chapter three voltage and current laws

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.



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.





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.



Fig. 3.25 (a) A circuit with N resistors in parallel. (b) Equivalent circuit.



(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



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:



Use voltage division to determine vx in the adjacent circuit.

vx 2

2 3 10 ||10 10V 2

2 3 510V 2V



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.



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

chapter three voltage and current laws

Documents

engineering circuit

b equivalent circuit

current sources

current lawsfig

multiple sources

addition of multiple

kirchhoffs laws

examples of circuits