Susan Brown

[Susan Brown]

Correct. Current flowing through a load will create a voltage difference, which we call a “voltage drop”. The main reason we have a slightly different term for that is that there’s a big functional difference between “potential” voltage and voltage drop.

You can think of voltage as a cause, and voltage drop as an effect.

Voltage causes current to flow (if there’s a complete circuit for the electrons)
Voltage drop is a result of current flowing through a load/resistance.

When you have voltage drop, you will have work being done (power is voltage AND current)

[Susan Brown]

Well, what I was talking about is just a wire. Two points along a wire, when there is no component between the two points you are measuring – you’ll measure 0 volts because there is no difference in electrical charge between those points.

If a switch is open, there will be a difference in charge because of the open in the circuit.

If a switch is closed, it will be just like you are measuring two points along the same wire, with no difference in charge.

[Susan Brown]

Well, that’s not always the case. It depends – sometimes you do want to measure voltage at a certain point with respect to N. But sometimes you are measuring voltage with your probes at two points in the circuit, such as across a load to check for voltage drop, or across an open switch.

Voltage is simply the difference in electrical charge between two points.

If you put your meter probes at two different points along the same wire, and there’s no open switch or load between those two points, what will you measure for voltage?

[Susan Brown]

That is talking about measuring voltage across a switch that’s either open or closed. If a switch is closed, it acts like a wire, correct? Do you measure a voltage difference at two points along a wire, when there is no load between those two measurement points?

[Susan Brown]

Hi Darron,

Unit 5 is describes various characteristics of series circuits and parallel circuits.

When two or more loads are in series, you can add the resistances of the loads together to get a “total” resistance. This can be useful when calculating the circuit current, since the “total resistance” is what determines that.

However, when loads are in parallel with each other, we talk about “equivalent resistance”. It’s a similar concept – it’s a way of describing the overall resistance that is present from all of the loads combined. It’s more complicated than simply adding them together, however, since they are not in series.

We show the formula for calculating equivalent resistance in Unit 5. But we also give the rule of thumb in the 3rd video in that unit (you can even see it in the thumbnail for the video!). That’s perhaps more important to know than doing the actual calculation, and it will help you to answer both of the questions that you posted

Take a look at that information again in Unit 5, and see if you understand it. Let me know!

[Susan Brown]

Hi Luke,

Good question!

This may be a 120/208v commercial system. You see this a lot in apartment buildings (you can do an internet search for 208 voltage supply to read more about this). If that’s the case, then your readings are to be expected.

A reminder – we always recommend you do measurements with respect to Neutral, not Ground. They should be the same potential, but that’s making an assumption that you may not know for sure.

[Susan Brown]

Hi Luke – let us know if you have any other questions. This is a great topic to make sure you understand. Also, I sent you an email earlier today about your Certification. Please look for it! (If you don’t see it in your inbox, check your spam folder.)

[Susan Brown]

You aren’t the first one to get tripped up on this one! 🙂

[Susan Brown]

Hi Samuel,

Good question. I’m glad you’re trying to really nail this down.

If we are talking AC current, then there isn’t technically a “first” load, since the current changes directions.

All the loads in the series circuit will affect what the circuit current is. In fact, you can just add the resistances of the loads together to get the total resistance for the circuit, and knowing that and the source voltage, you can calculate the current of that circuit.

The source voltage gets dropped across each load in proportion to the resistance of the load. If you had two identical loads in a circuit, then each would drop half of the voltage. If one load was twice as high as the other, then it would drop twice the amount of voltage as the other. But once you have calculated the current going through the circuit (which is the same throughout a series circuit), then you can also just calculate the voltage drop across each load with E = I x R.

Does that help? Let me know if you have any followup questions.

[Susan Brown]

Hi Abe,

I’m glad you brought this to my attention. The correct answer is supposed to be “all of the above” – and that used to be one of the options. Somehow that had gotten altered – I’ve fixed it now.

Thanks!
Susan

[Susan Brown]

Hi Dan,

You are correct that voltage and current are both required for a load to do work. And the combination of voltage and current is also known as “power”. (P = I x E)

That’s the answer we are looking for there.

[Susan Brown]

So it has to do with the sealed system, correct?

[Susan Brown]

Do you see where in the video he’s talking about seeing the ice ball? What type of failure is being described there?

[Susan Brown]

Just want to make sure – Did you notice the word “not” in the question?

[Susan Brown]

Hi Jim – thanks for the question! I’m going to have to consult with Mr. Samurai on this one, and he is out for the whole day. As soon as I have an answer for you I’ll get back to you.

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