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UES-FIA-EIE-AEL115 Ciclo I-2012
1.5 Tcnicas de Anlisisde Circuitos:
a) Divisor de Voltaje
b) Divisor de Corriente
UES-FIA-EIE-AEL115 Ciclo I-2012
Chapter 3 Voltage and Current Laws
Engineering Circuit Analysis Sixth EditionW.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin
Copyright 2002 McGraw-Hill, Inc. All Rights Reserved.
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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.
UES-FIA-EIE-AEL115 Ciclo I-2012W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright 2002 McGraw-Hill. All rights reserved.
1.5 a) Divisor de Voltaje:We may find v2 by applying KVL and Ohms law:
An illustration of voltage division.
UES-FIA-EIE-AEL115 Ciclo I-2012
For a string of N series resistors, we may write:
1 2 ...
kk s
N
Rv v
R R R
= + + +
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.Copyright 2002 McGraw-Hill. All rights reserved.
UES-FIA-EIE-AEL115 Ciclo I-2012W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright 2002 McGraw-Hill. All rights reserved.
Exercise: Use voltage division to determine vx in the adjacent circuit.
UES-FIA-EIE-AEL115 Ciclo I-2012W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright 2002 McGraw-Hill. All rights reserved.
1.5 b) Divisor de Corriente:The current flowing through R2 is
An illustration of current division.
UES-FIA-EIE-AEL115 Ciclo I-2012
1 2 ...
kk s
N
Gi iG G G
=
+ + +
For a parallel combination of N resistors, the current through Rk is
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.Copyright 2002 McGraw-Hill. All rights reserved.
UES-FIA-EIE-AEL115 Ciclo I-2012W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright 2002 McGraw-Hill. All rights reserved.
Exercise: Determine the current Ix if I1 = 100 mA.