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
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 o
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
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
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
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
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 t
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 p
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
containi
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
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 gr
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
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
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 t
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
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 t
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
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 o
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 V
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.
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, combinin
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
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 cl
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.
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 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 e
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
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 poten
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.
WW
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 circui