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

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1 Response so far »

  1. 1

    robert mac kay said,

    thank you for allowing people such as myself to log on to your site. I’m taking an electrical class at delta college, and am having problems wrapping my brain around parallel and combo circuits, and the way to work a circuit problem step by step.


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