m:i ELECTRIC CIRCUITS

Transcription

1 Ohm's Law OBJECTIVES After completing this chapter, the student should be able to: Identify the three basic parts of a circuit. Identify three types of circuit configurations. Describe how current flow can be varied in a circuit. State Ohm's law with reference to current, voltage, and resistance. Solve problems using Ohm's law for current, resistance, or voltage in series, parallel, and seriesparallel circuits. Describe how the total current flow differs between series and parallel circuits. Describe how the total voltage drop differs between series and parallel circuits. Describe how the total resistance differs between series and parallel circuits. State and apply Kirchhoff's current and voltage laws. Verify answers using Ohm's law with Kirchhoff's law. See accompanying CD for interactive presentations, tutorials and Ohm's Law examples in MultiSim, Chapter 5. Ohm's law defines the relationship among three fundamental quantities: current, voltage, and resistance. It states that current is directly proportional to voltage and inversely proportional to resistance. This chapter examines Ohm's law and how it is applied to a circuit. Some of the concepts are introduced in previous chapters. m:i ELECTRIC CIRCUITS As stated earlier, current flows from a point with an excess of electrons to a point with a deficiency of electrons. The path that the current follows is called an electric circuit. All electric circuits consist of a voltage source, a load, and a conductor. The voltage source establishes a difference of potential 49

2 SECTION 1 DC CIRCUITS that forces the current to flow. The source can be a battery, a generator, or another of the devices described in Chapter 3, Voltage. The load consists of some type of resistance to current flow. The resistance may be high or low depending on the purpose of the circuit. The current in the circuit flows through a conductor from the source to the load. The conductor must give up electrons easily. Copper is used for most conductors. The path the electric current takes to the load may be through any of three types of circuits: a series circuit, a parallel circuit, or a seriesparallel circuit. A series circuit (Figure 51) offers a single continuous path for the current flow, going from the source to the load. A parallel circuit (Figure 52) offers more than one path for current flow. It allows the source to apply voltage to more than one load. It also allows several sources to be connected to a FIGURE 51 A series circuit offers a single path for current flow. FIGURE 53 A seriesparallel circuit is a combination of a series circuit and a parallel circuit. FIGURE 54 L.J R 3 Current flow in an electric circuit flows from the negative side of the voltage source, through the load, and returns to the voltage source through the positive terminal. 1 I LOAD I E T = FIGURE 55 A closed circuit supports current flow. FIGURE 52 A parallel circuit offers more than one path for current flow. ~ R, = ~ R2 ~ J it L J J single load. A seriesparallel circuit (Figure 53) is a combination of the series and parallel circuit. Current in an electric circuit flows from the negative side of the voltage source through the load to the positive side of the voltage source (Figure 54). As long as the path is not broken, it is a closed circuit and current flows (Figure 55). However, if the path is broken, it is an open circuit and no current can flow (Figure 56).

3 CHAPTER 5 OHM'S LAW FIGURE 56 An open circuit does not support current flow. FIGURE 58 Current flow in an electric circuit can also be changed by varying the resistance in the circuit. tr lr I! It FIGURE 57 Current flow in an electric circuit can be changed by varying the voltage. IE!E It I! The current flow in an electric circuit can be varied by changing either the voltage applied to the circuit or the resistance in the circuit. The current changes in exact proportions to the change in the voltage or resistance. If the voltage is increased, the current also increases. If the voltage is decreased, the current also decreases (Figure 57). On the other hand, if the resistance is increased, the current decreases (Figure 58). This relationship of voltage, current, and resistance is called Ohm's law. 51 QUESTIONS 1. What are the three basic parts of an electric circuit? 2. Define the following: a. Series circuit b. Parallel circuit c. Seriesparallel circuit 3. Draw a diagram of a circuit showing how current would flow through the circuit. (Use arrows to indicate current flow.) 4. What is the difference between an open circuit and a closed circuit? 5. What happens to the current in an electric circuit when the voltage is increased? When it is decreased? When the resistance is increased? When it is decreased? DB OHM'S LAW Ohm's law, or the relationship among current, voltage, and resistance, was first observed by George Ohm in Ohm's law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in a circuit. This may be expressed as: or where: voltage current = resistance E 1= R I = current in amperes E = voltage in volts R = resistance in ohms

4 SECTION 1 DC CIRCUITS FIGURE 5 9 FIGURE 511 R,1 "Of E i L IT=2A R'?f 1~OV FIGURE 510 ET IT = RT ET 0.02 = 1200 ET = 24 volts Whenever two of the three quantities are known, the third quantity can always be determined. EXAMPLE: How much current flows in the circuit shown in Figure 59? ET = 12 volts ~ = 1000 ohms EXAMPLE: What resistance value is needed for the circuit shown in Figure 511 to draw 2 amperes of current? ~ = 2 amps ET = 120 volts ET IT = RT 120 2= RT 12 IT = 1000 ~ = amp or 12 milliamps EXAMPLE: In the circuit shown in Figure 510, how much voltage is required to produce 20 milliamps of current flow? ~ = 20 rna = 0.02 amp E =? T. ~ = 1.2 kd.= ohms 60 ohms = R, 52 QUESTIONS 1. State Ohm's law as a formula. 2. How much current flows in a circuit with 12 volts applied and a resistance of 2400 ohms? 3. How much resistance is required to limit current flow to 20 milliamperes with 24 volts applied?

5 CHAPTER 5 OHM'S LAW 4:. How much voltage is needed to produce 3 amperes of current flow through a resistance of 100 ohms? FIGURE 513 In a parallel circuit, the current divides among the branches of the circuit and recombines on returning to the voltage source. DIll APPLICATION OF OHM'S LAW In a series circuit (Figure 512), the same current flows throughout the circuit. IT = IR\ = IR2 = IR,... = IRn The total voltage in a series circuit is equal to the voltage drop across the individual loads (resistance) in the circuit. ET = ER\ +ER2 +ER,... +ERn The total resistance in a series circuit is equal to the sum of the individual resistances in the circuit. Ry = R] +R2 +R3... +Rn In a parallel circuit (Figure 513), the same voltage is applied to each branch in the circuit. ET = ER\ = ER2 = ER,... = ERn The total current in a parallel circuit is equal to the sum of the individual branch currents in the circuit. FIGURE 512 In a series circuit, the current flow is the same throughout the circuit.... The reciprocal of the total resistance is equal to the sum of the reciprocals of the individual branch resistances. 1 1 III = RT R] R2 R3 Rn The total resistance in a parallel circuit will always be smaller than the smallest branch resistance. Ohm's law states that the current in a circuit (series, parallel, or seriesparallel) is directly proportional to the voltage and inversely proportional to the resistance. E 1= R In determining unknown quantities in a circuit, follow these steps: 1. Draw a schematic of the circuit and label all known quantities. 2. Solve for equivalent circuits and redraw the circuit. 3. Solve for the unknown quantities. E = T IT... R3 T'T J R. ~ 24 NOTE: OHM'S LAW IS TRUE FOR ANY POINT IN A CIRCUIT AND CAN BE APPLIED AT ANY TIME. THE SAME CURRENT FLOWS THROUGHOUT A SERIES CIRCUIT AND THE SAME VOLTAGE IS PRESENT AT ANY BRANCH OF A PARALLEL CIRCUIT

6 SECTION 1 DC CIRCUITS FIGURE 514 FIGURE 516 R1 = 560 Q R1 = 1.2 k.q =E T =12V FIGURE 515 FIGURE 517 RT = 2240 Q EXAMPLE: What is the total current flow in the circuit shown in Figure 514? ET = 12 volts R] = 560 ohms R2 = 680 ohms R3 = I kg = 1000 ohms First solve for the total resistance of the circuit: ~ = R] + R2 + R3 ~ = R, = 2240 ohms Draw an equivalent circuit. See Figure 515. Now solve for the total current flow: ET IT = RT 12 IT = 2240 ~ = amp or 5.4 milliamp EXAMPLE: How much voltage is dropped across resistor R2 in the circuit shown in Figure 516? ET = 48 volts R] = 1.2 kd = 1200 ohms R2 = 3.9 kd = 3900 ohms R3 = 5.6 kd = 5600 ohms First solve for the total circuit resistance: ~ = R] + R2 + R3 R, = R, = 10,700 ohms Draw the equivalent circuit. See Figure 517. Solve for the total current in the circuit: ET IT = RT 48 I T 10,700 ~ = amp or 4.5 milliamps Remember, in a series circuit, the same current flows throughout the circuit. Therefore, IR, = IT' ER, IR =, R2 ER = ' 3900 E R, = volts

7 CHAPTER 5 OHM'S LAW FIGURE 518 gt= 120V TL. I T = 200 ma.. R? 2 R3= 5.6 kq EXAMPLE: What is the value of R2 in the circuit shown in Figure 518? First solve for the current that flows through R[ and R 3. Because the voltage is the same in each branch of a parallel circuit, each branch voltage is equal to the source voltage of 120 volts. ER IR! =~ Resistor R2 can now be determined using Ohm's law. IR2= amp ER2= 120 volts R =? 2. ER! = 120 volts R j = 1000 ohms 120 I = R! 1000 IR! = 0.12 amp = R2 ER3= 120 volts R3 = 5600 ohms ER3 IR3=~ 120 I R IR3= amp In a parallel circuit, the total current is equal to the sum of the currents in the branch currents. Ir = amp IR! = amp IR2=? IR3= amp IT = IR, + IR2+ 1R = IR = IR = IR2 R2 = ohms EXAMPLE: What is the current through R3 in the circuit shown in Figure 519? First determine the equivalent resistance (R A ) for resistors R[ and R 2. R =? A. R[ = 1000 ohms R2 = 2000 ohms =+ RA RA = ohms (adding fractions requires a common denominator) Then determine the equivalent resistance (R B ) for resistors R4' R5' and R 6 First, find the total series resistance (Rs) for resistors R5 and R 6.

8 SECTION 1 DC CIRCUITS FIGURE R1 = 1 k.q...= R2 = 2 k.q R4 = 4.7 kq... R3 = 5.6 k.q ~ R =? 5. Rs = 1500 ohms R6= 3300 ohms R =? B. R4 = 4700 ohms Rs = 4800 ohms..... ' ,.. R5 = 1.5 k.q Rs = x, + R6 Rs = Rs = 4800 ohms 111 RB R4 s, 1 1 = 1 + (In this case, a common denominator would be too RB large. Use decimals!) RB = ohms Redraw the equivalent circuit substituting RA and RB' and find the total series resistance of the equivalent circuit. See Figure 520. RA = ohms R3 = 5600 ohms RB = ohms R6 = 3.3 kq ~ = RA + R3 + RB R, = ~ = ohms Now solve for the total current through the equivalent circuit using Ohm's law. 4=? ET = 120 volts RT = ohms ET IT = RT 120 IT = = amp or 13.9 milliamps In a series circuit, the same current flows throughout the circuit. Therefore, the current flowing through R3 is equal to the total current in the circuit. IR,= 4 IR3= 13.9 milliamps FIGURE QUESTIONS RA = Q RB = Q R3 = 5.6 kq 1. State the formulas necessary for determining total current in series and parallel circuits when the current flow

9 CHAPTER 5 OHM'S LAW FIGURE R1 = 500 Q. I...: through the individual components is known. 2. State the formulas necessary for determining total voltage in series and parallel circuits when the individual voltage drops are known. 3. State the formulas for determining total resistance in series and parallel circuits when the individual resistances are known. 4. State the formula to solve for total current, voltage, or resistance in a series 01' parallel circuit when at least two of the three values (current, voltage, and resistance) are known. 5. What is the total circuit current in Figure 52l? 4 =? Er = 12 volts R] = 500 ohms R2 = 1200 ohms R3 = 2200 ohms 11II KIRCHHOFF'S CURRENT LAW In 1847 G. R. Kirchhoff extended Ohm's law with two important statements that are referred to as Kirchhoff's laws. The first lawknown as Kirchhoff's current lawstates: T FIGURE 522 r,b r tl = ~ R1 R2 ~2t..2J : : 11 L J A The algebraic sum of all the currents entering and leaving a junction is equal to zero. Another way of stating Kirchhoff's current law is: The total current flowing into a junction is equal to the sum of the current flowing out of that junction. A junction is defined as any point of a circuit at which two or more current paths meet. In a parallel circuit, the junction is where the parallel branches of the circuit connect. In Figure 522, point A is one junction and point B is the second junction. Following the current in the circuit, 4 flows from the voltage source into the junction at point A. There the current splits among the three branches as shown. Each of the three branch currents (II' 12' and 13) flows out of junction A. According to Kirchhoff's current law, which states that the total current into a junction is equal to the total current out of the junction, the current can be stated as: 4 = I] Following the current through each of the three branches finds them coming back together at point B. Currents 1],12' and 13 flow into junction B, and 4 flows out. Kirchhoff's current law formula at this junction is the same as at junction A: I] I) = 4

10 . SECTION 1 DC CIRCUITS FIGURE = 3mA _ 54 QUESTIONS 1. State Kirchhoff's current law. 2. A total of 3 A flows into a junction of three parallel branches. What is the sum of the three branch currents? 3. If 1 ma and 5 ma of current flow into a junction, what is the amount of current flowing out of the junction? 4. In a parallel circuit with two branches, one branch has 2 IDA of current flowing through it. The total current is 5 ma. What is the current through the other branch? 5. Refer to Figure 523. What are the values of 12 and I3? KIRCHHOFF'S VOLTAGE LAW Kirchhoff's second law is referred to as Kirchhoff's voltage law, and it states: The algebraic sum of all the voltages around a closed circuit equals zero. Another way of stating Kirchhoff's voltage law is: The sum of all the voltage drops in a closed circuit will equal the voltage source. FIGURE 524 In Figure 524 there are three voltage drops and one voltage source (voltage rise) in the circuit. If the voltages are summed around the circuit as shown, they equal zero. ET E j E2 E3 = a Notice that the voltage source (E ) T has a sign opposite that of the voltage drops. Therefore the algebraic sum equals zero. Looking at this another way, the sum of all the voltage drops will equal the voltage source. ET = E j + E2 + E3 Both of the formulas shown are stating the same thing and are equivalent ways of expressing Kirchhoff's voltage law. The key to remember is that the voltage source's polarity in the circuit is opposite to that of the voltage drops.

11 CHAPTER 5 OHM'S LAW FIGURE 525 ET =? 1V R1 Rs 4V 55 QUESTIONS 2V R2 R4 3V R3 1. State Kirchhoff's voltage law two different ways. 2. A series resistive circuit is connected to a 12volt voltage source. What is the total voltage drop in the circuit? 3. A series circuit has two identical resistors connected in series with a 9volt battery. What is the voltage drop across each resistor? 4. A series circuit is connected to a 12volt voltage source with three resistors. One resistor drops 3 V and another resistor drops 5 V.What is the voltage drop across the third resistor? 2V 5. Refer to Figure 525. What is the total voltage applied to the circuit? SUMMARY An electric circuit consists of a voltage source, a load, and a conductor. The current path in an electric circuit can be series, parallel, or seriesparallel. A series circuit offers only one path for current to flow. A parallel circuit offers several paths for the flow of current. A seriesparallel circuit provides a combination of series and parallel paths for the flow of current. Current flows from the negative side of the voltage source through the load to the positive side of the voltage source. Current flow in an electric circuit can be varied by changing either the voltage or the resistance. The relationship of current, voltage, and resistance is given by Ohm's law. Ohm's law states that the current in an electric circuit is directly proportional to the voltage applied and inversely proportional to the resistance in the circuit. E 1= R Ohm's law applies to all series, parallel, and seriesparallel circuits. To determine unknown quantities in a circuit: Draw a schematic of the circuit and label all quantities. Solve for equivalent circuits and redraw the circuit. Solve for all unknown quantities. Kirchhoff's current law: The algebraic sum of all the currents entering and leaving a junction is equal to zero; it may be restated as the total current flowing into a junction is equal to the sum of the current flowing out of that junction. Kirchhoff's voltage law: The algebraic sum of all the voltages around a closed circuit equals zero; it may be restated as the sum of all the voltage drops in a closed circuit will equal the voltage source.

12 SECTION 1 DC CIRCUITS Using Ohm's law, find the unknown value for the following: 1. I =? E = 9 V R = 4500 ohms 2. I = 250 IDA E =? R = 470 ohms 3. I = 10 A E = 240 V R =? 4. Find the current and voltage drop through each component for the circuits shown below. 5. Use Kirchhoff's laws to verify answers for question 4. (A) (B) R, =50Q R, = 150 Q R2 = 300 Q (C) 75Q 75Q 75Q