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Solar Power Class: Ohm’s Law Wheel

August 5, 2008 · Filed under class, solar power · Tagged , , , , ,

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.

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Solar Power Class: Formulae Roundup

August 5, 2008 · Filed under class, experiment, solar power · Tagged , , , , , , , , , , , , ,

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!

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Solar Power Class: Kirchoff’s Laws

August 4, 2008 · Filed under class, experiment, solar power · Tagged , , , , , , , , , , , , ,

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~

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Solar Installation Class - Ohm’s Law

July 18, 2008 · Filed under class, experiment, solar power · Tagged , , , , ,

Okay, for all of you waiting for notes from the solar panel installation class, here’s the scoop. The first class was cool, there’s a big machine shop there, and as a bonus, we’re also going to learn about machine controls for wind systems. Nice! There promises to be lots of applied action working with actual solar panels, so I should be able to share some real knowledge of what works and what doesn’t. For those of you who’ve been with me since the beginning, we’re going to repair my solar panel (the inspiration for this blog!), too. Finally!

But of course, as with most things in life (and especially with electronics!), it makes sense to understand the science behind things before getting your hands dirty, or shocked in this case! So We spent the first class learning the basics of voltage, current, power, and resistance. Ever heard of Ohm’s Law? (Those of you having evil science class flashbacks, don’t worry, I wouldn’t share it unless it was quite necessary for your safety.) Here is Ohm’s Law for DC circuits.

P = I * E and E = I * R (”pie” and “ear” phonetically, to help you remember)

See? Simple! Only four letters! What does it all mean? Well, in a circuit, you have four things at work. First, Power. Power (the P) is measured in terms of watts. It’s the work being done using the energy in the circuit. So if you baked six pies, and your clown friend used all of them to cream the faces of his fellow friends, the power would be those six pies you made available for him to throw. When you hear talk about kilowatt hours, this is a measurement of how much power is produced and available to do work by a system. Or how much you are using, as reflected on your power bill. So, once again, that’s POWER (P).

Next up, Current. Current is a measurement of the electrons flowing through the circuit. You see, in order for the energy to get from one atom to the next in the wire, it pushes electrons down the line, carrying a charge (since electrons are negatively charged particles). This is what we think of as “flow”. Current is measured in amperes and is represented by the letter I (that’s “i”). CURRENT (I).

Third, you have resistance, which is measured in Ohms (it is Ohm’s Law, after all!). Usually, you will see resistance represented by the logical symbol R, but Ohms also have their own symbol, Ω. For the purposes of this discussion, I’ll stick to R. Resistance is what it sounds like, a measure of how hard it is for the electrons to jump from one atom to the next. It’s a measurement of volts per ampere (or how hard it is to push one electron from one atom to the next). It’s not a 100% rule, but usually, you will find that resistance stays pretty much the same in your circuit calculations. It is most affected by distance traveled, the width of your wire, and by temperature.  RESISTANCE (R).

And, lastly, there’s voltage, tying it all together. Voltage is also known as ElectroMotive Force (EMF), giving rise to the symbol for voltage, E. Sometimes you’ll see V for voltage, but here I’ll be using E. Voltage is measured in Volts (whew, an easy to remember one!). Solar panels, wind turbines, etc, are generally hooked up to 12 Volt batteries, either in series or parallel (we’ll get to that in a second), allowing you to store the energy produced by the panel/turbine. Common setups will run at either 12V or 24V, depending on application. So, rounding it all up, we have VOLTAGE (E).

POWER (P) - CURRENT (I) - RESISTANCE (R) - VOLTAGE (E)

Got it? Good. Now let’s start putting it all together. Using the formulas above, you will see that Voltage = Current * Resistance. So, when the voltage goes up, the current goes up also… they are directly proportionate. This is a VERY IMPORTANT CONCEPT in circuits. It’s the equation E = I * R, or “ear”.

The other equation, P = I * E or “pie”, is for figuring out how much power is produced by a circuit, and is equally important. Written out, it’s Power = Current * Voltage. In any given situation you’re probably going to know two out of four variables and you’re trying to figure one of the other two. With these two equations, you can always solve for the ones you don’t have already. Here’s an example:

If you know that you have a 12V battery and you need to get 72 watts of power to run your iPod, how much current do you need? By the way, current is usually adjusted by using bigger or smaller wires, because the size of the wire affects the number of electrons that can easily flow through at any one time. This is due to resistance, but I’m getting quite ahead of myself.

So using our equations, E = I * R and P = I * E, you know that E = 12 volts and P = 72 watts. So, the second equation can easily be solved: 72 = I * 12, or I = 72/12. This works out to I = 6. Since I is measured in amperes, the answer would be I = 6 amperes. If you then wanted to figure out the resistance in the circuit, you can now plug these numbers into the first equation: 12 = 6 * R, or R = 12/6, which works out to R = 2 ohms. Using two variables, you’re able to figure out everything happening along those wires!

Now let’s say you wanted to use a battery system supplying 24 volts instead. Simply substitute and do the above calculations in the same way. What did you get? Here’s the breakdown. Since E = 24 volts now, and P = 72 watts, then you should arrive at I = 3 amperes via 72 = I * 24, or I = 72/24, or 3! From there you can figure out the resistance of this circuit: 24 = 3 * R, and therefore R = 8 ohms. There is more resistance along this circuit than along the circuit running on 12 Volts, and less current.

Okay, so that’s Ohm’s Law (of DC circuits) for you! I’m a student too, so this is about as deep as I can go into the subject right now, but expect more detailed explanations and the reasons for knowing all this junk soon. Stay tuned! In the meantime, if you want to read more about Ohm’s Law and why it’s important, read the Wikipedia article here.

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