Matakuliah
Tahun
Versi
: H0042/Teori Rangkaian Listrik
: 2005
: <<versi/01
Pertemuan 3
Basic Nodal and Mesh Analysis
1
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.
2
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.
3
Chapter 4 Basic Nodal and Mesh
Analysis
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. 4.10
(a) The set of branches identified by the heavy lines…
Fig. 4.12
“Determine the two mesh currents, i1 and i2, in …”
Fig. 4.16
“Find the three mesh currents in the circuit below.”
Fig. 4.19
Circuit from Practice Problem 4.8.
Engineering Circuit Analysis Sixth Edition
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin
Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved.
4
Obtain values for the unknown voltages across the
elements in the circuit below.
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.”
At node 1
v1
v1  v2

 3.1
2
5
At node 2
v2
v v
 2 1  - (-1.4)
1
5
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
5
(a) The circuit of Example 4.2 with a
22-V source in place of the 7-W
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:
v1  v2 v1  v3
83 

3
4
At the “supernode:”
3  25 
v2  v1 v3  v1 v3 v2

 
3
4
5 1
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
6
Determine the node-to-reference voltages in the circuit below.
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.
7
Examples of planar and nonplanar networks; crossed wires without
a solid dot are not in physical contact with each other.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
8
(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.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
9
Determine the two mesh currents, i1 and i2, in the circuit below.
Fig. 4.12
“Determine the two
mesh currents, i1
and i2, in the circuit
below.”
For the left-hand mesh,
-42 + 6 i1 + 3 ( i1 - i2 ) = 0
For the right-hand mesh,
3 ( i2 - i1 ) + 4 i2 - 10 = 0
Solving, we find that i1 = 6 A and i2 = 4 A.
(The current flowing downward through
the 3-W resistor is therefore i1 - i2 = 2 A. )
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
10
Find the three mesh currents in the circuit below.
Creating a “supermesh” from meshes 1 and 3:
-7 + 1 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0
[1]
Around mesh 2:
1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0
[2]
Finally, we relate the currents in meshes 1 and 3:
i1 - i3 = 7
[3]
Rearranging,
i1 - 4 i2 + 4 i3 = 7
[1]
-i1 + 6 i2 - 3 i3 = 0
[2]
i1
[3]
- i3 = 7
Solving,
i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
11
Find the voltage v3 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.
12
RESUME
• Perhitungan arus beban RL
disederhanakan dengan persamaan
tegangan berdasarkan metode mesh.
• Perhitungan tegangan beban RL
disederhanakan dengan persamaan arus
berdasarkan metode node.
13