Susan Brown

[Susan Brown]

Great! Glad to help.

[Susan Brown]

I just went to look at your quiz, and it was questions 6 and 7 that you missed. 6 is above, here is 7:

Question #7 – In order to calculate the voltage drop across a load in series with another load, we first need to calculate the ______ .

I’ll deal with this one first. The choices are total circuit resistance, current, or voltage. We can rule out total circuit voltage, because that is not calculated, it’s given.

To calculate voltage drop across a load, we need to know both the current in the circuit (which is constant throughout the circuit) and the resistance of the load, right? E=I x R.

The resistance of the load is what it is. In other words, it’s an intrinsic value of the load and must be specified on the tech sheet.

Current, however, has to be calculated knowing the circuit supply voltage and the total resistance of the circuit. I = E/R. Does that make sense? So, the first step in eventually being able to calculate the voltage drop across a particular load is to calculate the total circuit resistance so that you can then calculate the circuit current.

For Question 6, the possible answers are:
The current flow through each load will be proportional to their resistance.
The current flow through each load will be proportional to their voltage drop.
The voltage drop across each load will be proportional to their intrinsic resistance.
(Note – we clarified the wording just a bit, changing “them” to “each load”)

We are talking about each load in the series. The current is the same throughout the circuit, right? So do the first two answers make sense? The current flow in the entire circuit is a function of the TOTAL resistance and supply voltage, but does not vary through each load.

E = IxR. Current is constant, but each load has it’s own resistance, so the voltage drop for a load will be proportional to the resistance of that load.

Hope that helps! Let me know if you have any follow-up questions.

[Susan Brown]

Hi Tyler,

Don’t feel bad – this is supposed to be a challenging course! The struggle is just part of the learning process. And we’re here to help ya.

First, let me just verify that these are the two questions you are asking about:

Question #5: What are loads in series?
Question #6: What is the true statement about loads in series?

Are those them?

[Susan Brown]

It’s hard to write out equations here, but I’ll try to show it.

It’s 1/(1/58 + 1/320)

1/58= 0.017
1/320= 0.003
Add those together = 0.02
Then 1/0.02 = 50

I rounded the decimals a bit, which is fine. To do the fractions on your calculator, you just enter 1 and then divide by the denominator.

Is this what you needed?

[Susan Brown]

Hi Robert,

That particular question was set up so that you just needed to know the rule of thumb about equivalent resistance: that it will be something less than the smallest resistance in the parallel circuit.

You can calculate the exact value of the equiv. resistance by using the formula that is shown in the last presentation in that unit (starting at about 2 1/2 minutes in), but you didn’t need to for this question.

Does that answer your question?

[Susan Brown]

Hi John,

Since March, we’ve added 9 new presentations to Fundamentals, including the one you are asking about. We almost immediately decided that the Ohm’s Law presentation went beyond the scope of Fundamentals, which focuses on basics, and was advanced enough that we should include it among the presentations in a new course on Advanced Schematic Analysis and Troubleshooting. We left the presentation in the Fundamentals course for the time being while we built the ASAT course, but moved it over to the ASAT course last week when that course was launched.

So you were one of the few students who happened to catch it during that brief window that it appeared in Fundamentals and got a bonus lesson (and a preview of the ASAT). Lucky you! 🙂

We’ve noted this and all other updates in the Course Updates log at the site.

[Susan Brown]

Hi Glenn,

I’m not sure what the “ugh of ohms law” is that you mentioned – perhaps that was a typo? 🙂

To cover all the bases, I’ll go over the basic math functions that are in the Ohm’s law pie chart that first makes its colorful appearance in Unit 3 of Basic Electricity.

1. Multiplication: as in, E = I x R
2. Division: as in, R = E ÷ I (written in the pie chart as E / I, which is just another way to show a division problem)

Those two are easy to do on any calculator.

3. Squared: as in, P = I² x R. A number that is “squared” is simply that number times itself. For example, 2 squared is the same as 2 x 2 which is 4. So, if your calculator doesn’t have a squared function button (usually labelled as x²) just enter the number and multiply it by the same number.

4. Square root: as in, E = √ P x R 
A square root is the opposite of a square. You are basically saying, what number was squared to get this number? If you want to know the square root of 4, you ask “what number times itself will equal 4?” The answer is 2. For all but the simplest numbers, you’ll need a calculator that has a square root button to calculate this. If you are using the calculator on an iPhone, turn the phone sideways to get a lot of extra buttons, including the square root. It looks something like this: ²√ x 

I hope that helps! Lemme know if you have any other questions.

[Susan Brown]

That’s some wise advice, Smashy!

You can retake the quiz when you are ready.

[Susan Brown]

Listen to the video starting at about the 6:20 mark 🙂

It’s great that you’re going back and studying new material, even though you finished the course a while back. Have you seen all of the new screencasts that were added to the Basic Electricity module over the last month? You should check them out if you haven’t – it’s killer stuff!

[Susan Brown]

Hi Smashy!

Thanks for the suggestion. That would be a good thing to be able to show in the results. I played around with the quiz settings in the software we use, and unfortunately it doesn’t work quite the way we need it to for our particular style of quizzes. I’m putting in a ticket with the software company to see if they can make this possible in the next update. If we can do this, I’ll let you know!

By the way – props to you for going back and studying the quizzes!

[Susan Brown]

Hi – sorry for the delay, there was a glitch in the system that notifies me of posts!

No, when new content is added to a course you will have to take any new quizzes. If you have already done that and you still have problems accessing the units, please let us know. There have been a couple of students who took the business class early on who have had problems getting back in, but we have a way to reset you if that happens. If you do end up needing to retake any quizzes, just think of it as a good review! Fortunately, they are short.

[Susan Brown]

Thanks for the heads-up! Yep, we made that change because we realized how long that sidebar was going to be as we kept adding courses. Now you only see the modules/units for the course you are in.

[Susan Brown]

You got it! And, no, cavemen did not figure out how to use mathematical formulas to represent real-world phenomena. Don’t be hard on yourself! 🙂

[Susan Brown]

That’s not quite it!

Look at the “pie chart” near the end of Unit 3. The bottom right quadrant shows three different formulas for calculating resistance depending on what information you have. And the quiz problem gave you power and voltage data.

[Susan Brown]

Hi Kevin,
Ohm’s law says V=I*R. Power is not part of that equation.

But we learned that P = I*V (“*” means times)

This problem doesn’t give you the current, I, so you either need to first calculate the current (from power and voltage) and then use that result to calculate the resistance, or you can use the pie chart in the lesson to find the formula that calculates R from just knowing volts and power. (Remember that E is the same as Volts.)

Does that make sense? Let me know if that helps you figure it out.

~Mrs. Samurai

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