to make the op-amp output positive. Also, the voltage across the op-amp
input terminals is not approximately
zero. For both of these reasons, the assumption is confirmed that op amp 1 is
saturated at the positive
saturation level. Therefore, V, = 14 V and

V - I_-x 8 6 . 4D = _ 33.51- 63' 1 12.27 Derive expressions for the conductance and the susceptance of an
admittance in terms of the
resistance and reactance of the corresponding impedance.
In general,
Rationalizing,
Since Y = G + jB,
Notice from
both fun

6.9 In the circuit of Fig. 6-15, assume that R, = 6 kR and then determine the
values of the other
resistors required to obtain U , = 2 ( v , + u2 + u3).
From the solution to Prob. 6.8, the multiplier of the voltage sum is
1
-3( I +;)=*
the solution to whi

120
- = 2.65 A
120
1'
/ = - = = s 2n(60)(120 x 1 0 - 3 ) 45.24
so pnlJX 1'1 = lZO(2.65) = 318 W
The reactive power absorbed is
Q = I ' X = 2.65'(45.24) = 318 VAR
which has the same numerical valuc ;is the peak power absorbed by thc
inductor. This is gener

Rf c, = - ) (< P a + - t'b Rf Rf + - cc
R b Rc
For the special case of all the resistances being the same, this formula
simplifies to
L', = -(LIa + t'b + Oc)
There is no special significance to the inputs being three in number. There
can be two, four, or

one loop current flowing through it. The number of these loops equals the
number of meshes if the
circuit is planar-that is, if the circuit can be drawn on a flat surface with no
wires crossing. In general,
the number of loop currents required is B - N +

i st quadrant 1 -z
/
I
I
I
I
I
I\II\I7
h (0)
I
I
I
4th quadrant
Fig. 12-6
3rd quadrant I
a vertical reactance axis designated by j X . Both axes have the same scale.
Shown is a diagram
of Z, = 6 + j 5 = 7.81/39.8 R Z2 = 8 - j6 = IO/-36.9 51 for a capaciti

resistor that obeys Ohm's law-a linear resistor. Any other type of voltagecurrent relation ( c = 4i2 + 6 ,
for example) is for a nonlineur resistor. The term "resistor" usually
designates a linear resistor. Nonlinear
resistors are specified as such. Figur

this equation, AQ is the change in stored charge and A V is the
accompanying change in boltage. From this,
Since C = Q / V is a linear relation, C also relates changes in charge and
voltage: C = AQiAV In
-1c
= 128V
-8 x l O 9 - t A s M m r
-x
C 10 x l o p

6 R Iv 6 R Ic
5.23 For the circuit shown in Fig. 5- 18, use superposition to find VTh
referenced positive on terminal a.
Clearly, the 30-V source contributes 30 V to VTh because this source, being
in series with an open circuit,
cannot cause any currents

I, = I,
14.1 Find ZTh, VTh, and I, for the Thevenin and Norton equivalents of the
circuit external to the load
impedance Z, in the circuit shown in Fig. 14-4.
6 R -j4R a
The Thevenin impedance Z T h is the impedance at terminals U and h with
the load impe

lOV1 - 6Vz = 42.
Similarly, for node 2 the self-conductance is
of the input currents from current sources is
-6Vl + 14Vz = 54.
principal diagonal as a result of the same mutual conductance coefficient in
both equations:
1OVI - 6V2 = 4 2
-6Vl + 14V2 = 54
T

zero value at and the fact that the waveform is going from negative to
positive then, just as a
sine wave does for a zero argument, the argument can be zero at this time:
104.7(-5 x 10-3) + 8 = 0,
from which 0 = 0.524 rad = 30 . The result is
Now consider

15.7 What power is absorbed by a circuit that has an input admittance of 0.4
+j0.5 S and an input
current of 30 A?
The formula P = V2G can be used after the input voltage I is found. I t is
equal to the current
divided by the magnitude of the admittance:

Find the equivalent resistance of four parallel resistors having resistances of
2, 4, 6, and 8 R.
Ans. 0.96 R
Three parallel resistors have a total conductance of 2 mS. If two of the
resistances are 1 and 5 kR, what is
the third resistance?
Ans. 1.25 kR
T

j8 Q
W
Fig. 14-14 b
the circuit is different. The voltage V , can be found from the current I
flowing through the 6-52 resistor
across which V , is taken. Since the 6- andj8-R impedances are in parallel,
and since 14 A from the current
source flows into t

Fig. 13-43
13.47 Find the node voltages in the circuit shown in Fig. 13-44.
Ans. V , = 1.17/-22.1 V, V, = 0.675/-7.33 V
12/-10" A
8/40" A
A
V-I
'
8s jl0 SZT
rd
12 s T S - j14 S
Fig. 13-44
13.48 Solve for the node voltages in the circuit shown in Fig. 13-4

15.83 At 400 Hz, what is the power factor of the circuit shown in Fig. 15.11?
What capacitor connected across
the input terminals causes the overall power factor to be 0.9 lagging?
A m . 0.77 lagging, 8.06 CIF
5 mH
Fig. 15-1 1
15.84 For a load energized b

more than this.
90 - 10 = 80 V,
10h(2 x 10-6) = 2 s. The proportionality is
t = 1601990 = 0.162 s. If an exact analysis is made, the result is 0.168 52 s.
In summary, by approximations the period is T= 0.162 + 0 = 0.162 s,
8.35 Repeat Prob. 8.34 with the

1. The ratio of the output variable to the specified source quantity. For
example, in the case in which the
independent source provides an input voltage and the output is the output
voltage, this ratio is the
voltage gain of the circuit.
2. The second is

Ans. c = 5 0 V f o r O s < r < 1 m s ; O V f o r 1 m s < t < 2 m s ; - 1 6 . 6 V f
or2ms<t<5ms
Find the total inductance of four parallel inductors having inductances of 80,
125, 200, and 350 mH.
Ans. 35.3 mH
Find the total inductance of a 40-mH inductor

30". This means that the peaks, zeros, and other values of r 2 occur earlier
than those of r1 by a time
corresponding to 30". Another but less specific way of expressing this phase
relation is to say that P~
and t i 2 have a 30" phase diflerence or that t

Ans. 760 R
l e I kfl C
*A0 2 -+ , B
10 +
0.003Vc. 20 kR V C
EI ' 0 0 - 0 2'
Fig. 5-52
5.51 Find the output resistance at terminals 2 and 2' of the transistor circuit
shown in Fig. 5-52 if a source with
a 50042 internal resistance is connected to terminals

in Fig. 15-2 if the load absorbs average power. Notice that, for the associated
voltage and current
references, the reference current enters the k current terminal and the
positive reference of the voltage
is at the a
source of average power -then one coi

the ground node. The other nodes are preferably identified by positive
integers, but these integers need
not be sequential.
of the component in ohms, volts, or
amperes, whichever applies. The resistances must be nonzero. Note that the
values must not cont

2 kQ to the output audio stage so that there is maximum power transfer to
the 6-R speaker. Since, in general,
the reflected resistance R, is equal to the turns ratio squared times the
resistance RI, of the load connected
t o the secondary ( R , = a 2 R ,

Parts (c) and ( d ) show that an angular difference of 180" corresponds to
multiplying by - 1. And parts
A/O & 180' =
A,& = A cos 0 + j A sin 0. With this approach
( c ) and (e) show that an angular difference of 360' has no effect. So, in
general, -Ab an

has a total reactance of 7 R. Since 5 R of this is in the mutual branch, there is
remaining
that could be from a single inductor that is not in the mutual branch. The
-8@' on the right-hand side
is the result of a total of 8/30' V of voltage source drops

in Fig. 17-4a, or primed terminals can be connected to unprimed terminals to
form the A (delta) shown
in Fig. 17-4b. The primed letters are included this once to show these
connections. But since the terminals
at which they are located also have unprimed