These equations can be written as 2l2+4I3=6 Solving these equations using any convenient method yields I1=1 A 1=8A, and 1, =11/2 A Any voltage in the network can now be easily calculated,e.g, V2=(I2-Iy)(2)=5V and V3=I(2)=11 V. Just as in the node analysis discussion, we now expand our capa bilities by considering circuits which contain current sources. In this case, we will show that for mesh analysis, the presence of current 4A sources makes the solution easier The network in Fig. 3.9 has four meshes which are labeled I1, I2, I3, and I4. However, since two of these currents, i.e., I, and L4, pass directly through a current source, two of the four linearly independent equa- FIGURE 3.9 A four-mesh network con- ons required to solve the network are 3=4 The two remaining KVL equations for the meshes defined by I and I2 are (1-12)(1)-(l1-l3)(2)=0 (l2-1)(1)-l2(2)-(l2-I4)(1)=0 Solving these equations for I and I, yields I,=54/11 A and 12=8/11 A. A quick check will show that KCL satisfied at every node. Furthermore, we can calculate any node voltage in the network. For example, V=(I )(1)=6 V and v=V3+(1-l)(1)=112/11V Both node analysis and mesh analysis have been presented and dis- cussed. Although the methods have been presented within the frame work of dc circuits with only independent sources, the techniqu are applicable to ac analysis and circuits containing dependent To illustrate the applicability of the two techniques to ac circuit1n analysis, consider the network in Fig. 3.10. All voltages and currents are phasors and the impedance of each passive element is known. In the node analysis case, the voltage V, is known and the voltage between V2 and V, is constrained. Therefore, two of the four required FIGURE 3. 10 A network containing five equatons are V4=12 V2+6/0°=V3 KCL for the node labeled Vi and the supernode containing the nodes labeled V2 and V3 c 2000 by CRC Press LLC
© 2000 by CRC Press LLC These equations can be written as 2I1 – I2 = –6 –I12 + 3I2 – 2I3 = 12 – 2I2 + 4I3 = 6 Solving these equations using any convenient method yields I1 = 1 A, I2 = 8 A, and I3 = 11/2 A. Any voltage in the network can now be easily calculated, e.g., V2 = (I2 – I3)(2) = 5 V and V3 = I3(2) = 11 V. Just as in the node analysis discussion, we now expand our capabilities by considering circuits which contain current sources. In this case, we will show that for mesh analysis, the presence of current sources makes the solution easier. The network in Fig. 3.9 has four meshes which are labeled I1, I2, I3, and I4. However, since two of these currents, i.e., I3 and I4, pass directly through a current source, two of the four linearly independent equations required to solve the network are I3 = 4 I4 = –2 The two remaining KVL equations for the meshes defined by I1 and I2 are +6 – (I1 – I2)(1) – (I1 – I3)(2) = 0 –(I2 – I1)(1) – I2(2) – (I2 – I4)(1) = 0 Solving these equations for I1 and I2 yields I1 = 54/11 A and I2 = 8/11 A. A quick check will show that KCL is satisfied at every node. Furthermore, we can calculate any node voltage in the network. For example, V3 = (I3 – I4)(1) = 6 V and V1 = V3 + (I1 – I2)(1) = 112/11 V. Summary Both node analysis and mesh analysis have been presented and discussed.Although the methods have been presented within the framework of dc circuits with only independent sources, the techniques are applicable to ac analysis and circuits containing dependent sources. To illustrate the applicability of the two techniques to ac circuit analysis, consider the network in Fig. 3.10. All voltages and currents are phasors and the impedance of each passive element is known. In the node analysis case, the voltage V4 is known and the voltage between V2 and V3 is constrained. Therefore, two of the four required equations are V4 = 12 /0° V2 + 6 /0° = V3 KCL for the node labeled V1 and the supernode containing the nodes labeled V2 and V3 is FIGURE 3.9 A four-mesh network containing current sources. FIGURE 3.10 A network containing five nodes and four meshes
V-V 2|0° 2 V-VV-V Solving these equations yields the remaining unknown node voltages 0.88=11.93/-4.22°V V2=3.66-j1.07=3.91-1634°V v3=9.66-j1.07=9.72/=634V lysis case, the currents I, and I, are constrained to be I1=20° I4-1 4/0 The two remaining KVL equations are obtained from the mesh defined by mesh current I2 and the loop which encompasses the meshes defined by mesh currents I, and L 2(l2-1)-(-1)l2-n2(l2-14) (1l3+6/0°-j2(14-12)-12/0。=0 Solving these equations yields the remaining unknown mesh currents I2=0.88/-634°A I4=1.13/235°A As a quick check we can use these currents to compute the node voltages. For example, if we calculate V2=-1(I3) V1=-l(12)+12/0° As a final point, because both node and mesh analysis will yield all currents and voltages in a network, which technique should be used? The answer to this question depends upon the network to be analyzed. If the network contains more voltage sources than current sources, node analysis might be the easier technique. If, however, the network contains more current sources than voltage sources, mesh analysis may be the easiest approach
© 2000 by CRC Press LLC Solving these equations yields the remaining unknown node voltages. V1 = 11.9 – j0.88 = 11.93 /– 4.22° V V2 = 3.66 – j1.07 = 3.91 /–16.34° V V3 = 9.66 – j1.07 = 9.72 /–6.34° V In the mesh analysis case, the currents I1 and I3 are constrained to be I1 = 2 /0° I4 – I3 = – 4 /0° The two remaining KVL equations are obtained from the mesh defined by mesh current I2 and the loop which encompasses the meshes defined by mesh currents I3 and I4. –2(I2 – I1) – (–j1)I2 – j2(I2 – I4) = 0 –(1I3 + 6 /0° – j2(I4 – I2) – 12 /0° = 0 Solving these equations yields the remaining unknown mesh currents I2 = 0.88 /–6.34° A I3 = 3.91 /163.66° A I4 = 1.13 /72.35° A As a quick check we can use these currents to compute the node voltages. For example, if we calculate V2 = –1(I3) and V1 = –j1(I2) + 12 /0° we obtain the answers computed earlier. As a final point, because both node and mesh analysis will yield all currents and voltages in a network, which technique should be used? The answer to this question depends upon the network to be analyzed. If the network contains more voltage sources than current sources, node analysis might be the easier technique. If, however, the network contains more current sources than voltage sources, mesh analysis may be the easiest approach. VV VV V VVVV 13 14 2 31 34 2 1 2 0 1 2 0 2 2 40 0 - + - - = ° + °+ - + - = °= j j / / /
ENGINE-STARTING DEVICE Charles F. Kettering Patented August 17, 1915 #】,150,523 E arly automobiles were all started with a crank, his backbreaking process was difficult for everyone, especially women. And it was dangerous. Backfires the person cranking. Numerous deaths and injuries were reported. gear, the car c ften resulted in broken wrists. Worse yet, if accidentally left in gear, the car could advance In 1910, Henry Leland, Cadillac Motors president, commissioned Charles Kettering and his Dayton Engineering Laboratories Company to develop an electric self-starter to replace the crank. Kettering had to overcome two large problems: (1)making a motor small enough to fit in a car yet powerful enough to crank the ne,and(2)finding a battery more powerful than any yet in existence. Electric Storage Battery of Philadelphia supplied an experimental 65-lb battery and Delco unveiled a working prototyp electric"self-starter"system installed in a 1912 Cadillac on February 17, 1911. Leland immediately ordered 12,000 units for Cadillac. Within a few years, almost all new cars were equipped with electric starters Copyright o 1995, DewRay Products, Inc. Used with permission. 53 6. 4 /62 04 4 c 2000 by CRC Press LLC
© 2000 by CRC Press LLC ENGINE-STARTING DEVICE Charles F. Kettering Patented August 17, 1915 #1,150,523 arly automobiles were all started with a crank, or arm-strong starters, as they were known. This backbreaking process was difficult for everyone, especially women.And it was dangerous. Backfires often resulted in broken wrists. Worse yet, if accidentally left in gear, the car could advance upon the person cranking. Numerous deaths and injuries were reported. In 1910, Henry Leland, Cadillac Motors president, commissioned Charles Kettering and his Dayton Engineering Laboratories Company to develop an electric self-starter to replace the crank. Kettering had to overcome two large problems: (1) making a motor small enough to fit in a car yet powerful enough to crank the engine, and (2) finding a battery more powerful than any yet in existence. Electric Storage Battery of Philadelphia supplied an experimental 65-lb battery and Delco unveiled a working prototype electric “self-starter” system installed in a 1912 Cadillac on February 17, 1911. Leland immediately ordered 12,000 units for Cadillac. Within a few years, almost all new cars were equipped with electric starters. (Copyright © 1995, DewRay Products, Inc. Used with permission.) E
Defining Terms ac: An abbreviation for alternating current. dc: An abbreviation for direct current Kirchhoff's current law(KCL): This law states that the algebraic sum of the currents either entering or leaving a node must be zero. Alternatively, the law states that the sum of the currents entering a node must be equal to the sum of the currents leaving that node. Kirchhoffs voltage law(KVL): This law states that the algebraic sum of the voltages around any loop is zero. A loop is any closed path through the circuit in which no node is encountered more than once Mesh analysis: A circuit analysis technique in which KVL is used to determine the mesh currents in a network. A mesh is a loop that does not contain any loops within it. Node analysis: A circuit analysis technique in which KCL is used to determine the node voltages in a network. Ohms law: A fundamental law which states that the voltage across a resistance is directly proportional to the g through it. Reference node: One node in a network that is selected to be a common point, and all other node voltages are measured with respect to that point. Supernode: A cluster of node, interconnected with voltage sources, such that the voltage between any two nodes in the group is knor Related Topics 3. 1 Voltage and Current Laws.3.6 Graph Theory erent J.D. Irwin, Basic Engineering Circuit Analysis, 5th ed, Upper Saddle River, N J: Prentice-Hall, 1996 3.3 Network Theorems auS Linearity and Superposition Linearity and observe that the system has an input designated by e(for excitation) and an output designated by r(for response). The system is considered to be linear if it satisfies the homogeneity and superposition condition The homogeneity condition: If an arbitrary input to the system, e, causes a response, t, then if ce is the input, the output is cr where c is some arbitrary constant. The superposition condition: If the input to the system, ej, causes a System esponse, r, and if an input to the system, e,, causes a response, T2, then a response,r, r2, will occur when the input is 4, +e If neither the homogeneity condition nor the superposition condi FIGURE 3.11 A simple system. tion is satisfied, the system is said to be nonlinear. The Superposition Theorem While both the homogeneity and superposition conditions are necessary for linearity, the superposition con- dition, in itself, provides the basis for the superposition theorem If cause and effect are linearly related, the total effect due to several causes acting simultaneously is equal to the sum of the individual effects due to each of the causes acting one at a time. c 2000 by CRC Press LLC
© 2000 by CRC Press LLC Defining Terms ac: An abbreviation for alternating current. dc: An abbreviation for direct current. Kirchhoff’s current law (KCL): This law states that the algebraic sum of the currents either entering or leaving a node must be zero. Alternatively, the law states that the sum of the currents entering a node must be equal to the sum of the currents leaving that node. Kirchhoff’s voltage law (KVL): This law states that the algebraic sum of the voltages around any loop is zero. A loop is any closed path through the circuit in which no node is encountered more than once. Mesh analysis: A circuit analysis technique in which KVL is used to determine the mesh currents in a network. A mesh is a loop that does not contain any loops within it. Node analysis: A circuit analysis technique in which KCL is used to determine the node voltages in a network. Ohm’s law: A fundamental law which states that the voltage across a resistance is directly proportional to the current flowing through it. Reference node: One node in a network that is selected to be a common point, and all other node voltages are measured with respect to that point. Supernode: A cluster of node, interconnected with voltage sources, such that the voltage between any two nodes in the group is known. Related Topics 3.1 Voltage and Current Laws • 3.6 Graph Theory Reference J.D. Irwin, Basic Engineering Circuit Analysis, 5th ed., Upper Saddle River, N.J.: Prentice-Hall, 1996. 3.3 Network Theorems Allan D. Kraus Linearity and Superposition Linearity Consider a system (which may consist of a single network element) represented by a block, as shown in Fig. 3.11, and observe that the system has an input designated by e (for excitation) and an output designated by r (for response). The system is considered to be linear if it satisfies the homogeneity and superposition conditions. The homogeneity condition: If an arbitrary input to the system, e, causes a response, r, then if ce is the input, the output is cr where c is some arbitrary constant. The superposition condition: If the input to the system, e1, causes a response, r1, and if an input to the system, e2, causes a response, r2, then a response, r1 + r2, will occur when the input is e1 + e2. If neither the homogeneity condition nor the superposition condition is satisfied, the system is said to be nonlinear. The Superposition Theorem While both the homogeneity and superposition conditions are necessary for linearity, the superposition condition, in itself, provides the basis for the superposition theorem: If cause and effect are linearly related, the total effect due to several causes acting simultaneously is equal to the sum of the individual effects due to each of the causes acting one at a time. FIGURE 3.11 A simple system
240+j0V 409 40+j0 j40Ω 409 FIGURE 3.12 (a)A network to be solved by using superposition; (b)the network with the current source nulled; and (c)the network with the voltage source nulle theorem. thea Consider the network driven by a current source at the left and a voltage source at the top, as 3. 12(a). The current phasor indicated by I is to be determined According to the superposition current I will be the sum of the two current components Iy due to the voltage source acting alone Fig. 3. 12(b)and Ic due to the current source acting alone shown in Fig 3. 12(c) Iv+Ic Figures 3.12(b)and(c)follow from the methods of removing the effects of independent voltage and current sources. Voltage sources are nulled in a network by replacing them with short circuits and current sources are nulled in a network by replacing them with open circuits. The networks displayed in Figs. 3. 12(b)and(c)are simple ladder networks in the phasor domain, and the strategy is to first determine the equivalent impedances presented to the voltage and current sources. In Fig 3. 12(b), the group of three impedances to the right of the voltage source are in series-parallel and posses an lI c 2000 by CRC Press LLC
© 2000 by CRC Press LLC Example 3.1. Consider the network driven by a current source at the left and a voltage source at the top, as shown in Fig. 3.12(a). The current phasor indicated by ^ I is to be determined. According to the superposition theorem, the current ^ I will be the sum of the two current components ^ IV due to the voltage source acting alone as shown in Fig. 3.12(b) and ^ IC due to the current source acting alone shown in Fig. 3.12(c). Figures 3.12(b) and (c) follow from the methods of removing the effects of independent voltage and current sources. Voltage sources are nulled in a network by replacing them with short circuits and current sources are nulled in a network by replacing them with open circuits. The networks displayed in Figs. 3.12(b) and (c) are simple ladder networks in the phasor domain, and the strategy is to first determine the equivalent impedances presented to the voltage and current sources. In Fig. 3.12(b), the group of three impedances to the right of the voltage source are in series-parallel and possess an impedance of FIGURE 3.12 (a) A network to be solved by using superposition; (b) the network with the current source nulled; and (c) the network with the voltage source nulled. ˆ ˆ ˆ I I I V C = +