Susan Brown

[Susan Brown]

We are measuring voltage that switch, from one side with respect to the other. Remember that a voltage reading represents a difference in charge between two points.

With the switch open, we would be measuring L1 on one side, and L2 on the other. Because L1 and L2 are out of phase with each other, this would result in a 240v reading. (See Basic Electricity, units 6 and 9 for a review of 240v service.)

A closed switch is like a wire – essentially zero resistance. When that switch is closed, L1 will continue on through the closed switch until in encounters either another open of some kind, or a load. So, the measurement that Scott gets in the video of (essentially) zero volts means that he is measuring L1 on both of the wires.

One thing that is helpful to keep in mind is that there are two things that will create a voltage difference. One is an open condition, the other is a resistance/load. In the first case you are measuring voltage potential, in the second is voltage drop.

Does that all make sense?

[Susan Brown]

The two things required for current flow is voltage and a complete circuit.

Yes. We have some voltage present, so we obviously don’t have a complete circuit. Something is open somewhere.

Now that I look at it, there is no power coming from L1 with respect to Neutral but there is power coming from L2 with respect to Neutral.

This is Figure 2. These measurements tell us which side has the open fault.

[Susan Brown]

What happens when you measure the voltage source and it should read nominally 120v, but it’s reading 0? Does that mean there is an open or beak in the wiring?

That would usually be the explanation – that there is something open between L1 at the power supply and where you have one of your probes (assuming the other probe is on N).

[Susan Brown]

Hi, I am having trouble figuring out Midterm question 9. I understand that it is a 240 VAC series circuit and that no current is flowing through the load due to the 0 reading across the heater. No voltage drop. It should read 240 volts.

All of that is correct. There is a simple explanation for what would cause this, in terms of a circuit fault.

Let’s work at the answer by thinking about current.

What two things are required to have current flow? (See the beginning of Unit 3 if you aren’t sure.)

[Susan Brown]

That is correct.

In general, with parallel circuits, a failure in one circuit has no effect on other circuits that are in parallel with it.

As you say – the voltage doesn’t change, so why would the current?

[Susan Brown]

Hi John,

There is often more than one way to go about these types of calculations, and we don’t require one particular way as long as you do something that is valid and arrives at the correct answer.

It is good to show your work as much as you can in case you made a mistake partway through, that way we can give partial credit (if you are talking about the Midterm Exam) and feedback about what the mistake was.

[Susan Brown]

🙂

[Susan Brown]

Correct!

[Susan Brown]

Yep! Although you don’t often run into that many loads in series in appliances.

[Susan Brown]

R1 is 10 ohms and R2 is 30 ohms.

So Rtotal is 10 + 30 = 40 ohms.

Divide each resistance by the total to find the fraction or percentage that resistor represents.

R1: 10/40 = 1/4 = 0.25 or 25%
R2: 30/40 = 3/4 = 0.75 or 75%

Then multiply that fraction or percentage times the source voltage to get each voltage drop.

[Susan Brown]

Sure, you can think of it that way.

Furthermore, the amount of voltage dropped across each load depends on the resistance of each load.

If you happened to have two loads of equal resistance, then each one would drop half of the source voltage.

Otherwise, you have to figure out the voltage drop in one of two ways.
1. By using the relative proportion of each load, or
2. By using E = I x R

An example of the first way is if you knew that your first load is 10 ohms and your second load is 30 ohms.

Rtotal is 40
R1 is 1/4 of the total, so would drop 1/4 of the source voltage. 120v x .25 = 30v
R2 is 3/4 of the total, so would drop 3/4 of the source voltage. 120v x .75 = 90v

Or, method 2 is to find the circuit current:
I = 120/40 = 3 amps
E1 = 3 x 10 = 30 volts
E2 = 3 x 30 = 90 volts

Method 2 tends to be easier when your resistances are not as simple as the example above.

Does that make sense?

[Susan Brown]

Hi John – you posted this in the Forum for the “Troubleshooting” Module, which is part of the confusion here. I don’t think you have reached that Module yet.

The individual lessons are called “Units”. Are you talking about Unit 6 in the Basic Electricity Module?

[Susan Brown]

Correct! I x E = P (The “PIE” equation)

[Susan Brown]

Ah! You accidentally posted in the wrong Forum. You are talking about the Advanced Washer, Dryer, and Dishwasher course. I will move the topic to the correct forum.

[Susan Brown]

Hi Christopher,
It also appears that you found the correct answers – is that so? Do you still have questions?

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