13.10 How Series and Parallel Circuits Differ

Transcription

1 13.10 How Series and Parallel Circuits Differ In Activity 13.2, you observed that when the two lamps were connected in series, the brightness of the lamps was less than when the lamps were connected in parallel. By using knowledge of voltage, current, and resistance, it is possible to explain these results. Let s start by examining loads in series. Loads in Series less resistance results in more current When you connect an energy source to a series circuit, a voltage or potential EJĒ FSFODF JT DSFBUFE BOE FMFDUSPOT CFHJO UP ë PX /P NBUUFS IPX NBOZ MPBET are connected in series, there is only one that the current can follow. Current through Loads in Series If you have a circuit with one load (Figure 1(a)), the total resistance of the circuit will be different than if you have two or more of those loads connected in series (Figure 1(b)). The have only one to follow and with two or more loads, they have more bumps to deal with along the way. Because of this, the current flowing through the circuit in Figure 1(b) will be less than the current flowing through the circuit in Figure 1(a). The next sample problem illustrates this result. SAMPLE PROBLEM 1 more resistance results in less current (b) Figure 1 From Ohm s relationship, the current in a circuit depends on the resistance in that circuit. Comparing the Current in Two Series Circuits The same type of lamp is used in two series circuits. The first circuit has two identical lamps (Figure 2). The second circuit has three identical lamps (Figure 3). The potential difference across the battery is 10 V. The circuit on the left has a total resistance of 10 Ω. The circuit on the right has a total resistance of 15 Ω. Use the total resistance given to calculate the current through each circuit. Figure 2 Figure 3 = 10 Ω V = 10 V electric current (II ) V I = Statement: 10 V I = 10 Ω I=1A = 15 Ω V = 10 V electric current (II ) V I = 10 V I = 15 Ω I = 0.7 A The current through each lamp in Figure 2 is 1 A, while in Figure 3 the current is about 0.7 A. A series circuit has four identical lamps. The potential difference of the energy source is 60 V. The total resistance of the lamps is 20 Ω. Calculate the current through each lamp. /& How Series and Parallel Circuits Differ 571

2 WRITING TIP Supporting the Main Idea A main idea, expressed clearly and directly, is a powerful communication tool. But often a main idea needs supporting details such as examples, reasons, or data to help readers understand what it means. For instance, when writing a critical analysis about voltage across loads in series, use an example like several lamps connected in series, or even a diagram, to help explain the main idea. DID YOU KNOW? Uneven Loads When the loads in a series circuit are not identical, the load with the greatest resistance causes the greatest decrease in electric potential energy for an electron passing through it. The reason is that the electron has to convert more electric potential energy into kinetic energy just to get through all the bumps of the load. Voltage across Loads in Series In Activity 13.2, you observed that the brightness of each lamp in a series circuit decreased as you connected more lamps. To understand why this happens, let s examine the electric potential energy of the circuit. The battery converts chemical energy into electric potential energy. This potential energy gets transferred to all the that leave the battery and creates the current flow in the circuit. As the move through the circuit, the electric potential energy gets converted into kinetic energy. Suppose the circuit has only one load a lamp. When an electron passes through the load, all the electric potential energy that the electron received from the battery gets converted into light and heat. After the electron completes one loop, it returns to the battery, only to be given more electric potential energy. This allows the electron to continue moving through the loop of the circuit over and over again. Since voltage is related to electric potential energy, the voltage drop across the one load will be the same as the voltage drop across the battery. When two identical loads such as two lamps are connected in series, an electron leaving the battery will have the same amount of electric potential energy as if there were only one load. The reason is that the battery can only supply so much potential energy. For the electron to go through each of the lamps, only half of the electric potential energy gets converted into light and heat. Since only half of the potential energy gets converted through each lamp, the voltage drop across each lamp is half the voltage drop across the battery. More identical lamps connected in series means less of an electron s electric potential energy gets converted into heat and light. The voltage drop across each load decreases. This result can be written mathematically: V load = V source where V load is the voltage drop across each identical load (in volts), V source is the voltage drop across the energy source (in volts), and is the total number of identical loads. Sample Problem 2 shows how to apply this relationship. SAMPLE PROBLEM 2 Calculating the Voltage in a Series Circuit A series circuit contains three identical lamps (Figure 4). The potential difference of the battery is 30 V. Calculate the potential difference across each lamp. V source = 30 V voltage drop across each lamp (V load ) V load = V source V load = 30 V 3 V load = 10 V Statement: The potential difference across each lamp is 10 V. Figure 4 A series circuit contains four identical lamps. The voltage across the energy source is 96 V. Calculate the voltage across each lamp. 572

3 Loads in Parallel When an energy source is connected to a parallel circuit, a potential difference is created across the two terminals of the energy source. The potential difference causes to flow. Unlike a series circuit, the have different (a) s to follow. The number of s depends on the number of loads connected in parallel. Current through Loads in Parallel some some the rest of the some of the other the rest of the In Figure 5(a), two identical loads are connected in parallel. Electrons leaving the battery have two possible s to follow. Since each of those s has the same load, the (b) current splits in two. Figure 5 The more loads that are connected in parallel, the more In Figure 5(b), three identical loads are connected in parallel. This time, leaving the battery have three s have to follow. possible s. Each of those s has the same load, so the current splits in three. The splitting of the current is exactly what happens when a river flows into two or more tributaries. The quantity of water that flows in each tributary is less than the quantity in the original river. This result can be written mathematically: I source I load = where I load is the current through each identical load (in amperes), Isource is the current coming out of the energy source, and is the total number of identical loads. Sample Problem 3 shows how to apply this relationship. SAMPLE PROBLEM 3 Calculating the Current in a Parallel Circuit The total resistance in the circuit in Figure 5(b) is 2 Ω. The potential difference of the battery is 18 V. Calculate the current through each lamp. = 2 Ω V = 18 V current through each lamp (I load) V (to calculate the current coming out of the energy source) and I source = I source I load = (to calculate the current through each identical load) 18 V I source = 2Ω I source = 9 A 9A I load = 3 I load = 3 A Statement: The current through each lamp is 3 A. A parallel circuit contains four identical lamps. The potential difference across the energy source is 48 V. The total resistance of the lamps is 12 Ω. Calculate the current through each lamp. /& How Series and Parallel Circuits Differ 573

4 WRITING TIP Conducting Research for Critical Analysis In a critical analysis of voltage across loads in parallel circuits, you might ask yourself how many appliances you can safely plug into an electrical outlet. You might conduct research to fi nd out that you can divide the number of watts of power on a circuit by the voltage in the house to determine the current used. You might conclude that if the current used is greater than the circuit can handle, the fuse will blow or the circuit breaker will trip. Voltage across Loads in Parallel An interesting difference between series and parallel circuits involves total resistance. In both parallel circuits shown in Figure 5 on the previous page, the that flow through any one branch go through only one load. The resistance that the in that branch experience will be less than if the loads were connected in series. This means that when you connect loads in parallel, the total resistance will be less than if the loads were connected in series. Another difference between series and parallel circuits involves the voltage drop across loads. As stated above, an electron that flows through any one branch of the parallel circuits in Figure 5 only passes through one load. All the electric potential energy that the electron receives from the battery gets converted into light and heat. Since voltage is related to electric potential energy, the voltage drop across each parallel load will be the same as the voltage drop across the battery. In other words, for any parallel circuit, the voltage drop across each parallel branch will be the same as the voltage drop across the energy source. That is why, in Activity 13.2, the brightness of the lamps did not change when you connected more lamps in parallel. Sample Problem 4 shows how to apply this result to a parallel circuit. SAMPLE PROBLEM 4 Calculating the Voltage in a Parallel Circuit A parallel circuit contains three identical lamps (Figure 6). The current coming out of the energy source is 2.5 A. The total resistance of the circuit is 6.0 Ω. Calculate the voltage across the energy source and across each lamp. I source = 2.5 A R T = 6.0 Ω voltage drop across the energy source ( V source ) and across each lamp ( V load ) V source = V load (loads are in parallel to energy source) V source = ( I source )( R T ) V source = (2.5 A)(6.0 Ω) V source = 15 V Figure 6 V load = 15 V Statement: The voltage across the energy source is 15 V, and the voltage across each lamp is also 15 V. A parallel circuit contains 10 identical lamps. The current through the energy source is 3.0 A. The total resistance of the circuit is 15 Ω. Calculate the voltage across the energy source and across each lamp. Resistance, Current, and Voltage in Circuits Table 1 summarizes how voltage, current, and total resistance differ in series and parallel circuits. Table 1 The Relationships of Loads in Series and Parallel Circuits Quantity Series circuits Parallel circuits total resistance of circuit (R T ) increases decreases current through loads (I load ) I source decreases as more loads are added I source splits among loads based on the number of branches in parallel voltage across loads (V load ) V source splits based on the number of loads voltage of each parallel branch is the same as V source 574

5 UNIT TASK You can apply what you have learned about series and parallel circuits to the Unit Task described on page 586. IN SUMMARY than if the loads were connected in series. lower the current will be in the circuit. lower the current will be through each branch. drop across the energy source is split equally among the loads. parallel branch will be the same as the voltage drop across the energy source. CHECK YOUR LEARNING 1. Which did you fi nd easier to understand in a parallel circuit, voltage or current? What would you ask your teacher to help you understand the more challenging concept? C 2. Compare the total resistance of loads connected in series with those loads connected in parallel. K/U 3. Why is it a bad idea to connect too many devices in parallel to an energy source, such as a wall outlet? K/U 4. (a) What would happen to the voltage drop across each lamp if you kept adding lamps to a series circuit? (b) What do you think you would observe in terms of the brightness of the lamps? K/U 5. The total resistance of the circuit in Figure 7 is 25 Ω. The voltage drop across the battery is 6.0 V. T/I (a) Calculate the current in the circuit. (b) Calculate the voltage drop across each lamp. 6. A house has a lamp in every room. The circuit for the lamps is shown in Figure 8. The voltage drop across the energy source is 120 V. The total resistance is 10 Ω. T/I (a) Calculate the current through each lamp. (b) Calculate the voltage drop across each lamp. Figure 8 7. A battery-powered set of fi ve patio lanterns is connected in series. An ammeter measures the current through the battery as 0.75 A. The total resistance of the circuit is 52 Ω. T/I (a) Calculate the voltage drop across the battery. (b) Calculate the voltage drop across each load. Figure How Series and Parallel Circuits Differ 575