power over the transmission lines, not simply voltage or current. The Ohm’s law equation showing the relationship between power, voltage, and current is P=I*E. According to this equation, you can jack up the voltage and reduce the current and still transmit the same amount of power in watts.
This is desirable because it reduces the required ampacity of the transmission line, meaning smaller diameter wires (lower cost, easier to work with). And it reduces I^2*R losses (actually called “I-squared R” losses) due to heating the conductors.
There are a lot of different Ohm’s law equations that show a lot of different relationships, so you need to make sure you’re using the correct equation to express what you’re thinking about since, in the end, electricity can only be understood through the math.
July 7, 2017 at 1:35 pm in reply to: Voltage drop conundrum (that may or may not be spelled right!) #12474Yes, that’s right. To illustrate this further, you can go back to the I = E / R equation that you brought up earlier. If you plug in the circuit’s source voltage for E and the circuit’s total resistance for R, then you get the circuit current.
Hi Boyd,
You’re certainly allowed to ask about your own appliances, but it must be phrased as a proper troubleshooting question, with a schematic provided. You’ll learn about this and all kinds of troubleshooting strategies that will really help you out in the Troubleshooting module of the Fundamentals course.
Once you have worked through the course to the end of the Troubleshooting module, please feel free to ask this question again as a new topic in the Troubleshooting forum, using all the tools you have learned. We would be more than happy to work through it with you!
July 7, 2017 at 11:43 am in reply to: Voltage drop conundrum (that may or may not be spelled right!) #12471Hi Grasshoppah,
That’s a pretty common question. To understand the answer, it helps to visualize how current works. Current is electrons physically moving through a conductor from one side of the power supply to the other. They’re like a bunch of billiard balls being pushed through a tube, all crowded together and knocking each other forward. If you think of it that way, it makes sense that current can’t lessen as it progresses through a circuit — if the electrons earlier in the circuit are moving, then the ones later in the circuit have to keep moving as well, because they’re still being pushed along.
Essentially, the current in a series circuit is decided as soon as power is applied to it. The power source instantly “sees” every load in the circuit and how much resistance it has, then supplies a certain amount of current accordingly.
Sorry for the confusion — I misread the table. Yes, that table is correct, and in fact would be regardless of which side is line and which is neutral. I edited my previous reply to avoid any further confusion.
I think this video from the Fundamentals course explains the formula better than I could through text:
If it still doesn’t make sense after watching that again, let me know and I’ll try to explain it.
Yes, that’s correct.
If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch.
A quick correction to what I said here — because the door switch is open and therefore not providing a path to neutral, you would most likely not read a full 120 VAC across the push to start switch with your standard VAC function. This is because, with that wire not being connected to neutral, it’s just floating there with some amount of ghost voltage sitting on it. So when you make your non-loading measurement, you would get a weird, indeterminate voltage reading — less than 120, but more than 0, probably with some fluctuation.
Here are a couple videos by the Samurai that illustrate this concept very well:
Hi George,
We’re dealing with a pretty simple circuit here: just the two switches in series with each other. Let’s say that you’re measuring across the push to start switch with it not actuated (and therefore open), and the door switch has failed, going permanently open.
In this case, the reading you would get depends on the meter you use. If you use the normal VAC function on your multimeter, you would read 120V across that push to start switch — even though you have no connection to neutral because of the open door switch. However, if you make the measurement with a loading meter, you would read 0V across the push to start switch, since even with the 120V present at one side of the switch, there is no valid path to neutral. That’s a much more useful measurement, since it shows you that something in the circuit is interrupting your power supply.
This is one of many cases where using a loading meter instead of a normal voltage meter is essential. The standard VAC function would lead you to believe that everything about the circuit is fine. But the loading meter clearly shows that there is something wrong with the circuit.
June 14, 2017 at 11:07 am in reply to: 1.3 Fundamentals -Basic Electricity end of Module exam #12147Yes, that’s the correct answer. Good job!
When someone says “there are 0 volts from L1 to neutral”, that’s just a shorthand way of saying “when I take a voltage measurement with one lead on L1 and the other on neutral, I read 0 volts.” You could also say “there are 0 volts between L1 and neutral.” It all means the same thing — that is, that you’re measuring L1 with respect to neutral. And yes, even though neutral isn’t involved in this 240V circuit, you still use it as a reference point for measuring the presence (or absence) of voltage.
As for question #6, your answer is a bit off. Could you show me your calculations?
Sure thing — I’ve sent them to you.
Yes, I am the Samurai’s son! I doubt I can give him a run for his money yet, since everything I know I learned from him, but maybe someday…
And yes, the capacitor would be charging and discharging at 60hz, in time with the voltage’s reversing polarity. That’s how capacitors behave in any AC circuit.
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