DC Circuits–Series, Parallel, and Combination Circuits

Purpose

• •
  To investigate resistors wired in series and parallel as well as combinations of the two.
• •
  To examine how current behaves at junction points in a circuit and how its flow is influenced by circuit resistances and emfs.
• •
  To study how power is affected by current, voltage, and resistance.
• •
  To study the effect of the internal resistance of a battery on the power available to a circuit.
• •
  To study the behavior of series-parallel combinations of resistors and learn how to analyze them using equivalent resistance.

Equipment

Simulation and Tools

Open the DC Circuits Apparatus simulation to do this lab.

Explore the Apparatus

The large blue area with the small dots (circles) is a circuit board where we'll create our circuits. In the lab toolbox shown in Figure 1, we see our choices of resistors, batteries, switches, wires, voltmeters, ammeters, bulbs, and diodes. Each of the circuit elements can be dragged and dropped onto the circuit board. Give it a try. The current flows out of the positive end of the battery, through the switch, into the positive terminal of the ammeter, and then exits the negative terminal. It then enters a junction or node where most of the current flows through a resistor, but a tiny amount enters the positive terminal of the voltmeter, exits its negative terminal, and rejoins the main flow of the current at a second node. The full current then enters the negative end of the battery. Chemical forces in the battery then move the charges through to the positive terminal where the cycle begins again. (Remember this is a fictitious conventional current. The flow of electrons actually moves in the opposite direction around the circuit.) The meters record currents and voltage drops in the circuit. When you add them to a circuit, this actually makes a small change in the currents and voltages you're interested in measuring. This is the case with any measurement. When you put a cool thermometer into a beaker of hot water, heat flows from the water into the thermometer, thus changing the water's temperature. A good measuring tool reduces these influences as much as possible.

Series and Parallel Circuits

Consider the "life" of an electron in your car's electrical system. Each time it leaves the negative pole of your car battery, it has a bewildering variety of routes to choose from. Just in your radio alone, there are many routes it might take before it returns to the battery. This complex arrangement allows each component of the electrical system to get just the current it needs. The analysis of such complex systems is beyond the scope of an introductory physics class, but many of the principles involved can be discovered using simple batteries, resistors, bulbs, and meters. By observing the brightness (Power = IV) of a simple bulb, we can learn how current and power are distributed in a complex circuit. The first part of this lab is an exploration. Your goal is to observe and organize your observations into models of the behavior of simple circuits. If you're working with a team, be sure to take turns doing the wiring. Feel free to go off the path. When you do, just be careful to avoid situations where current can flow through a path with little resistance, that is, one where there is no light bulb. Also, save your batteries by opening the circuit whenever you don't really need to see the bulbs glow. Well, not really. This is a virtual apparatus. You can't hurt it. But in a real circuit, shorting out a battery means you'll have to buy a new real one.

Procedure

Please print the worksheet for this lab. You will need this sheet to record your data.

I. Explore Series, Parallel Circuits, and Combination Circuits

Let's explore and see what's ahead. The circuits you'll need, shown in Figure 4, are available in the "Pick a circuit" pull-down menu. Select "Four 3 Bulbs." The gap at the bottom of each circuit will be filled with a battery later.

A. Initial Observations

Let's make some observations. In what follows you'll be guided to make various observations, but you should be sure not to leave it at that. This apparatus provides you the opportunity to explore, develop your own models, and to do your own tests. After this activity, you should never look at a circuit diagram in a book or test and fail to "see" how it would behave. We can calculate the power dissipated in terms of the current through a bulb, I, the resistance of a bulb, R, and potential difference across a bulb, V. From the similarities and differences in the brightnesses of the bulbs, it would appear that there must be significant differences in the currents and voltages in these circuits. Since all four circuits are made of identical components, it appears that their arrangement is key to their electrical behavior.

B. Current

We'll first explore how the currents are determined by the circuit structure. We'll rely on the little moving current dots as an indicator of current flow. They're not perfect, but they do give a pretty good idea of what's happening. Their speed is proportional to the current through them. From equation 2

P = IV = I²R =

V 2

R

, specifically,

P = IV,

you should see the price you pay for the bright bulbs in our original bulb circuit (b). The larger current will discharge the battery more quickly. Since the battery chemistry limits the total amount of charge that it can provide, the larger current will use up this charge in a shorter time

(q = It)

in circuit (b). Hopefully you observed that the current is the same throughout circuit (a). So, This is the nature of a series circuit. While the ammeters are not directly measuring the current within the resistors, if the current flowing into one end of a resistor is the same as the current flowing out the other end it would be hard to explain how the current, the number charges per second, could be different within the resistor. Otherwise the law of conservation of charge would be violated. So the statement about the nature of the series circuit above seems to hold true for this series circuit. So the fundamental statement about the current in a circuit like (b) is This is the nature of a parallel circuit. You now know equations describing the current flowing in series and parallel circuits: where I_s or I_p is the current flowing into or out of the series or parallel section of the circuit and I_1–3 are the currents in the three resistors. You've also (hopefully) created Kirchhoff's Point Rule describing the current flow at a point in a circuit. We use the term "point" to refer to the collection of points along a conductor that are at the same potential. The wires in a typical circuit generally have resistances that are so small relative to the other circuit elements that we can assume a negligible voltage change as currents flow through them. So, in Figure 8a, all the wire between the + battery terminal and the left end of the left bulb is one point electrically. Likewise, the right side is another equipotential "point." In Figure 8b, the left vertical section of wire is one "point" and the right section is another "point." Figure 8 shows three equivalent versions of series circuit (a) and three for parallel circuit (b). Regardless of how they look, all three versions of each type of circuit are equivalent. It's just that the "points" are more or less elaborate in different versions. We generally draw circuits as simply as possible. There are a couple more observations we need to make about series and parallel circuits before we move on to voltages.

C. Voltage

Your results above should have confirmed that power increases with current. What about voltage? The "V" terms in equation 2

P = IV = I²R =

V 2

R

represent the voltage drop across a bulb or voltage increase across a battery. So we can now relate power dissipation by our resistors to voltage just as we did with current. We don't have anything corresponding to the moving current dots to visually represent voltage changes so we'll need to start right away with voltmeters. Unlike ammeters, voltmeters don't require that you open a circuit to add them. Since they're simpler to work with, we'll look at both series and parallel circuits at the same time. We now have current and voltage equations for series and parallel circuits. Notice that the equations are similar. The + and = signs are just swapped. If you look back at the behavior of the circuits, it should be very easy to see why they behave this way. This should also help you to remember these equations. How about circuits (c) and (d)? Neither of these is simply a series or parallel circuit. Each of these contains a section of its circuitry that is either series or parallel.

D. Resistance

In both the series and parallel circuits you've been working with, the individual resistances and battery voltages have been the same. But the result, the power dissipation by the individual resistors and bulbs, has been very different. Somehow the arrangement of the resistors in the two circuits has influenced the current flowing through them and the voltage drops across them. In circuits 11a and 11b, you have a series and a parallel circuit made with identical parts. In each case a battery maintains a 60-V potential difference across a group of three resistors. Clearly, the currents through the batteries are very different in these circuits. You might say that from the battery's point of view, there is effectively a different resistance in each circuit. We say that each circuit has a different equivalent resistance, R_eq. There are two ways of finding out the equivalent resistance of each circuit—experimentally and mathematically. We'll start with the experimental approach. Your experimental answer for the series circuit probably seems reasonable. Adding more resistors should add more total resistance. So, for a series circuit, the equivalent resistance would be But in the parallel circuit, it seems to work in the other direction. One favorite analogy is with grocery store checkout lines. The checkout is the resistance to the flow of shoppers. You never add more checkout lanes in series! Well, at certain times of the year—Girl Scout cookie sales, etc. But we do like to see more checkout lanes in parallel. Clearly, doubling the number of checkouts in parallel would halve the resistance to shopper flow. But why do you get exactly 10 Ω for your R_eq in parallel? If Circuits 11(b) and 12(b) are equivalent, the current through the battery, I_b, in each circuit must be the same. In Circuit 12(b) this current is given by From equation 4

I_p = I₁ + I₂ + I₃    Parallel Current

, we know that the current, I_b, in Circuit 11(b) must be From equation 6, we know that the voltage across each parallel resistor is V_b. So from Ohm's Law, we can write the currents in the individual branches as Thus, for Circuit 11(b) we have Combining our equations for I_b in Circuits 11(b) and 12(b), we have Dividing each term by V_b, we have so R_eq = 10 Ω as we found experimentally! We can now complete our chart of series and parallel relationships.

II. Case Studies with Series, Parallel, and Combination Circuits; Power

We'll now look at a few situations involving simple circuits.

A. Internal Resistance, Terminal Voltage, and "Dead Batteries"

Everyone is familiar with the gradual dimming of a flashlight over time. Is this some sort of design feature to alert you when it's time to replace the batteries, or recharge them if they're rechargeable? You've built the ideal flashlight. (If it was real, you'd be rich.) A real battery is made up of more than just a couple of electrodes and an electrolyte. As soon as it's made it begins to deteriorate, with new, undesirable materials created by the reactions that make the battery function. These materials generally resist the flow of current through the battery, providing internal resistance. When you turn on the flashlight the rate of buildup of these materials rapidly increases. Let's add some R_int to our flashlight. Current passing through the battery is increased in voltage by the value of the emf and decreased by an amount IR_int. The net voltage change is referred to as the terminal voltage, V_t, which is given by equation 9. The voltmeter in Figure 14 is measuring the terminal voltage. Note that the emf is fairly constant at 60 V, but the terminal voltage, which is what is available for use by the circuit, decreases as the current increases. A second resistance in the circuit is the bulb resistance. So our realistic flashlight behaves like an ideal battery in series with two resistors. But it's actually more complex than that, as you'll see. Let's see how the power provided by the battery is being used in the flashlight at this point. To find the power dissipated by the bulb and by the internal resistance, we need to measure the current, I, through the circuit, which is the same everywhere since everything is in series. Note the total change in voltage around the loop. As given by Kirchhoff's Voltage (Loop) Rule, the total change must be zero. For our circuit, Subtracting the battery's internal voltage drop, V_ir, reflects the internal voltage loss that's not available to the external circuit that the battery is powering. It's going to get worse! When we turned on our ideal flashlight, we "took a walk" and found that no harm was done. We all know that that's not how it works. What really happens is that you can't find your way back to the campsite, or you have to call Triple-A. Batteries have a bad habit of "running down."

B. Simplifying Circuits, Power Dissipation, Power Supplied by a Battery

In the circuit in figure 15, a 12-V battery supplies current to a network of four resistors. If this was a section of a circuit in your car hooked up to your car's 12-V battery, you might want to know what size fuse should be provided to protect it from overload. Or if the 1.0-Ω resistor represented a bulb, you might want to know the proper wattage of the bulb that should be used. Many students find solving this sort of problem perplexing. This is partly because it can be quite tedious and partly because without direct experience with circuits, it's difficult to see how to visualize a solution. We'll now look at two different circuits. We'll answer a few questions about each one, first using direct measurement from our lab simulations, and then using an algebraic technique that requires that you reduce the circuit step by step to its simplest form and then use these circuits to create the necessary equations to solve for unknowns. Circuit 1 Figure 15a will be our first circuit. The two tasks are listed below. Task 1: Determine the current supplied by the battery in order to select the minimum fuse current. Task 2: Determine the power dissipated by the 1.0-Ohm resistor. That was pretty simple, but we had to create a prototype in the lab and that's a bit inconvenient. How could we make the same determinations just from the circuit diagram given that none of the values used above are readily available in the circuit diagram? In some simple cases, the necessary values can be determined by reducing the circuit step-by-step and drawing successively simpler equivalent circuits. We do this by finding groups of resistors that are either in series or parallel and creating a new diagram with these groups replaced by their equivalent resistances. This is repeated until no further simplification is possible. It is essential that each of the diagrams is fully labeled. Many students find this process puzzling, given that the ultimate use of this collection of circuits is unclear. You can think of it as a process similar to writing down what you know and want to know before beginning to think about how to solve a homework problem. What you're doing is clarifying what you know at the start. Often, after writing down what you initially know, you can see that there are other quantities you can readily find from these initial values. The drawing of equivalent circuits is a similar technique. Let's try it. But first, let's recall what we mean by the term point in an electric circuit. We use the term "point" to refer to the collection of points along a conductor that are at the same potential. In Figure 15, everything beneath the lower ends of the battery and two resistors is a "point." This is because we assume that our wires have negligible resistance, so there would be no voltage changes as current flows from one point to another in this section of the circuit. We're interested in replacing groups of resistors that are either in series or parallel. Let's first remind ourselves of the meanings of these two terms. Two or more resistors between points (a) and (b) are in series if they are connected end to end with no branching along the way so that the current flows through each resistor in succession. The current is the same through each resistor, but the voltage drop across each depends on the current. Two or more resistors between points (a) and (b) are in parallel if they are connected to common end points (a) and (b) with each branch carrying a current that depends on its resistance. The total current between (a) and (b) equals the sum of the currents through all the branches. Geometrically, we might say that the 6.0-Ω and 1.0-Ω resistors are in parallel. However, while they do have one common end point, (d), on the other end they are connected to points (b) and (c) which are at opposite ends of the 2.0-Ω resistor. If the 6.0-Ω and 1.0-Ω resistors are in parallel, the potential difference across each, V_bd and V_cd, should be the same. Let's measure these voltages and see. The potential difference between (b) and (d) is not the same as between (c) and (d), so the 6.0-Ω and 1.0-Ω resistors are not in parallel. You would never actually have to make a measurement like this since points (b) and (c) would not be at the same potential if there was any kind of circuit element such as a battery or resistor between them. Similarly, the 4.0-Ω and 2.0-Ω resistors, while connected-to-end, are not in series because there is a branch between them at (b). So be careful in determining when resistors are in series or parallel. Notice how helpful it is to add these labeled nodes a–d. It's a good habit to develop. The 2.0-Ω and 1.0-Ω resistors in Figure 16 do meet our definition of series resistors; thus, we can use equation 7 to determine their equivalent resistance and draw a new circuit with just the one 3.0-Ω equivalent resistor. This resistor could be anywhere within the right section of the circuit. Be sure to include all labels and current arrows just as in the figure. In our new Figure 17, we can see another pair of resistors that can be reduced to one resistance. The 3.0-Ω and 6.0-Ω resistors have common end points at (b) and (d). Thus, they are in parallel. Note that all the points from the right end of the 3-Ω resistor down and around to the negative terminal of the battery are all at the same potential. They might all be represented by point (d). Wouldn't the 4.0-Ω and 6.0-Ω resistors be in parallel between points (b) and (d) too? No! The battery is in one of these branches. We must have only resistors in the parallel or series circuits that we want to simplify. We can create a new circuit in Figure 18 with one 2.0-Ω resistor between points (b) and (d) that is equivalent to the parallel 3-Ω and 6-Ω resistors. The 2.0-Ω value of that resistor is given by equation 8. Finally, we can see that the 4.0-Ω and 2.0-Ω resistors in Figure 18 are in series between points (a) and (d). Their equivalent resistance is

6.0 Ω

. This gives us our final equivalent circuit. Err. That was interesting but what were we doing? We were looking for a way to

• •
  determine the current supplied by the battery in order to select the proper fuse.
• •
  determine the power dissipated by the 1.0-Ohm resistor, P_{1.0 Ω}.

And we want to do this without having access to the actual circuit. That is, we'll have to derive the data we need from the four circuit diagrams we've created. Whew! That was a lot of work. Hopefully you're now a believer in carefully documenting what you know and want to know. Circuit 2 It's your turn now. You'll work with the circuit below in Figure 20. Your task is to determine the power dissipated by each of the two 4-Ω resistors. You'll find the power dissipated by the left one, P_{4 Ω left}, by measurements taken from the apparatus. You'll find P_{4 Ω right} using algebra. Determine the power dissipated by the 4-Ω resistor in the right-hand circuit, P_{4 Ω right}, using algebra. Show the following calculations. (Feel free to check your work using meters.)

Copyright © 2013-2014 Advanced Instructional Systems, Inc. and Kentucky Educational Television | Credits