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 sho
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,.
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,
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' ) c
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
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
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
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
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 ba
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
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
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)(12x
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
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-a
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-circuite
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
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 bes
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
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
para
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
(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
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 remin
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 P
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 mot
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
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 co
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 resi
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
Ther
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
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