Susan Brown

[Susan Brown]

Hi Tyler,
I moved this to be its own topic, by the way.

One quick correction. 120v and 240v are “voltage”, not “power.”

Power is both voltage AND current and is measured (usually) in watts.

Let’s look at this one:
A heating element needs to produce 5000 watts of heat. If you have a 120v power supply, how much current will need to be flowing through the element?

We are asked to find Current, I
We are given power (watts) P, and Voltage, E

If you look on the Ohm’s Law chart of equations, which one starts with “I = ” and uses P and E?

[Susan Brown]

Good question.

Let’s put it in context, as that is always important when working with these equations.

Let’s say we have a 120v circuit (L1-N) with one load in it – a heating element, for example.

So, we know the voltage drop across that element is 120v. (Don’t worry too much yet about what voltage “drop” is – we’ll cover that more as you continue on through Basic Electricity. All this means is that if you will have line voltage on one side of the element and neutral on the other in a 120v circuit.)

So, the voltage is a set quantity, which means that the product of current times resistance (I x R) must equal 120v.

If the resistance of the element is 100 ohms, then the current will be I = E/R = 120v/100ohms = 1.2 amps.

If the resistance of the element is 200 ohms, then the current will be I = E/R = 120v/200ohms = 0.6 amps.

As the resistance increased, the current decreased, as you would expect since Resistance *resists* or inhibits current flow. So in this context, resistance and current are inversely proportional.

Now let’s say you have a different scenario. You have an element with a resistance of 500 ohms. What would be the current flow if this element were in a 120v circuit compared to a 240v circuit (L1-L2)?

I = E/R = 120v/500 = 0.24 amps
I = E/R = 240v/500 = 0.48 amps

As voltage increased, current increased. They are directly proportional.

Does that help?

[Susan Brown]

Hi Derrick,

In order to see all of the answer choices again, I would have to set you back to that unit so you could get more attempts on the quizzes. That would erase all the other quizzes after that point and you’d have to retake them. If you want to do that, we can, but we can also just work on the questions you are unsure of here.

For example, here’s one of the questions you missed.

You have a circuit with two lightbulbs in series. If one of the bulbs burns out, what change, if any, would you expect to see on an ammeter (current/amps) reading?

With two bulbs in series, and one burns out, what happens to the current?

(Burned out bulb means that the filament is open.)

[Susan Brown]

Hi Ray – I sent you an email with feedback on that question. Did you not receive it? Check your spam folder if it isn’t in your inbox.

[Susan Brown]

How many loads are in each circuit?

If it is a single load, then Kirchhoff’s Law tells you the answer: the sum of the voltage drops must equal the source voltage. This means if you just have one load in a circuit it will drop the whole 120v.

This is where we encourage folks to use the “Zen trick” when you have parallel circuits, particularly if it is a series-parallel configuration. If you “become the load” and reach out to touch L1 and N, you can figure out if that load is in series with any others or not. (Be on the lookout for shunts!)

[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.

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