Chapter 3 Voltage and Current Laws Engineering Circuit Analysis Sixth Edition W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved. User Note: Run V iew Show under the Slid e Show menu to enable slide selection. Fig. 3.1 “A circuit containing three nodes and five branches …” Fig. 3.2 Example node to illustrate the application of KCL. Fig. 3.19 “Addition of multiple voltage or current sources.” Fig. 3.20 Examples of circuits with multiple sources, ... Fig. 3.22 (a) Series combination of N resistors. Fig. 3.25 (a) A circuit with N resistors in parallel. Fig. 3.30 An illustration of voltage division. Fig. 3.32 Circuit for Practice Problem 3.12. Fig. 3.33 An illustration of current division. Fig. 3.100 “Determine the current I x if I 1 = 100 mA.”
Chapter Three Voltage and Current Laws
Dec 30, 2015
DC Electrical Circuits
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Chapter 3 Voltage and Current Laws
Engineering Circuit Analysis Sixth Edition
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin
Copyright © 2002 McGraw-Hill, Inc. All Rights Reserved.
User Note:
Run View Show under the Slide Show menu to enable slide selection.
Fig. 3.1 “A circuit containing three nodes and five branches …”
Fig. 3.2 Example node to illustrate the application of KCL.
Fig. 3.19 “Addition of multiple voltage or current sources.”
Fig. 3.20 Examples of circuits with multiple sources, ...
Fig. 3.22 (a) Series combination of N resistors.
Fig. 3.25 (a) A circuit with N resistors in parallel.
Fig. 3.30 An illustration of voltage division.
Fig. 3.32 Circuit for Practice Problem 3.12.
Fig. 3.33 An illustration of current division.
Fig. 3.100 “Determine the current Ix if I1 = 100 mA.”
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
(a) A circuit containing three nodes and five branches.
(b) Node 1 is redrawn to look like two nodes; it is still one node.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Figure 3.2
(a) Series connected voltage sources can be replaced by a single source. (b) Parallel current sources can be replaced by a single source.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Examples of circuits with multiple sources, some of which are “illegal” as they violate Kirchhoff’s laws.
(a) Series combination of N resistors. (b) Electrically equivalent circuit.
Fig. 3.22 (a) Series combination of N resistors. (b) Electrically equivalent circuit.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Fig. 3.25 (a) A circuit with N resistors in parallel. (b) Equivalent circuit.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
(a) A circuit with N resistors in parallel. (b) Equivalent circuit.
Beginning with a simple KCL equation,
or
Thus,
A special case worth remembering is
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
An illustration of voltage division.
We may find v2 by applying KVL and Ohm’s law:
so
Thus,
or
For a string of N series resistors, we may write:
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Use voltage division to determine vx in the adjacent circuit.
vx 2
2 3 10 ||10 10V 2
2 3 510V 2V
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
An illustration of
current division.
The current flowing through R2 is
For a parallel combination of N resistors, the current through Rk is
or i2 G2
G1 G2
i
ik
GkG1 G2 L GN
i
Determine the current Ix if I1 = 100 mA.
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
I x 15
15 30100mA 33.33 mA
Vs v1 v2 v3 0 KVL
v1 iR1
v2 iR2
v3 iR3
Ohm's Law
i R1 R2 R3 VsLet Vs 10V , R1 8R2 13 and R3 5
Then i 10V
260.3846 A
and v1 3.0769 V , v2 5 V
v3 1.9231 V.
Vs v1 v2 v3 0 KVL
v1 iR1
v2 iR2
v3 iR3
Ohm's Law
i R1 R2 R3 VsLet Vs 10V , R1 8R2 13 and R3 5
Then i 10V
26 0.3846 A
and v1 3.0769 V , v2 5 V
v3 1.9231 V.
Single-Loop Circuit Analysis
I s i1 i2 i3 KCL
i1 v / R1 , i2 v / R2 , i3 v / R3 , Ohm's Law
I s v 1 / R1 1 / R2 1 / R3 Let I s 20 mA , R1 1k , R2 3k , R3 7k
Then v0.02A
0.001476 S13.548 V
i1 13.548 mA , i2 4.516 mA , i3 1.936 mA
I s i1 i2 i3 KCL
i1 v / R1 , i2 v / R2 , i3 v / R3 Ohm's Law
I s v 1 / R1 1 / R2 1 / R3 Let I s 20 mA , R1 1k , R2 3k , R3 7k
Then v0.02A
0.001476 S13.548 V
i1 13.548 mA , i2 4.516 mA , i3 1.936 mA
Single-Node-Pair Circuit Analysis
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