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?

[Susan Brown]

Hi Danny – I will send you an email with your options, so keep an eye out for that.

[Susan Brown]

1. That’s correct – the answer is “It does not have a supply voltage, it switches 120 VAC at its working contacts”
2. We have never seen an NTC thermistor that didn’t have a 5vdc supply

[Susan Brown]

Okay, I had stated that the Resistances were R1 = 100, RA = 80 and RB = 480.

You are trying to do a similar calculation using different resistances?

If R1 is 84, RA is 83, and RB is 423:

The Equivalent resistance of RA and RB would be 69.4 ohms [1/(1/83 + 1/423)]

If you put that in series with R1 you have a total resistance of 69.4 + 83 = 152.4 ohms.

So I = 240/152.4 = 1.6 amps

Note – you won’t have to do series-parallel calculations for the Midterm Exam. You just have to understand how voltage and current work in series, parallel, and series-parallel circuits.

[Susan Brown]

Hi John,

What is the scenario you are talking about? Is this from one of the videos or a quiz question?

[Susan Brown]

A couple of volts plus or minus is fine.

The reason you got 119 is just because 120v/85 is 1.4118, not just 1.4. If you preserved all the decimal points, you would get exactly 120v

[Susan Brown]

Hi John,

Yes, I = E/Rtotal, so in a 120v circuit with a total of 85 ohms, the current would be 1.4 amps.

Then, if you want to calculate voltage drop across a load, you would use E = I x R, where R is the resistance of the load in question.

If there is just a single load in the circuit, then you know that it will drop the full source voltage without having to do any calculations. You actually showed that in your example by ending back at 119v for the voltage drop calculation.

[Susan Brown]

Amps is the one property/measurement where it is usually best to preserve the digit after the decimal point rather than rounding up. Volts, Ohms, and Watts *usually* do not require this level of precision.

For example, on gas valve specifications, you will often see ranges that look like 2.5-3.0 amps or 3.2-3.6 amps, etc. So, several 1/10s of an amp can be significant.

You will encounter this on the Midterm Exam. You’ll be asked to calculate the circuit current for a series circuit, then later use that as part of your calculation of voltage drop across each load. If you round the current up to the nearest whole number, you would end up with voltage drops that add up to exceed the source voltage, which is not possible.

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