Solar Power Class: Combined Circuits

The next installment of our electrical theory class is about combined circuits. Though studying all the simple circuits of the past few posts (see Ohm’s Law, Kirchoff’s Laws the Formulae Roundup if you’d like a review) is well worth the effort so that you understand the reasons why your power system is or is not working, most often in a power generation setting, you’ll be dealing with things hooked up in combined circuits, which means in combinations of series and parallel that produce the required output wattage to run a house circuit.

So how do you deal with the laws we just learned when you see a circuit that doesn’t seem to fit either category?

Solving Combo Circuits:

For the above example, can you show the remaining measurements for each resistor, and also calculate the total resistance for the circuit? To do this, we must reduce the circuit to a simple circuit of only one type. Before we get started, let me share one more little formula with you. Don’t worry, this one is designed to make your life easier. In a parallel circuit with only two resistors, instead of solving for total resistance in the usual way, you can find it using the following formula, also known as the “product over sum” method:
Rt = (R1 * R2) / (R1 + R2)

First, let’s work on the total resistance measurement. For a combination circuit, you use all the knowledge you’ve been storing up about simple series and parallel circuitry, and simply solve from the outside in, treating each circuit as though it is independent of the rest.

The outermost two resistors are hooked up in parallel (R5 and R6). Because there are only two resistors in this circuit, we can solve for the “equivalent” resistance by using the formula above. Therefore, R5-6 = 30*60 / 30 + 60 = 20Ω.

The next step “into” the center of the circuit is a series circuit composed of R3, R4, and the equivalent value for R5-6. Since in series circuits, we can simply add the partial resistnaces to get the total, we arrive at: 20 + 10 + 20 = 50Ω

This leaves us with a simple series circuit with three resistors – R1, R2, and the equivalent of R4-5-6. Again, we have a series circuit, so we simply add up the partial resistances to arrive at our final Total Resistance for the entire combination circuit. Therefore, 50 + 30 + 50 = 130Ω = Rt

Now that we know the total resistance, finding the total current through the circuit is as easy as using Ohm’s Law: E = I * R. Therefore, 240w = I * 130Ω, and I = 1.846 amps. Now we can use Ohm’s Law’s other formula to figure out the total power. P = I * E = 1.846 * 240 = 443.1 watts.

Now we know all of the totals for the circuit. They are as follows:
Rt = 130Ω
Et = 240v
It = 1.846 A
Pt = 443.1w.

From here you could go on to solve the current or voltage drops at any particular point along the circuits, or similarly figure out the work done at any point along the circuit’s path.

There you have it: you can now solve practically any circuit you’ll encounter.  And remember, practice makes perfect, so keep doing examples until you feel thoroughly comfortable with all these past few lessons.  If you’d like another explanation of combination circuits, check out the links below:

Physics Classroom Tutorial

Understanding and Calculating Combination Circuits

Electrical Engineering Training Series

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~