5.63
Fig. 5-54
In the circuit of Fig. 5-55, what resistor connected across terminals LI and h
will absorb maximum power,
and what is this power?
Ans. 100 kR, 62.5 pW
6kR1
Fig. 5-55
For the circuit shown in Fig. 5-41, use superposition to find the contribu
16.13 Find i , . i, and I , for the circuit shown in Fig. 16-13. The
transformers are ideal.
A good procedure is to find i, using reflected resistances, then find i, from i,
and last find i, from i,.
The 8 R reflects into the middle circuit as making a to
This current is equal to the conductance times the voltage at the node at
which the current enters the
resistor minus the voltage at the node at which the current leaves the resistor.
The quantity (V, - V2) is,
of course, just the resistor voltage referen
given that copper has 1.38 x 1024 free electrons per cubic inch and that the
cross-sectional area of
No. 14 AWG wire is 3.23 x 10-3 in2.
S
1.1 I
1.12
1.13
1.14
1.15
The a~w-agder ift ~cloci t( 1y' ) cqu:ils the current di\,idcd by the product
of the cross
across an inductor or capacitor with reactance X , then Q = V 2 / X . So, Q =
V2/(oL for an inductor
and Q = -coCV2 for a capacitor.
COMPLEX POWER AND APPARENT POWER
There is a relation among the real power of a load, the reactive power, and
another power
12 x 24 288
12 + 24 36
_ _ _ _- - = 8 R
This combined with the 32-R resistance gives a total resistance of
8 x 32 256
8 + 32 40
R -=-=(j6.4R T A 60-W, a 100-W, and a 200-W light bulb are connected in parallel across a
120-V line. Obtain
the equivalent hot
coil across lines A and B. The k terminal of the current coil is toward the
source, and the k
terminal of the potential coil is at line A.
cos (ang V, ang lA), for which 1,- znd the angles of V, and I, are needed. Since no
phasors are specified in the pro
selected, then Fig. 17-1 Id can be used, which has these currents for an ACB
phase sequence. I t can be seen
that I, IBc, and I, lag I, I, and I, respectively by 30'. The magnitude of
each load phase current is,
of course, 34.6/$ = 20 A. Thus,
I, = 20/-60
the arctangent of the ratio of the imaginary part to the real part. With 0
known, A can be found by
substituting 8 into either x = A cos 0 or into
Another popular way of finding A is from a formula based on squaring both
sides of A cos 8 = s
and of A sin
currents immediately after a switching operation are the same as
immediately before the operation. This
is an important fact for RL (resistor-inductor) circuit analysis.
TOTAL INDUCTANCE
The total or equivalent inductance (LT or Leq) of inductors connecte
The approach is to find di,/cit, the slope, from the graph and insert it into t' =
Ldi,dt
t' = L dildt = (400 x 10-3)( -40) = - 16 V.
[20 x 10-3 - (-40 x 10-3)]/(3 x 10-3) = 20 A,s and the voltage is
CHAP. 91
i (mA)
20
10
0
- 10
-m
-30
-4
INDUCTORS, INDUC
10.34 Find the reactances of a 120-mH inductor at ( U ) 0 Hz (dc), ( h ) 40
rad s. (c) 60 Hz, and ( t l )
30 kHz.
From X , = COL= 27cfL
(U)
( h ) X I , = 40(120 x 10- ') = 4.8 SZ
(c) X I *= 2~(60)(12x0 10 ') = 45.2 SZ
(d)
X,< = 27c(0)( 120 x 10 3 ) = 0 !2
VI
DC C'IKC'UIT ANAl.YSIS
4s
176 A
V1
4.63
4.64
4.65
4.66
4.67
32 A 0 ZJ S
100 A
>8S
IFig.
4-46
24 A
48 A
Fig. 4-47
Repeat Prob. 4.62 with the three current-source changcs of 176 to 108 A.
112 to 110 A, and 48 to 66 A.
Am. C; = 3 V, I', = 4 V, v3 = 5 V
Fo
6.41-51.3
12.46 Determine V, and I, in the circuit of Fig. 12-24.
1 3 kR j 4 kR I
2/-30 V c, v
4kR
1 v0
j4 kR
I *A1 -1 0
Fig. 12-24
Because this circuit has the same configuration as the inverter op-amp circuit
of Fig. 6-4, the same
formula applies, with
CHAP. 161 TRANSFORMERS 367
16.27
16.28
16.29
16.30
Since phasors are not specified or mentioned, presumably the electric
quantities specified and wanted
are rms. Because the secondary is open-circuited, I , = OA, which means
that o M 1 2 = 0 and
toL,I, =
impedance can be found by using the input current. Lvhich from
Because this is a series circuit, inipcdancc should be used t o find thc
resistance and capacitance. The
P = 1.1 x PF is
The magnitude of the impedance is equal to the voltage di\.ided by the
determination of R T , , it is necessary to apply a source and calculate the
ratio of the source voltage to the
source current. Any independent source can be applied, but often a particular
one is best. Here, if a 12-V
voltage source is applied positive a
1.50 Find the total energy available from a rechargeable 1.25-V flashlight
battery with a 1.2-Ah rating.
Ans. 5.4 kJ
1.51 If all the energy in a 9-V transistor radio battery rated at 0.392 Ah is
used to lift a 150-lb man. how high in feet
will he be lifte
4.23 Find the node voltages in the circuit shown in Fig. 4-26.
65 A
n
1,
Fig. 4-26
+
Fig. 4-27
One analysis approach is to transform the voltage source and series resistor
to a current source and
parallel resistor, as shown in the circuit of Fig. 4-27.
Th
Applying the quadratic formula,
- 7.68 & 7.6S2 - 4( -48) - 7.68 15.84
R = - ~- 22
The positive sign must be used to obtain a physically significant positive
resistance. So,
R = -7.68 + 15.84 _._~_=_ 4_.0 8 R
-I
L
Note in the solution that the Thevenin and
(h)
(c.)
Ans.
7.21 sin oit + I 1.2 cos (of
-8.63 sin 377f - 4.19 cos 377r
4.12 sin (64t - 10-) - 6.23 sin (64t - 35 ) + 7.26 cos (64t - 35 ) - 8.92 cos (64f
+ 17 )
(a) 13.3 sin (to? + 57.2 ), ( h ) -9.59 sin (377r + 25.9 ), (c) 5.73 sin (64r +
2.75 )
In F
If the voltage source in the circuit of Fig. 14-25 is replaced by a short circuit,
the circuit is as shown
in Fig. 14-27, where V" is the component of V from the independent current
source. As a reminder, the
current to the left of the parallel resistor a
13.52 Solve for I in the circuit shown in Fig. 13-49.
Ans. -253/34" A
0.2 n -j0.4 fl
0.25 fl
Fig. 13-49
292
13.53
13.54
13.55
13.56
MESH, LOOP, NODAL, AND PSPICE ANALYSES OF AC CIRCUITS
[CHAP. 13
In Probs. 13.53 through 13.58, given the specified PSpice c
Because the added capacitor provides the change in susccptance. its
capacitance is
Naturally, less capacitance is required to improkc the power factor to 0.8s
lagging than to 1.
15.39 An induction motor draws 50 kW at a 0.6 lagging power fxtor from a
480-
R
Fig. 3-36
3.82 In the circuit shown in Fig. 3-36, let
new current in the R , resistor.
Ans. 1.33 A
R , = 6 R and R , = 12 Q. Then use current division to find the
52 SERIES AND PARALLEL DC CIRCUITS [CHAP. 3
3.83 A 60-A current flows into a resistor netw
capacitor voltage. find the capacitance.
Ans. 0.5 jtF
What change in voltage of a 20-pF capacitor is produced bq ;i moicincnt of 9
x 10" electron\ bctLtcen
plates '?
Ans. 7.21 V
A tubular capacitor consists of two sheets of aluminum foil 3 cm wide and I
m
3.15 Resistors R I , R , , and R , are in series with a 100-V source. The total
voltage drop across R , and
R2 is 50 V, and that across R, and R , is 80 V. Find the three resistances if the
total resistance
is 50 SZ.
The current is the applied voltage div
so R , = 8 kR, the multiplier of i'b becomes
R,. 4
-(I RC + p ) = 4 or Rb + Rc R , + R , 5
CHAP. 63 0 PERATIONA L-A M PLI FI ER CIRCUITS 123
Inverting results in
-R, - -1 5 -R+b I = Rc 4 R c 4
or
Therefore,
of the design process. So, if R , is selected as
If the secondary circuit contains no independent sources and no current paths
to the primary circuit,
i t is possible to reflect impedances in a manner similar to that used for ideal
transformers. For an
understanding of this reflection, consider the circ
8
9
0.1
0.0 1
COLOR CODE
The most popular resistance color code has nominal resistance values and
tolerances indicated by
the colors of either three or four bands around the resistor casing, as shown
in Fig. 2-3.
First Second Number of zeros
digit digit o