dependent source expressed as usual in terms of the branch variables.
We also need to summarize the behavior of the input port to completely
characterize the dependent source. Since our idealized VCCS does not
require any power to be supplied at its input
that i2 = i1 = I. In other words, the current from the source flows
entirely through the resistor. Next, from the element law for the resistor,
it follows that v2 = Ri2 = RI. Finally the application of KVL to the one
loop yields v1 = v2 = RI to complete t
some function of a branch variable i as shown in the figure. Our method
of applying node analysis to a circuit containing dependent sources
begins by assuming that we know the value of each dependent source.
This assumption allows us to treat each depende
chapter, we will introduce methods of analyzing general circuits
containing nonlinear elements, trying whenever possible to use analysis
methods already introduced in the preceding chapters. Chapter 7 will
develop further the basic ideas on nonlinear anal
W of power. (In other words, the resistor will overheat if the power
dissipation is greater than 0.5 W.) Determine the current through the
resistor. Further, determine whether the power dissipation in the resistor
exceeds its maximum rating. The current t
other words, GP = I v = G1 + G2. (2.94) Hence, the equivalent
conductance of the two parallel resistors is the sum of their individual
conductances. This is consistent with the physical derivation of
resistance in Equation 1.6 since placing resistors in p
stated as 1. The rate of change of magnetic flux linked with any portion
of the circuit must be zero for all time. 2. The rate of change of the
charge at any node in the circuit must be zero for all time. A node is any
point in the circuit at which two or
circuits. 30 CHAPTER ONE the circuit abstraction The same set of ideal
two-terminal elements serve to build either information processing or
energy processing systems as well. Information and energy processing
includes the communication, storage, or trans
circuit is the potential FIGURE 2.14 Voltages in a closed loop in the
network. i4 a b c + v1 - + v3 - + v4 - V + - d 2.2 Kirchhoffs Laws
CHAPTER TWO 61 vM1 vM v1 vN + + + + FIGURE 2.15 A loop
containing N branches. difference between the two nodes. The po
FIGURE 3.39 Circuit with two independent sources. 1 V + - 1 A e 2 2
2 I 3.5 Superposition CHAPTER THREE 151 1 V + - ev 2 2 2
1 V + - ev 2 2 FIGURE 3.40 Subcircuit corresponding to the
voltage source acting alone. 1 A ei 2 2 2 1 A ei 1 2 FIGURE
3.41 Subc
relatively constant at about 0.6 volts (due to the nature of the
exponential), so the voltage across the resistor will be approximately vI
0.6 V. This kind of insight is the principal value of the graphical
method. Second, in contrast to all previous exa
will continue to light the bulb indefinitely. A similar model might apply
for a microphone. When an information processing system such as an
amplifier is connected to the microphone, its output might drop from a
1-mV peak-to-peak signal to a 0.5-mV peak-t
making the simplification, where the sourced current is now specified in
terms of i1. Notice that there is a lot less clutter in the latter figure. e x a
m p l e 2.30 current-controlled current source Consider next the circuit
shown in Figure 2.69b. This
source is the opposite of the total power into the two resistors, energy is
conserved in the two-resistor current divider. That is, the power
generated by the current source is exactly dissipated in the two resistors.
Resistors in Parallel Resistors in pa
for convenience. Following Step 4, the solution of Equation 3.38 is e =
(G1 + G2)V1 + G1V2 G1V3 G1 + G2 + G3 . (3.39) Finally, to
complete the node analysis, the solution for e could be used in Step 5 to
determine the branch voltages and then the branch c
a network that obeys KVL and KCL. Therefore calculate the quantity
vnin. It should be zero. e x e r c i s e 2.10 A portion of a larger network
is shown in Figure 2.81. Show that the algebraic sum of the currents into
this portion of the network must be ze
across its control input port determines the value of the current iOUT
through its output port. In principle, such a dependent source can
provide power, but for simplicity the power terminals inherent to the
source are not shown. The diamond shape of the
2 2 a RTH a FIGURE 3.64 Measuring RTH. 2 2 a 2 A 10 I2
a FIGURE 3.65 Circuit to further illustrate the power of the Thvenin
method. a 10 I2 VTH RTH + - a FIGURE 3.66 Circuit with network
to the left of the aa terminal pair replaced with its Thvenin equiva
voltage - divider circuit; (b) shorthand notation. R1 R2 R3 RL VO VS =
5 V Gnd (b) R1 R2 R3 RL VO + - VS = 5 V (a) + - is supplying power
to the rest of the circuit. Further, assume that the voltage VO is of
interest to us. Notice also that the two voltag
useful, complex systems. The set of abstractions that transition from
science to engineering and insulate the engineer from scientific minutiae
are often derived through the discretization discipline. Discretization is
also referred to as lumping. A disci
application of KVL around the loop now yields v1 + v2 = 0. (2.32) Note
that Equations 2.31 and 2.32 differ from Equations 2.19 and 2.20
because of the polarity reversal of i2 and v2. Finally, combining
Equations 2.29 through 2.32 yields i1 = i2 = I (2.33)
where 1 m is 106 m. If the resistivity of its poly-crystalline silicon
ranges from 106 m to 102 m, what is the range of its resistance? The
cross-sectional area of the resistor is A = 1011 m, and its length is l =
104 m. Thus l/A = 107 m1. Using Equation
In a manner analogous to KCL, Kirchhoffs voltage law can be stated as:
1 A 2 A 1 A 2 A 3 A 1 A 3 A 2 A FIGURE 2.13 Circuits that violate KCL.
KVL The algebraic sum of the branch voltages around any closed path
in a network must be zero. Alternatively, it
the calculation of isc for this particular topology easy. Because of the
short circuit, R2 can have no voltage across it, hence has no current
flowing through it. Now by superposition, isc = I + V/R1. (3.125) The
calculation of voc from Figure 3.89b is st
electromagnetic wave launched by the signals. 12 CHAPTER ONE the
circuit abstraction effects, the lumped element approximation becomes
analogous to the pointmass simplification, in which we are able to
ignore many physical properties of elements such as t
expression: All denominator terms are of the same sign. Thus the
denominator cannot be made zero for any nonzero values of
conductances. (If the denominator could be made zero, we could get
infinite e1 for finite sources values, a violation of conservatio
transformations p r o b l e m 3.11 A student is given an unknown
resistive network as illustrated in Figure 3.140. She wishes to determine
whether the network is linear, and if it is, what its Thvenin equivalent
is. Resistive network Unknown network FIGUR
c and b, respectively. To find vc, let us follow the path g f c.
Accordingly, there is a 1-V increase in potential from g to f, and a
further 2-V increase from f to c resulting in an accumulated potential
of 1 V at c. Thus vc = 1 V. Similarly, because the
dependent current source. 3.4 Loop Method * CHAPTER THREE 145
By simplifying Equation 3.62, we obtain: vO = 1 RP RI RL RP + RL
vI. (3.63) We have thus expressed vO as a function of vI when iO = iI.
WWW e x a m p l e 3.12 a m o r e c o m p l e x d e p e n
in matrix form: source terms on the left, unknown variables on the right.
184 CHAPTER THREE network theorems FIGURE 3.129 V + - R1 R2
R2 R3 e x e r c i s e 3.27 Find v1 by superposition for the circuit in
Figure 3.130. FIGURE 3.130 V R1 R2 v1 R3 + - I PRO