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1 Pertemuan 3 Basic Nodal and Mesh Analysis Matakuliah: H0042/Teori Rangkaian Listrik Tahun: 2005 Versi: <

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Presentasi berjudul: "1 Pertemuan 3 Basic Nodal and Mesh Analysis Matakuliah: H0042/Teori Rangkaian Listrik Tahun: 2005 Versi: <

1 1 Pertemuan 3 Basic Nodal and Mesh Analysis Matakuliah: H0042/Teori Rangkaian Listrik Tahun: 2005 Versi: <

2 2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Merumuskan berdasarkan metode simpul (node) dan mesh (loop) Menghitung variabel V dan I pada elemen Resistor.

3 3 Outline Materi Materi 1: identifikasi node di RL Materi 2: menganalisa tegangan yang diserap oleh beban dengan metode analisa node. Materi 3 : identifikasi mesh di RL Materi 4 : menghitung arus yang diserap oleh beban dengan metode analisa mesh. Materi 5 : aplikasi analisa RL dengan metode node dan mesh.

4 4 Chapter 4 Basic Nodal and Mesh Analysis Engineering Circuit Analysis Sixth Edition W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved. Fig. 4.1 “Obtain values for the unknown voltages …” Fig. 4.5(a) The circuit of Example 4.2 with a 22-V source... Fig. 4.7 “Determine the node-to-reference voltages …” Fig. 4.9 Examples of planar and nonplanar networks... Fig (a) The set of branches identified by the heavy lines… Fig “Determine the two mesh currents, i 1 and i 2, in …” Fig “Find the three mesh currents in the circuit below.” Fig Circuit from Practice Problem 4.8.

5 5 Fig. 4.1 (a) A simple three-node circuit. (b) Redrawn circuit to emphasize nodes. (c) Reference node selected and voltages assigned. (d) Shorthand voltage references. If desired, an appropriate ground symbol may be substituted for “Ref.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Obtain values for the unknown voltages across the elements in the circuit below. At node 1 At node 2

6 6 Fig. 4.5 (a) The circuit of Example 4.2 with a 22-V source in place of the 7-  resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons. W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. (a) The circuit of Example 4.2 with a 22-V source in place of the 7-  resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons. At node 1: At the “supernode:”

7 7 Fig. 4.7 “Determine the node-to-reference voltages in the circuit below.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Determine the node-to-reference voltages in the circuit below.

8 8 W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Examples of planar and nonplanar networks; crossed wires without a solid dot are not in physical contact with each other.

9 9 W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. (a) The set of branches identified by the heavy lines is neither a path nor a loop. (b) The set of branches here is not a path, since it can be traversed only by passing through the central node twice. (c) This path is a loop but not a mesh, since it encloses other loops. (d) This path is also a loop but not a mesh. (e, f) Each of these paths is both a loop and a mesh.

10 10 Fig “Determine the two mesh currents, i 1 and i 2, in the circuit below.” W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Determine the two mesh currents, i 1 and i 2, in the circuit below. For the left-hand mesh, i ( i 1 - i 2 ) = 0 For the right-hand mesh, 3 ( i 2 - i 1 ) + 4 i = 0 Solving, we find that i 1 = 6 A and i 2 = 4 A. (The current flowing downward through the 3-  resistor is therefore i 1 - i 2 = 2 A. )

11 11 W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Find the three mesh currents in the circuit below. Creating a “supermesh” from meshes 1 and 3: ( i 1 - i 2 ) + 3 ( i 3 - i 2 ) + 1 i 3 = 0 [1] Around mesh 2: 1 ( i 2 - i 1 ) + 2 i ( i 2 - i 3 ) = 0[2] Rearranging, i i i 3 = 7[1] -i i i 3 = 0[2] i 1 - i 3 = 7[3] Solving, i 1 = 9 A, i 2 = 2.5 A, and i 3 = 2 A. Finally, we relate the currents in meshes 1 and 3: i 1 - i 3 = 7[3]

12 12 W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition. Copyright ©2002 McGraw-Hill. All rights reserved. Find the voltage v 3 in the circuit below.

13 13 RESUME Perhitungan arus beban RL disederhanakan dengan persamaan tegangan berdasarkan metode mesh. Perhitungan tegangan beban RL disederhanakan dengan persamaan arus berdasarkan metode node.


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