## Photovoltaic Class

To all my fellow travelers on the road to solar living! If you’re following along with the solar power class, be sure to note that the class now has its own page on the site (it’s at the top of the page on the right-hand side), where you can find all the lessons in one easy to remember place. If you are following along, stop by and say hello to your other classmates by commenting on that page, so we can facilitate discussion and learning for all involved. Also, if you find great resources that everyone will want to know about, post them there!

For my fellow Los Angeles classmates who are checking in, though all the information now posted is publicly available to all who visit the site, there will eventually some information and resources posted which are specific to our class that will be available only via password. By commenting and saying hello, I will know your email address to send you a password when such information is posted.

See you there!

## Solar Power Class: Ohm’s Law Wheel

Here’s a quickie for those of you following along with the solar power class. Remember Ohm’s Law, which we’ve been discussing? Well, using the two formulae that I gave you earlier (E = I * R and P = I * E ), you can make 12 different formulas that will solve for any two out of three variables. Rather than break down the total chart, just look at this nice graphic depiction of the ways you can solve for each. Whether you prefer to use the original formulas or memorize all twelve of these, this will give you a better idea of the interrelation between the various components of electric circuitry.

## Solar Power Class: Formulae Roundup

Okay, so back to the classroom, while it’s fresh on your mind. Yesterday we talked about Kirchoff’s Laws, as well as a few other formulas that you’ll use when trying to figure out your electrical generation setup and capacity. To review, here’s a listing of formulas:

Voltage = “Electromotive Force” = E
Current = “Amperes” = I
Resistance = “Ohms” Ω = R
Power = “Watts” = P

Ohm’s Law (for DC power): voltage = current * resistance ( E = I * R ),
power = current * voltage ( P = I * E )

Total Circuit Resistance: SERIES: Rt = R1 + R2 + … + Rn
PARALLEL: 1/Rt = 1/R1 + 1/R2 + … + 1/Rn

Voltage Divider Formula (Series): Erx = Et ( Rx / Rt ) (solve for resistor “x”)
Current Divider Formula (Parallel): Irx = It ( Rt / Rx ) (solve for resistor “x”)

Kirchoff’s Voltage Law (Series): Et – E1 – E2 – … – En = 0 (circuit w/ “n” resistors)
Kirchoff’s Current Law (Parallel): It – I1 – I2 – … – In = 0 (circuit w/ “n” resistors)

Whoa, boy, that’s a lot of formulas! Using these, you can pretty much figure out whatever you want about a simple series or parallel circuit. Of course, the circuits designed for power generation are rarely simple like the above, but we’ll get to that in a second!

So now that you’ve beat these formulas in your head, how do you use them? Here’s a sample problem:

“A 220V series circuit has three resistors. The first has a resistance of 25Ω, the second has a resistance of 50Ω, and the third has a resistance of 35Ω. What is the current and resistance of the circuit, and what is the voltage drop across each resistor? How much power will it produce?”

First, you diagram everything to see what you’re working with. You see that it’s a series circuit, which means that the current is common and the total resistance is the sum of all the partial resistances.

Rt = 25Ω + 50Ω + 35Ω = 110Ω

and since E = I * R, 220v = I * 110Ω, or I = 220/110 = 2 amps.

So now you know the voltage (220 V), the resistance (110 Ω), and the current (2 amps). To figure out the power produced, you simply use Ohm’s Law, (P = I * E), or P = 220 V * 2 amps = 440 watts. We’ll add that to the list: power (440 W).

The final part of the work is to figure out the voltage drops across the circuit at each resistor. This will require the voltage divider formula, since it’s a series circuit. ( Ex = Et ( Rx / Rt ) )

Here are the values plugged in for resistor 1: Er1 = 220V ( 25Ω / 110Ω ) = 50V.
Here are the values plugged in for resistor 2: Er2 = 220V ( 50Ω / 110Ω ) = 100V.
Here are the values plugged in for resistor 2: Er3 = 220V ( 35Ω / 110Ω ) = 70V.

You can check your work on this by adding all the voltage drops together: they should equal the voltage given in the problem: 50V + 100V + 70V = 220V. Good! Now for extra credit you can figure out the power produced at each resistor by using your power formula and the voltage and resistance of each resistor in the circuit. If you do, post your answer in the comments section!

This is basically the same process as for parallel circuits. Let’s use the same problem as above, but now the circuit is hooked up in parallel:

In this case, to get total resistance, you use the formula 1/Rt = 1/R1 + 1/R2 + 1/R3. This looks scarier than it is. to get a reciprocal (one divided by a number) on a calculator, you simply plug in the number, hit the divided by sign, then the equals sign. The calculator will do all the dirty work for you, but you should try doing them in your head sometimes, ’cause you never know when the calculator might stop working when you need to figure something out! So:

1/Rt = 1/25Ω + 1/50Ω + 1/35Ω = .04 + .02 + 0.0286 = 0.0886

This is the reciprocal of the total resistance (this is also called the “conductance”), so we take the reciprocal of this and get a total resistance: 11.287Ω. A total resistance of less than each of the parts? Yes, this example illustrates two things about parallel circuits: first, the numbers are rarely as “pretty” as in series circuits. Secondly, the total resistance of a parallel circuit is always less than the lowest partial resistance. If it’s not, then you calculated something incorrectly.

So now, as before, we know the voltage (220V) and the resistance (11.287Ω). To figure out the total current, we use Ohm’s Law ( E = I * R ): 220V = I * 11.287Ω so I = 19.49 amps. You can now easily figure out the power produced: P = 19.49A * 220V = 4287.8 watts of power (or 4.3 kW).

Using the same resistors and voltage, you’re generating more than ten times the power by hooking things up in parallel!

Rounding out the lesson, you’re now trying to figure out the current drop across the resistors (remember, in a parallel circuit, current is divided, not voltage). Use the Current Divider Formula: Irx = It ( Rt / Rx )

Here are the values plugged in for the first resistor: Ir1 = 19.49A ( 11.28Ω / 25Ω ) = 8.794 A.
Here are the values plugged in for the second resistor: Ir2 = 19.49A ( 11.28Ω / 50Ω ) = 4.397 A.
Here are the values plugged in for the third resistor: Ir3 = 19.49A ( 11.28Ω / 35Ω ) = 6.281 A.

To check your numbers, use Kirchoff’s Current Law, and add up the current drops:
8.794 A + 4.397 A + 6.281 A = 19.472 A, which is the same as above, save my rounding. This is why it’s important to keep numbers true to the highest number of decimal places you can… your work will be that much more accurate.

Again, if you want to figure out the power available at any point, use the same process as for the series circuit.

Congratulations! You’ve now mastered simple series and parallel circuits. That’s a lot to know. Next time, we’ll take a look at how to deal with more complex circuits, like the ones you might encounter in your alternative energy generating system, and also talk about why all this math is important to you when all you want to do is drop off the grid and tune out. I found that this stuff didn’t really cement in my head until I tried figuring out practical problems with them, so I’d recommend you do a little practice session, perhaps using the problems found at Open DNS: Ohm’s Law Practice 1, and Ohm’s Law Practice 2. Or try this site if you like to “play and learn”: Ohm’s Law Battleship Game, where your shots won’t land unless you can solve some problems along the way. See you next time, captain!

## Solar Power Class: Kirchoff’s Laws

Whew! It’s been a busy couple of weeks in my photo-voltaic class. I was starting to fear that I’d have to take a math class to keep up with all the formulas! When I posted the first week’s lesson, I realized later that I’d given out misinformation, which is the danger of posting about something you don’t yet understand! So, from this point out, I’ll just post the lessons after I’ve been tested in the contents, that way you’re always getting the information someone has TOLD me I understand. Since we had a test this past weekend, here’s a new dose of mathematical fun!

This week: Kirchoff’s Laws. Last time, I discussed Ohm’s Law of DC power, which interrelated voltage, current, amperage, and power, and provided several formulas you can use to figure out any of the above for a circuit. If you’d like to review, check out the original post here. Now, let me repeat a few pertinent facts: a series circuit is when you basically hook everything up in a big loop, positive end to negative end in a chain. See the diagram below:

A parallel circuit is one in which the positive and negative ends are “shunted” together (parallel circuits are sometimes called shunts) creating a ladder effect. Again, see the diagram below:

Series circuits are called voltage divider circuits, because though a common current flows across the wire, at each stop along the way, voltage is dropped. These are two important concepts: 1. current is common. 2. voltage is divided along the circuit. Parallel circuits are the opposite. Though a common voltage flows through all the wires, the current is divided between the different potential paths. Therefore, in a parallel circuit, 1. voltage is common, and 2. current is divided along the circuit. Series circuits are called “voltage dividers” and parallel circuits are called “current dividers”. VERY IMPORTANT is you want to know how to manipulate these circuits later on.

There are even two formulae which will help you to calculate a voltage or current at any particular point along a circuit. Say you have three resistors along your circuit. In a series circuit, a voltage divider, if you want to know the voltage of resistor “b”, you would use the voltage divider formula: Erb = Et (Rb/Rt), where t represents total and Er is the voltage drop. Here’s an example:

In a series circuit with a total resistance of 100 Ohms, and a voltage of 120V, resistor b has a resistance of 25 Ohms. The total voltage drop across resistor b would be:

Erb = 120 v ( 25 Ohms / 100 Ohms ) = 30 V

Now, if you aren’t sure what the resistance of a particular resistor on the circuit is, then Kirchoff’s Law of Voltage (for series circuits only!) comes into play. His law states that the total voltage minus the voltage of each resistor, etc on the circuit will always equal zero. In other words, the Total Voltage equals the sums of all the voltage drops along the path. Here’s the official equation: Et – E1 – E2 – … – En = 0, where the circuit has n resistors. So if you know that one resistor has a voltage drop of 25 v and the third has a voltage drop of 50 v, and the total voltage is 130 v, then 130 – 25 – E2 – 50 = 0, and E2 = 55 v. Got it?

Now, on to parallel circuits, ones you’ll see a lot of in battery configurations. Because parallel circuits are current dividers, they need a separate formula for figuring out current drops around the circuit. This is called the Current Divider Formula (using Current at Resistor b): Ib = It ( Rt / Rb ). As with the voltages of a series circuit, if you need to know the current drop at a particular point, Kirchoff had a law for that, too. It’s called Kirchoff’s Current Law (for parallel circuits), and it states that the total current minus the current drops along the way, equals zero. So It – I1 – I2 – … – In = 0, where the circuit has n resistors.

Now, this is a LOT of information to absorb, especially in practice, so let me stop here for now, and we’ll pick up here tomorrow with the rest of the lesson. It seems like way too formulas to ever be useful, but once you get to solving practical equations with them, it’s not too bad. But let’s save that for the next lesson, sleep on it, and I’ll see you in class tomorrow~

## America’s First Wind Powered City

It’s official: at least one city in the United States has finally ponied up for a wind powered station that will meet the entire city’s needs.  Meet Rock Port, Missouri, poised to take that trophy home for America.  Fortunately situated near a bluff and with a windy enough climate to sustain a projected 16 gigawatt hours of electricity per year, Missourans are about to get a healthy does of green in their power mix.  Annual consumption has historically only been around 13 gigwatt hours, so that power company will also be able to sell power across the grid to other places, as well as to supply electrical power when winds are down.  With this year’s tornado season as evidence, I don’t think that will be happening too often!

For more information, look up Loess Hill Wind Farm, the company pairing with the government to provide this service.

## Quickie: Renewable Energy Installation Database

For all you Earth Day surfers out there, here’s a good site to check out, courtesy of the Stella Group in Washington DC:

Geographic Database of Renewable Energy Installations

It’s a nicely compiled state by state listing of proposed and operational energy installations.  If you’re looking for a little “been there” inspiration or want to know what sorts of alternative energy are well-suited for your area, start here!

## PowerCube Energy: Solar in a Box

If the intricacies of setting up a home solar solution have you flummoxed, you may be looking for an out-of-the-box solution for your energy needs.  It’s not exactly portable (unless you own a forklift!), but the PowerCube 600 Energy system is just that… a box that you simply open and start harvesting light energyVisit the PowerCube site for pictures of the cube being set up to appreciate how easy it really is. The site and technology appear to be young, but the promise of a standalone power system in a box can hardly be overstated.

From what I can see, the box has a variety of power outs so that you can hook up various devices to the unit. And the site claims that you can increase your energy output by daisy chaining multiple units together, providing enough for off-grid applications and primary power-source situations. I like the box design, it looks sturdy and easy to ship, given its size, and it seems like a good fit for programs that offer solar power to remote communities across the globe. I haven’t been able to access the spec sheet yet, but the maker’s site, a yacht building company, shows the product in more operative detail.  All from Reluminati, an eco-concious design lab that sports several lines of solar powered products.  Be the first on your block to sever your ties to the grid when the PowerCube rolls off the assembly line this summer.