This was a tough but good week! I am happy with everyone’s progress on the material. Even if you think you are struggling, I think that everyone is actually doing pretty great. Keep up the good work!

**Computer Programming**

I am very impressed at how well everyone took to writing HTML this week. We started learning CSS and everybody is well on their way. The two most important concepts for CSS are (1) the idea of separating out *content* from *presentation*, and (2) using class and id attributes to specify special HTML elements on your page.

Note, however, that in class we put the stylesheet directly into our page for simplicity, but the book will have you store the stylesheet in a separate file, included by the “link” tag. These are basically equivalent, but the nice thing about the link tag is that if every page references the same stylesheet, you only have to change one file to change your whole website.

You all are doing great, and very soon we will be doing JavaScript.

**Electronics**

Many of you noticed that this was a math-heavy chapter. It’s obvious I need to rework the chapter a bit to make it more obvious how to work the last problems. Below is how to work the now-infamous “Problem 13”, but first, a quick summary of the method. It is time-consuming, but hopefully each individual step makes sense. The steps are simple, but combining them is more tedious than difficult, once you know the trick. Also, just to make everyone feel better about it, one of my Calculus students said, “this week the math in Electronics was harder than it was for Calculus”. Anyway, take heart. Once you understand how these things work, it will be second nature to you. And that is more my goal – once doing this sort of calculation is second nature, then the stuff that we do next semester with audio circuits will be *much* easier to understand. When you can look at a circuit and can tear it apart in your mind, then understanding what people put in a schematic will be much easier even if you don’t bother with the actual calculations. Knowing that it *can* be calculated is just as important as everything else.

The key to it is realizing that anytime you come up with an “equivalent resistance”, you can (at least temporarily) replace that whole subcircuit with a single resistor with that resistance. This is especially helpful when you have combinations of series and parallel resistances. You basically use the series and parallel resistance formulas until you have a value for the resistance of the whole circuit, then use Ohm’s Law to find the current. Then you go back through and use the newly-found current to find voltages for each segment, breaking your circuit back apart into components as you go. So, here is the play-by-play for problem 13:

First we need to get the resistance for the whole circuit.

Starting with the positive terminal of our battery, we notice that the circuit immediately branches into two parallel lines. One line is R1, the other line is R2, R3, and R4. R2, R3, and R4 are all in series with each other, so the total resistance of those resistors is found by simply adding them together. We will replace them with a new resistor, which we will call R2-3-4, which will be 250 + 125 + 100 = 475 ohms.

So, now, our parallel resistance is between R1 (300 ohms) and R2-3-4 (475 ohms). Using Equation 7.2, the total resistance of this will be 1/((1/300) + (1/475)) = 183.87 ohms.

So, now, we can replace that whole shebang with a single resistor of 183.87 ohms. We will call this replacement resistor, R1-2-3-4.

Now, since we have replaced that whole subcircuit with a single resistor (R1-2-3-4), we now have a series circuit between R1-2-3-4 and R5. So this resistance is found simply by adding them together. So the total resistance will be R1-2-3-4 + R5 = 183.87 + 150 = 333.87 ohms.

Therefore, the total resistance for the circuit is 333.87 ohms.

Now we can find out how much current is passing through the whole circuit using Ohm’s Law.

I = V / R = 9 / 333.87 = 0.02696A

Okay, now, to solve for voltage, we can either start at the top or the bottom of the circuit. Let’s start at the top. So, we have this parallel resistance (R1-2-3-4) which is 183.87 ohms. 0.02696A is how much total current is flowing through it. Therefore, we can use Ohm’s Law to find out the voltage drop across this circuit:

V = I * R = 0.02696 * 183.87 = 4.957V

Now, to find out how much current is flowing across each individual element, we can also use Ohm’s Law.

For R1, the current isn’t known, but the voltage is (we just calculated the voltage drop across the whole parallel circuit). So, using Ohm’s Law, we find:

I = 4.957 / 300 = 0.01652A

So, the voltage drop across R1 is 4.957 and the current is 0.01652A.

Now, the other branch, R2-3-4, is 475 ohms. We can either calculate the current using Ohm’s Law, or just use Kirchoff’s Current Law. The Current Law says that the total coming out must be equal to the total coming in, so since we have 0.02696A coming in and 0.01652A going out one way, that means that there is 0.01044A going through the other pathway. Ohm’s Law would give us the same answer:

I = V / R = 4.957 / 475 = 0.01044A

Now that we have the current, note that the current going through each resistor will be the same (since they are in series), so, we can use Ohm’s Law to find the voltage drop in each:

For R2: V = I * R = 0.01044 * 250 = 2.61V

For R3: V = I * R = 0.01044 * 125 = 1.305V

For R4: V = I * R = 0.01044 * 100 = 1.044V

The total voltage drop matches the expected = 2.61 + 1.305 + 1.044 = 4.954V (a little off from rounding – we will use our original number, 4.957 for the rest).

Now, there are two different ways of finding the voltage drop across R5. We can either use Ohm’s Law (since we know the current (0.02696A) or we can use the fact that we just ate up 4.957V to find the remaining voltage.

Using Ohm’s Law: V = I * R = 0.02696 * 150 = 4.044V

Or, using the fact that it is a 9V battery, and that R5 is the only resistor left, we know that R5 *must* eat up the rest of the voltage in the circuit. So, since it started at 9V, and the parallel circuit ate up 4.957V, the voltage before the resistor is 9 – 4.957 = 4.043V. After the resistor we are back at the negative terminal, so the voltage is zero, so the voltage drop across this resistor is 4.043 – 0 = 4.043V. This only differs from our Ohm’s Law version due to rounding.

This week we are working with diodes. I didn’t give out any diodes because we were running short on time, but the text doesn’t require you to have one. I will try to give out regular diodes and zener diodes next week if you want to play around with them, though I may not currently have enough for everyone.

Next week we are going to go over some basic resistor circuit patterns, and then the next week we are going to build our first semi-useful device — a darkness sensor.

Don’t forget to download the new version of the book!

**Calculus**

Today we covered and proved some of the basic rules of Calculus. I am not so interested in you all knowing the proofs as much as understanding the types of ways that rules are proven, because it helps you realize why we need special rules for different operations in Calculus.

You should memorize these rules: The power rule, the constant multiplier rule, the addition rule, and the exponent rule (i.e., all of the rules from the chapter). I will try to get the book updated tonight with exercises for Chapter 8 (if you download it and there are no exercises, that meant that I haven’t finished it yet).