Susan Brown

[Susan Brown]

Do you want me to reset that quiz so you can retake it?

[Susan Brown]

Hi Eugena,

You’re talking about how the LOI gets L1 and L2.

Good job being “Zen-like” 🙂

What you see is correct, although there is one additional way you can potentially reach L1.

(In other words, there are 2 correct answers to Question 2)

[Susan Brown]

The “inverse of the Rs” formula is for finding equivalent resistance of loads that are in parallel.

When you have loads in series, you want to find total resistance (in order to do things like calculate the circuit current), which is just the sum of the resistances.

[Susan Brown]

Good question!

No, you put the probes on either side of the load. You want to measure one side of the load with respect to the other side.

Rewatch the second video in Mod. 3, unit 8, and look for when he measures voltage across the light bulb that is lit up – that is an example of measuring voltage drop.

[Susan Brown]

Yes, you can calculate voltage drop using E=IxR, but you only have to do that when you have more than one load in series in a circuit.

What if you have only one load in a circuit – what would the voltage drop across that load be?

(This is a big hint that should help you get part of the way towards figuring out Question 8)

[Susan Brown]

Hi Nathan,

Start a new topic here in the Forums about this question and we can help you. I sent the student above an email because he had answers on his exam that I wanted to comment on privately.

[Susan Brown]

Did you look at the Midterm Help Page? We give you suggestions:

https://mastersamuraitech.com/midterm-exam-help-page/

For question 7, we say:

Make sure you understand the scenario. These parallel circuits are functioning normally, then the top branch (element) fails open. We want to know the change in current or voltage drop in the branches and then the overall current draw from L1. The videos at the end of Unit 4 are particularly helpful to review, but also Unit 5.

[Susan Brown]

I know the feeling! I also studied ChemE in college. This electrical stuff takes some work!

Check out this topic and see if it helps:

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/equivalent-resistance-in-parallel-circuits/

[Susan Brown]

Hi Jim,

The videos at the end of unit 4 are good to review for this one.

When you have circuits in parallel, and one of the circuits goes open, it will not affect any other of the parallel branches. It will affect the overall current and the equivalent resistance of the entire set of circuits.

The resistance of the bulb is what it is, and the voltage supply to the bulb doesn’t change, so the current through the branch that hasn’t failed will not change.

Does that help?

[Susan Brown]

Happens to us all! 🙂

[Susan Brown]

Hi John,
Electric dryers run on 240vac.

(We’ll get to your other two questions shortly!)

[Susan Brown]

Voltage drop is just the difference in voltage measured across a load that is caused by current flowing through the load, and is the same regardless if it’s AC or DC.

E = I x R whether it is AC or DC.

And all of the voltage of the power supply will be dropped across the load(s) in the circuit, regardless of AC or DC.

The amount of heat produced by a given resistance will vary between AC and DC (that’s what the “RMS value” is about). Is that what you are thinking of?

[Susan Brown]

correct!

[Susan Brown]

Hi Nathan,

First of all, 240 and 120 are voltages, not currents.

I’m not sure which question you are asking about. Your topic title says Q.3 ?

The voltages are indicated by whether the circuit is L1 – N or L1 – L2.

Revisit Unit 6 for a review of household power supply.

[Susan Brown]

Hi James,

In this question, R1 is 5 ohms and R2 is 5000 ohms.

We know that voltage drop is proportional to resistance (E = I x R)

We also know that work/power is proportional to voltage drop (P = I x E)

So, you can figure out logically that if R1 is 1000 times less than R2, it will also do 1000 times less work than R2.

The reason your calculation didn’t come up with the same conclusion is that when you are using P=E^2/R in a scenario with multiple loads in series, the “E” has to be the voltage drop across that particular load, not the source voltage.

You could try calculations again to confirm this, but first you would need to calculate the circuit current, I, then you could use P = I^2 x R. You could also calculate the voltage drops across each load and then P=I x E.

Does that make sense?

[Author]

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