CHAPTER 2
ENGINEERING CIRCUIT ANALYSIS
(d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz
SELECTED ANSWERS
1.
(a) 12 μs (b) 750 mJ (c) 1.13 k Ω
(g) 39 pA (h) 49 k Ω (i) 11.73 pA
3.
300 kW; 3.7 m; 25 mm; 71 kJ; 290 fs
5.
131 kW; 1.4 GJ; 1 battery
7.
13 GW; 100 mW
9.
290 kJ; 1.5 kJ
11.
6.2 A; 3.5 A; The current is never negative; 34 C
13.
12 MV; 0; -18.7 MV; -6.2 MV
15.
-6.4 mW; -120 W; 60 W; 12 W
17.
73 W; -36 W; 28 W
19.
5 mW, 0, -2 mW; 36 J; 22 J
21.
64 W, 256 W, -640 W, 800 W, -480 W
23.
-1 mV
25.
58 W; 4.8 A
27.
5.6 mA, 4.5 mA; 23 mW, 28 mW
29.
43.5 mW; 231 mW; 253 mW
31.
Since we know that the total power supplied is equal to the total power absorbed, 2 2 we may write: V s I = I R1 + I R2. Now invoke Ohm’s law.
33.
500 μA, 2.5 mW; -500 μA, 2.5 mW; -500 μA, 2.5 mW; 500 μA, 2.5 mW
35.
-2 V (at t = 0.324 s)
37.
2 km. Hmmmm….
39.
1.7 μΩ cm
41.
560 mΩ, 1.3 W
43.
266 mΩ; 514 mA
45.
Design. Many possible solutions. Hint: Start with finding resistivity, then choose geometry.
.
CHAPTER 3
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1.
Circuit diagram not shown.
3.
(a) 4 nodes; (b) 5 branches; (c) yes, path; no, loop.
5.
(a) 4; (b) 5; (c) yes,no,yes,no,no yes,no,yes, no,no
7.
(a) 3 A; (b) -3 A; (c) 0
9.
ix = 1 A; iy = 5 A.
11.
If the DMM appears as a short, then all 5 A flows through the DMM, and none through the resistors, resulting in a (false) reading of 0 V for the circuit undergoing testing.
13.
(a) 12 V; (b) -2.2 V
15.
R = 34 Ω; G = 90 mS
17.
Circuit I: i = 0; Circuit II: i = 1.1 A
19.
-23.5 V
21.
(a)
23.
(a) 8 V, V, -4 V, -12 -12 V; (b) (b) 14 V, 2 V, -6 V; (c) (c) 2 V, -10 -10 V, -18 -18 V
25.
(a) 25 W; (b) 24 W; (c) 16 W; (d) 18.4 W; (e) -600 W
27.
None of the conditions specified in (a) to (d) can be met by this circuit.
29.
5.0 A; 10.4 V
31.
(a) 2.4 k Ω; (b) R = 0
33.
-250 cos 5t mV
= 60 V v2 = 60 V v3 = 15 V v4 = 45 V v5 = 45 V v1
= 27 A i2 = 3 A i3 = 24 A i4 = 15 A i5 = 9 A i1
(b)
= -1.62 kW = 180 W = 360 W = 675 W = 405 W
CHAPTER 3
35.
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
(a) P5A
= –5 vx
= –1.389 kW
P100Ω
= (vx) / 100
2
=
771.7 W
P25Ω
= (vx) / 25
2
=
3.087 kW
Pdep
=
2
–vx(0.8 ix) = –0.8 ( vx) / 25
= –2.470 kW
(b) P5A
= –5 vx
P100Ω
= (vx) / 100
2
=
240.9 W
P25Ω
= (vx) / 25
2
=
963.5 963. 5 W
Pdep
=
P8A
= –8 vx
P6Ω
= (vx) / 6
=
P8A
= –7 vx
= –210 W
P12Ω
= (vx) / 12
P4Ω
= (vx) / 4
–vx(0.8 iy)
= –776.0 W
= –428.1 W
37. 2
= –240 W 150 W
2
=
75 W
2
=
225 W
39.
(a) – 50 mA; (b) Can set vS = 50 V.
41.
638 mW
43.
1.45E-3 miles
45.
(a) 1 A; (b) 9 A
47.
(a) 10 mA; (b) 3.8 A
49.
(a) 570 mA; (b) 0; (c) – 71 mA
51.
-515 V
53.
R = 1 k Ω
55.
Ω; (a) 10 k Ω || 10 k Ω; (b) 47 k Ω + 10 k Ω + 1 k Ω || 1k Ω || 1k Ω; (c) 47 k Ω || 47 k Ω + 10 k Ω || 10 k Ω + 1 k Ω
57.
5.5 k Ω
59.
60 Ω; 213 Ω; 52 Ω
eq
CHAPTER 3
ENGINEERING CIRCUIT ANALYSIS
61.
250 W; 188 W; 338 W; 180 W; 45 W
63.
(a) 850 mS; (b) 136 mS
65.
Proof
67.
607 mV
69.
22 A
71.
One possible solution: 11 mA, 1 k Ω, 1 k Ω
73.
139 μA; 868 μW
75.
18 μW
77.
(a) VS
+ R4) ; R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 ) R 1 (R 2 + R 3 + R 4 ) (b) VS ; R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 ) (c) VS
R 2 (R 3
R2 R 1 (R 2
+ R 3 + R 4 ) + R 2 (R 3 + R 4 )
79.
(a) 42 A; (b) 11.9 V; (c) 0.238
81.
VS ⎜⎜
83.
vout
⎛ ⎝ R 2 (R 3
⎞ ⎟ + R 4 + R 5 ) + R 3 (R 4 + R 5 ) ⎟⎠ R3 R5
= -56.02 sin 10t V
.
SELECTED ANSWERS
CHAPTER 4
ENGINEERING CIRCUIT ANALYSIS
1.
(a) -8.4 V; (b) 32
3.
(a) v1 = 264 V, v2 = 184 V and v3 = 397 V; (b) >> e1 = '4 = v1/100 + (v1 - v2)/20 + (v1 - vx)/50'; >> e2 = '10 - 4 - (-2) = (vx - v1)/50 + (vx - v2)/40'; >> e3 = '-2 = v2/25 + (v2 - vx)/40 + (v2 - v1)/20'; >> a = solve(e1,e2,e3,'v1','v2' solve(e1,e2,e3,'v1','v2','vx'); ,'vx'); >> a.v1
5.
-1.74 V
7.
172 V
9.
(a) 58.5 V, 64.4 V; (b) 543 W
11.
-28 V
13.
-8.1 V
SELECTED ANSWERS
15. v1
= 3.4 V
v5
= 1.7 V
v2
= 7.1 V
v6
= 3.8 V
v3
= 7.5 V
v7
= 3.5 V
v4
= 4.9 V
v8
= 2.4 V
17.
(a) 26 V, (b) 83 mW
19.
-3.25
21.
-91 V
23.
45 W
25.
v1
27.
(a) 143 mA; (b) 16 W
29.
(a) 3.1 A; (b) 370 W
31.
2.79 A
33.
-380 W
35.
i1
= -8.6 V, v2 = -3.9 V and
v3
= 6.1 V
= 239 μA, i2 = 1.08 mA, i3 = -1.20 mA mA and i4 = -480 μA
CHAPTER 4
ENGINEERING CIRCUIT ANALYSIS
37.
(a) -5700 Ω; (b) this value is unique.
39.
(a) 330 μA; (b) 330 μA; (c) units of resistance.
41.
P2mA P4V P6V PdepV PdepI
43.
-3.65 W
45.
-1.03 V
47.
5Ω
49.
(a) 0; (b) 96 V; (c) -38 V
51.
3.55 A; 1.69 A
53.
121 mA; 4.70 A
55.
Hint: i3 = 1.24 A and i4 = 1.42 A by mesh analysis.
57.
350 mA
59.
i1
61.
-4 mA
63.
-16 V
65.
3.14 V, 1.71 V, 714 mA, -143 mA, -2.14 A, 857 mA
67.
One possible solution:
= 5000(i1 – i2)(i1) = 4 (-i2) = 6 (-i3) = 1000 i3 (i3 – i2) = 10,000(i3 – i4)(0.5 i2)
SELECTED ANSWERS
= 5 mW = -6 mW = 9 mW = 4.5 mW = -5.6 mW
= 2.65 A, i2 = 3.20 A, i3 = -3.80 A, i4 = -1 mA
where R = 5/3 Ω = 1 Ω + 2/3 Ω = 1 Ω + 1 Ω || 1Ω || 1Ω + 1Ω || 1Ω || 1Ω. 69.
One possible solution: 9 V in series with 5 1- Ω resistors (R1) and 5 1- Ω resistors (R2 – R5). Take V1 across R2-R5, V2 across R3-R5, and V3 across R4-R5.
CHAPTER 5
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1.
Define percent error as 100 [ex – (1 + x)]/ ex. If we choose x < 0.1, we ensure that the error is less than 1%.
3.
4.7 V, 2.0 A
5.
4V
7.
10.8 V
9.
(a) 1.3 A; (b) 60 W, 18 W, -130 W, 32 W, 20 W
11.
(a) 200 V; (b) -143 V
13.
957 μW
15.
Impossible; 76 mW
17.
(a) 18 V
19.
2.46 V; 0.546 V, 1.91 V.
21.
(a) 42 V voltage source in series with 6 Ω and in series with 10 Ω; (b) 26 V; (c) Cannot remove the resistor across which v appears or v may become lost.
23.
10 mW
25.
33 μW
27.
(a) 12.8 mV
29.
764 nA
31.
Current source is 7.25 A, resistor is 2 ohms.
33.
1.57 V, 811 mΩ
35.
The final circuit is an 8.5 V voltage source in series with a 2.0 MΩ resistor.
37.
(a) An 8/5 A current source in parallel with 5 Ω, in parallel with R L.
→
40 V and 10 V
→
100 V.
CHAPTER 5
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
39.
-2 V
41.
(a) The Thévenin equivalent is a 9.3 V source in series with a 17 Ω resistor, which is in series with the 5 Ω resistor of interest; (b) 928 mW.
43.
(a) 25 Ω; (b) 303 Ω; (c) Increased current leads to increased filament temperature, which results in a higher resistance (as measured). Th is means the Thévenin equivalent must apply to the specific current of a particular circuit – one model is not suitable for all operating conditions.
45.
(a) 6.7 Ω, -300 mA, arrow upwards; (b) 6.7 Ω, -150 mA, arrow upwards.
47.
(a) 38.9 V, 178 Ω; (b) 1.96 W.
49.
VTH = 0, R TH TH = 192 Ω.
51.
15 Ω, 15 Ω
53.
VTH = 0; The Norton equivalent equ ivalent is 0 A in parallel with 1.3 Ω.
55.
VTH (and hence IN) = 0; R TH TH = R N = 198 mΩ.
57.
2 MΩ
59.
VTH =
vin Ri ( Ro − AR f ) R1 Ro + Ri Ro + R1 R f + Ri R f + R1 Ri + AR1Ri
; R TH TH =
Ro (Ri Rf + R1 Rf + R1 Ri) -------------------------------------------------------------Ri Ro + R1 Ro + Ri Rf + R1 Rf + R1 Ri + A R1 Ri. 61.
16 Ω, 6.3 W
63.
65 V, 15 Ω, 70 W
65.
(a) 200 V; (b) 125 W; (c) 80 Ω
67.
There is no conflict with our derivation concerning maximum power. While a dead short across the battery terminals will indeed result in maximum current 2 draw from the battery, and power is indeed proportional to i , the power delivered 2 2 to the load is i RLOAD = i (0) = 0 watts. This is the minimum, not the maximum, power that the battery can deliver to a load.
69.
Select R 1 = R TH TH = 8 k Ω
CHAPTER 5
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
71.
1.2 Ω, 0.54 Ω, 4.9 Ω
73.
9.9 Ω
75.
5.5 V, 1.0 Ω
77.
-13 V, 27 Ω
79.
Although the network may be simplified, it is not possible to replace it with a three-resistor equivalent.
81.
IS(max) = 224 mA
83.
1.4 Ω
85.
One possible solution of many:
87.
One possible current-limiting scheme is to connect a 9-V battery in series with a resistor R limiting limiting and in series with the LED; R limiting limiting = 220 Ω.
CHAPTER 6
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1.
(a) -30 V; (b) -2.5 V; (c) 1.4 V
3.
(a) vout = −10vin
5.
One possible design is to use a simple inverting op amp circuit with R f f = 9.1 k Ω and R in in = 5.1 k Ω.
7.
To get a positive output that is smaller than the input, the easiest way is to use inverting amplifier with an inverted voltage supply to give a negative voltage, where R f f = 1.5 k Ω and R in in = 5.1 k Ω
9.
(a) 1.7 V; (b) 3 V; (c) -2.4 V
11.
(a) vout = 2vin
13.
-2.2 V
15.
One possible solution of many: a non-inverting op amp circuit with the microphone connected to the non-inverting input terminal, the switch connected between the op amp output pin and ground, a feedback resistor R f = 133 Ω, and a resistor R 1 = 1 Ω.
17.
V1 = 21 V
19.
vout
21.
R f f = 236 k Ω and R 1 = 1 k Ω.
23.
(a) B must be the non-inverting input; (b) Choose R 2 = R B = 1 inverting input.
25.
vout(0.25 s) = 0.93 V
27.
4.2 V
=
N
29.
- R f
= 8 sin 10t ; (b)
-4 (1
+
vout = −10vin
vout = 2vin
= −10 − 5 sin 5t
= 2 + 0.5 sin 10t
sin sin 3t ) V ; -5.6 V
vi
∑ R i =1
31.
= −20 sin 5t ; (b)
i
Pick R 1 = 10 k Ω. Then vS = -0.21 V.
Ω;
(c) A is the
CHAPTER 6
ENGINEERING CIRCUIT ANALYSIS
33.
One possible solution of many:
35.
Set R = 10 k Ω:
SELECTED ANSWERS
Then connect several into:
after setting R f2 f2 = R f1 f1 = R in in = R =10 k Ω. 37.
1 kV
39.
-179 kV
41.
1.7 V
43.
R f f = 0, R in in = 100 k Ω, R 2 = 51 Ω.
CHAPTER 6
ENGINEERING CIRCUIT ANALYSIS
45.
R f f = 120 k Ω and R in in = 200 k Ω, R = 560 Ω.
47.
R = 400 Ω, R 1 = 82 Ω.
I
Is
49.
R = 91 Ω, R 1 = 560 Ω, 467 > R L > 67 Ω.
51.
(a) –3.7 mV; (b) 28 mV;
53.
55.
vout vin
=
- 100A 101 + A
vout = -16 mV
; A = 9999.
(c) –3.7 V.
SELECTED ANSWERS
CHAPTER 6
57.
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
(a)
(b) 5
-6
vout = 10 (-0.00004v2 - 9.99980×10 v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 – 1.99996 sin t 5 5 (c) vout = 10 vd = 10 × (v2 / 2 − va ) =0.99998v2-0.99998v1 = 1.99996 sin t
59.
(a) V3 = 27 V;
61.
Positive voltage supply, negative voltage supply, inverting input, ground, output pin.
63.
This is a non-inverting op amp circuit, so we expect a gain of 214.
65.
For vx = -10 mV, PSpice predicts vd = 6 μV, where the hand calculations based on the detailed model predict 50 μV, which is about one order of magnitude larger. For the same input voltage, PSpice predicts an input current of -1 μA, whereas the hand calculations predict 99.5vx mA = -995 nA (which is reasonably close).
67.
(a) Negative saturation begins at Vin = –4.72 V, and positive saturation begins at Vin = +4.67 V. (b) 40.6 mA.
69.
CHAPTER 6
71.
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
(a) 15 12 V 10 5 ) V ( t u o
0 -2
V
-1
0
1
2
-5 -10
-12 V
-15 V active (V)
73.
75.
⎛ R2 ⎞ R3 ⎟ ; (b) Vout = 0; (c) R = 4.3 k Ω and R = V ref ⎜⎜ − ⎟ ⎝ R1 + R2 R3 + RGauge ⎠ = 4.7 k Ω, gain of 5.39 for R = 4.7 k Ω, so R = 11.5 k Ω. (a) V out = V 1 − V 2
CHAPTER 7
ENGINEERING CIRCUIT ANALYSIS
1.
(a) 0; (b) -613sin120 πt mA; (c) -40e nA
3.
(a) 30 (1 − t ) e
5.
(a) 6.95 pF; (b) 17 kV; (c) 72
7.
Design problem: more than one solution. Hint:
9.
(a) 33.4 mV; (b) 33.4 mV; (c) 50.1 mV
11.
(a)
13.
(a) 2 k Ω; (b) 20 mJ
15.
(a) 0; (b) -613sin120 πt μV; (c)
17.
(a) 150 (1 − t ) e
19.
(a)
SELECTED ANSWERS
-t
− t
t mA ; (b) 4e−5 (100co 0cos10 s100t − 5si 5sin10 n100t )
100t 120sin 400 400t μ A ; (b) 6.4 μJ; (c) 400(1 − e )V ; (d) −120sin
− t
mA
vc
= 500 − 400e −100t V
−240e−6t pV
fV ; (b) 100e−5 ( 20co 0cos100t − sin1 sin10 00t ) t
pV
(b) 40 ms; (c) t = 20, 40 ms; (d) 2.5 J 21.
(a) 4t 2 + 4t V ; (b) 4t 2
+ 4t + 5 A
23.
(a) 2 A; (b) 5.6 J; (c) 1 A
25.
(a) 2.33 V; (b) 480 mA: (c) 1.1 A
27.
(a) 6.4 J; (b) 100 mJ; (c) Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V); (d) Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A)
29.
(a) 0, 400 mW
31.
4.3 μF
CHAPTER 7
ENGINEERING CIRCUIT ANALYSIS
33.
(a)
35.
Cequiv = 85 nF
37.
140 nF
39.
(a) 3 H; (b) N
41.
292 pH
43.
(a) 11.4 Ω; (b) 11.4 H; (c) 8.8 F
45.
(a)
SELECTED ANSWERS
(b) 3.6 V
(b)
(c)
-80t
mA; (b) 80e −80t − 60V ; (c) 20e −80 t + 60V
47.
(a) -6.4e
49.
9.2 V, 2.4sin10 t V
3
CHAPTER 7
ENGINEERING CIRCUIT ANALYSIS
1
idt + v ; (b) v ′ + c ∫
1+ A
SELECTED ANSWERS
+ Avs′ = 0
51.
(a) vs
=
53.
(a) vs
= 10.0sin1 10.0sin10 0t + 0.00 0.0005 05 − 0.00 0.0005co 05cos10 s10t ; (b) 10sin10t V
55.
(a) V out
57.
One possible solution of many (with C = 1 mF, R = 600 k Ω):
59.
One possible solution of many (with C = 1 μF, R = 1 MΩ):
61.
(a)
i
=
− R f
o
RC
vo
t
∫ v dt ' ; (b) Capacitor values are more readily available than s
L 0 inductor values.
(b) 20v20 + 1 5 ×10
1 5 × 10−6 t
−6
t
∫ (v o
c
∫ (v
20
o
− vc )dt + 12 = is
− v20 )dt − 12 + 10vc + 8 × 10−3 vc′ = 0
(c) iL
− is
20 ic − iL 10
+ 5 × 10−6 iL′ + +
1 8 × 10−3
t
iL
− ic
10
=0
∫ i dt + 2 = 0 o
c
CHAPTER 7
ENGINEERING CIRCUIT ANALYSIS
63.
65.
iout iS
=
G in (V2 - V1 ) 1 L1
t
∫ V d t ′ 0
1
+ G f V2
+ G in (V1 - V2 )
67.
32 J
69.
2.6 mJ
71.
221 μJ
73.
R = 1 Ω and L = 1 H
75.
558 pJ
SELECTED ANSWERS
CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS
1.
(a) 1.25 mA; (b) 740 mA; (c) -6.6 V; -6.6 V
3.
50 mH
5.
3.5 Ω
7.
(a) 4 A, 0 V; (b) 4 A, -48 V
9.
(a) 2e −400t A, t > 0 ; (b) 37 mA: (c) 1.7 ms
11.
(a) 6.9; (b) 2
SELECTED ANSWERS
13.
15.
6.3 k Ω, measuring to 5 τ
17.
(a) 100 s; (b) 366 nA
19.
(a) 4.999 V; (b) 4.998 mA; (c) 49.9 mJ
21.
(a) 9.95 Ω
23.
(a) 69 μs; (b) 35 μs
25.
20 V, 100 mA; 4.5 V, 0 A; 1 V, 0 A.
27.
(a) 2.7 A; (b) 1.9 A
29.
(a) 85 V; (b) 29 V; (c) 35 μs
31.
(a)
33.
5 A; 2.3 A; 1.9 A
35.
(a) 30 A; (b) 1.7 ms; (c) i(t ) = 30 30e −600t A ; (d)
37.
(a) 290 mA; (b) 200 mA; (c) 50 mA; (d) 277 mA; (e) 34 mA
iL (t )
0 .4 e−750 A, t > 0 = 0. t
−1440e−600 V ; (e) t
6e −600t + 14 A
CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS -100t
SELECTED ANSWERS
39.
(a) -6 mA; (b) 12e
mA
41.
(a) 20e
43.
(a) 99.8 V; (b) 88e
45.
(a) 100 V, 0; (b) 100 V, 100 V; (c) 80 ms; (d) 100e V; (e) 5e-12.5t mA; (f) −20e −12.5t + 80V , −80e −12.5t + 80V ; (g) 16 mJ, 100 mJ, 20 mJ
47.
(a) 20 mA; (b) 20e
49.
v (t )
51.
(a) 9; (b) 9; (c) 9; (d) 3; (e) 3
53.
1 A; 600 mA; 600 mA
55.
(a) 1; (b) 12; (c) 1.47
57.
2.5 A; 3 A; 2.5 A; 2 A; -2 A
59.
(a) 9.8 V; (b) 2 Ω
61.
(a)
iL (t )
−200000 = (2 ( 2 − 2e ) u (t ) mA; (b)
63.
(a)
iL (t )
= 4(1 − e−1000 )u (t )A ; (b) v1 (t ) = (100 − 80e−1000 )u(t )V
−250,000 t
V (t > 0 ) ; (b) 9.4 V -2539t
V -12.5t
−10000t
t
= 6u(t ) − 6u( t − 2) + 3u( t − 4)
V
t
6e
−200000t
u (t )V t
t
9
− 0.9e
−
6
10
t
65.
(a) i(t ) =
67.
2.5 V
69.
(a) 2 A; (b)
71.
(a) 80 mA; (b) 0.08(1 − e (d)
− 2e−5000 mA, t < 0
5
iL (t )
9
A; (b) 1.04 A
= 5 − 3e−40 A, t > 0 t
−25t
−25 > 0 ; (c) 0.16 − 0. 0.08e A, t > 0 ; 0.016co 6cos50 s50t + 0.032si 2sin50t − 0.016 e−25 A, t > 0
) A,
t
t
t
= 0.1 + (0.1 − 0.1e−9000 ) u (t )A t
73.
iL (t )
75.
(a) 3 A; (b) 2.4 A; (c) 2.6 A
77.
(a) 20(1 − e
−40t
) u(t )A ; (b) 10 − 8e
−40t
u (t ) A
CHAPTER 8 ENGINEERING CIRCUIT ANALYSIS
=
0.94
⎡⎣ −10e−0.1 + 10 cos 4t + 400 sin 4t ⎤⎦ 1601 t
79.
v (t )
81.
4.5 (1 − e −10t )
83.
iA
85.
(a)
87.
6.32 V; 15.7 V
89.
(a) 80 V; (b) 80 + 160e −100000t V,
91.
693 ns
93.
(a) 242 mV: (b) 3.11 mW; (c) 15 μJ
95.
1.0e
97.
vo(t )
99.
2.5 μF
101.
(a)
103.
8
= 10 + 7.5e−10 vc (t )
-t /10 /10
3
t / 10
5
= 10 + 7.5e−10
t
mA,
t
SELECTED ANSWERS
> 0 , i = 2.5mA t < 0 A
−500 = −8u (−t ) + (16 − 24 2 4e−500 ) u (t ) ; (b) −0.4u (t ) + (0.8 + 2.4e ) u (t )m A
u(t )
t
t
V -20×10 3t
= -0.2[1 + e
]u(t ) V
> 0 ; (c) 80 V; (d)
t
80 − 32e−20000t V,
>0
t
CHAPTER 9
ENGINEERING CIRCUIT ANALYSIS
1.
175 ×103 s −1 ; (b) 22.4 krad/s; (c) overdamped. (a) 175
3.
(a) 5×10 s ; (b) 32 Trad/s; (c)
5.
1.44 H; 14 mF; 4.9 Ω
7.
(a) 100 aF; (b) 1 MΩ; (c) 5 Gs-1; (d)
8
-1
−5 ×109 −
j 70.71×10
12
−0.5 ±
Grad/s ; (d) underdamped j32 Gra
−5 ×109 +
3
11.
−158.5t (a) 158 mΩ; (b) i (t ) = 4.169e
13.
(a)
15.
(a) v (t ) = −18 ⎡⎣e −0.069t
17.
2.025e −50
19.
v (t ) = 170e
21.
(a) 50 V; (b) –2 A; (c) vc (t ) = − 25 25e −2000t
23.
R < 160 ohms
25.
−3.2×10 t 3.2 ×10 6 t + 10 (a) 1.6 mΩ; (b) iL ( t ) = e
27.
(a) 8 mH: (b) 930 mA; (c) 24 ms
29.
160 ohms
31.
8.11 8.11×1013 cm
33.
e
35.
−5000 4 4 (a) e (20 (200cos1 0cos10 0 t + 100si 0sin10 n10 t ) V, t > 0 ;
4
− 0.169e−6.31×10 t
−20e −10t + 60e−40t V, t > 0 ; (b) 160e −10 t − − e −0.181t ⎤⎦
A
120 e −40t A
V ; (b) 8.6 s, -6.1 V
−8t
− 42e−2t , t > 0 + 75e −6000t , t > 0 ; (e) 270 μs; (f) 2 ms
(
)
(−2co 2cos200 s2000t + 4si 4sin 2000 t )A, )A, t > 0 t
−100t
−1
s ,
− 0.025e−450t A, t > 0
5
(b) 10 − e
12
-1
(a) 800 rad/s (b) 954×10 s ; (c) 95300%
−4000t
j 70.71×10
7.1 ×10−5 s -1 ; (e) 7.1
9.
t
SELECTED ANSWERS
−5000 t
(10 co cos 10 4 t − 7.5 sin 10 4 t ) mA mA, t > 0
sin 1000t mA, t > 0
37.
0.6e
39.
R = 10.4 ohms; 2.15 s
CHAPTER 9
ENGINEERING CIRCUIT ANALYSIS
41.
(a,b) e (4 cos 5t + 0.8 sin 5t ) A; (c) 4.7 s
43.
e
45.
(2.25e
47.
t t (a) 0.5e −10 A, t > 0 ; (b) 100e −10 V t > 0
49.
4.7 kV
51.
1.5 ohms; 23 J
53.
vC (t ) = e
−2500 t
55.
vC (t ) = e
−0.21t
57.
iL(t ) = 10 - e
59.
e
61.
− t 12 − e (t + 2) V, t > 0
63.
(a) 2.5e −500
65.
(a) 30 V; (b) 51 V; (c) 44 V; (d) 44 V
67.
(a) 0; (b) 0; (c) 920 mA; (d) -1.03 A
69.
1003 ohms
71.
one possible solution:
SELECTED ANSWERS
-t
−4 t
(10co 0cos2 t + 20si 0sin 2t )A, t > 0 −200 t
− 0.25e−6000t ) u(t ) + 2u(−t )V )V
⎡100 cos (1.6 ×105 t ) + 1.6 sin (1.6 ×105 t ) ⎤ ⎣ ⎦
cos18t + 0.14 si sin 18t ] [13 co -4t
−4000 t
V
V for t > 0 and 13 V, t < 0
(20 sin 2t + 10 cos 2t ) A, t > 0
(2co (2cos200 s2000t − 4si 4sin 2000 t )A, t > 0
t
− 22.5e−1500t mA, t > 0 ; (b)
25e −500
t
+ 22.5e−1500t mA, t > 0
CHAPTER 9
73.
(a)
75.
(a)
ENGINEERING CIRCUIT ANALYSIS
1
dv
=
-
diL
=
- 4iL ;
v 3.3 dt (b) one possible solution:
dt
(b) one possible solution:
SELECTED ANSWERS
CHAPTER 10
1.
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
5sin (29 (290.9t + 325.0°) ; (b) 8.5 cos(290. os(290.9 9t − 125 125 °) ; (a) f (t ) = 8.5sin
(c)
4.875 5 + cos cos 290. 290.9 9 t + 6.96 6.963sin 3sin 290. 290.9 9 t −4.87 o
3.
(a) 58, 57; (b) 134
5.
85 Mrad/s, 39 V, pi
7.
600t + 9 ) lags 6 cos (2π 6600t – 9 ) by 360 – 9 – 189 = 162 ; (a) -6 cos (2π 6 o o o (b) -cos ( t - 100 ) lags cos (t - 100 ) by 180 ; o (c) -sin t lags sin t by 180 ; o o (d) 7000 cos ( t – π ) lags 9 cos (t – 3.14 ) by 180 – 3.14 = 176.9 .
9.
(a) 800 mV; (b) 771 mV; (c) 814 mV; (d) 805 mV
11.
13.3 cos (5t – 89.6 ) μV
13.
743 cos (500 t – 22 ) mA
15.
(a) 26 μs; (b) 10 or 26 μs; (c) 16 or 26 μs
17.
12.5cos(500 t – 0.11 ) mA
19.
1.4cos(4 4cos(40 00t − 45 °) + 1.3co 3cos(20 s(200 t − 27 °) A
21.
(a)
23.
(a) 16.8 – j 5.9; (b) – j 204; (b) 0.31 + j 1.7
25.
(a) 18.7 ∠ -16 ; (b) 3.2 ∠ 46
27.
(a) 39∠ − 76° ; (b) 4∠ − 70° ; (c) 2.4 + j8.9 ; (d) 0.67 + j 0.21
29.
65e
31.
7.6∠113 113 ° A ; (c) 3.9∠ − 108° A ; (d) -65 V; (d) 54 V (a) 12∠20°A ; (b) 7.6
33.
35 mV
35.
o 6cos (5000t + 79 °) V ; (a) 18.3 cos (5000 t – 41 ) V; (b) 76cos (c) 58 cos (5000t + 118 °) V
37.
9.9cos 9cos (40 (400t + 79°) V
o
o
o
o
o
o
−ω Vm sin ω t = Ri′ +
o
(10 t +126 126° ) j (10
1
i ; (b) C
ω CVm
1 + ω 2 C2 R 2
⎛ ⎝
cos ⎜ ω t + tan
−1
⎞ ⎟ ω CR ⎠ 1
o
A
CHAPTER 10
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
39.
(a) –j292 Ω; (b) –j2.92 Ω; (c) –j292 mΩ; (d) –j292 nΩ
41.
(a) 478 + j176 Ω; (b) 588 + j120 Ω
43.
212 cos (800 t – 46 ) mA
45.
(a) 196∠ −11°Ω ; (b) 72 μF; (c) 11.3 and 444 rad/s
47.
R2
49.
(a) 10.6 – j1.9 Ω; (b) 10 + j0.25 Ω
51.
(a) 1 Ω + 4 H; (b) 5 Ω + 2 H + 500 mF; (c) 1.2 Ω + 69 mH; (d) 5 Ω
53.
(a) j88 mS; (b) j8.8 S; (c) j880 S; (d) j8.8 GS
55.
2 Ω, 2 H
57.
(a) 10 rad/s; (b) 10 rad/s; (c) 102 krad/s; (d) 52 krad/s, 134 krad/s
59.
(a) 250 μF; (b) 100 μF
61.
(a) 1 S + 250 mH; (b) 5 Ω || 1 F || 1 H; (c) 820 m Ω || 69 mF; (d) 5
63.
34∠23° V
65.
70cos( 70cos(10 1000 00t − 45°) V
67.
1.2cos( 2cos(1 100t − 76 °) A
69.
(a)
o
= 4.3 Ω, R1 = 3.2 Ω
5
Vo Vs
5
=−
71.
16 mW
73.
4.9 F
jω C1R f A
1 + A + jω C1R f
; (b)
Vo Vs
=
− jω C1R f A (1 + A) (1 + jω C f R f ) +
jω C1R f
⎛ ⎛ ⎞ ⎜ − ω ⎜ g C + C + C + C ⎞⎟ ⎟ ⎜ ⎜ m ′ ′ ⎟ RL R S ⎟⎠ ⎟ ⎝ −1 ⎜ tan ⎜ -1 ⎟ 2 2 ω (2C + C C ) + ⎜ ′ ′ ⎟ ⎜ RS R L ⎟ ⎝ ⎠ μ
75.
⎛ −
−1 (a) ang( Vout) = tan ⎜⎜
⎝
jω C μ ⎞
g m RS2
⎟⎟ ⎠
Ω
π
μ
μ
-
μ
μ
π
CHAPTER 10
ENGINEERING CIRCUIT ANALYSIS
ω
77. ω
+
(
2
j 2ω
− 1)
SELECTED ANSWERS
, 2 Ω, 2 H
79.
2.5 Ω, 1.25 H, 0.89∠ − 63 ° A
81.
158 158∠108 108 ° V , j150 Ω
83.
(a) 88 cos (t – 107 ) mV
o
85.
87.
89.
(a)
(b)
v1(t ) = 3.2×10 cos (2×10 t – 87 ) + 310×10 -3
4
o
-12
Vout VS
4
o
1 + 6.4 ×10
5
-12
o
cos(2 ×10 t – 93 ) V 5
o
91.
57∠ − 77° , 26∠ − 140° , 51∠ − 50° , 143 143∠13° , 51∠ − 140° , 51∠ − 140°
93.
-40.5 , 28
o
o
−24
2
ω
cos (2×10 t + 177 ) V and
v2(t ) = 31×10 cos(2×10 t – 177 ) + 116×10 -9
0.802
=
CHAPTER 11
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1. 117 W; 137 W; -19.7 W 3. -8 W; -0.554 W; W; 0.422 W 5. -23.5 W; 4.31 W; 32.1 W; -12.9 W 7. 54 kW; 7.31 kW; 134 W 9. 226 mW; 294 K, representing temp increase of 111 mK 11. 297 W; 0; 186 W; 0 13. 10.9 W; 20.8 W 15. 26 W 17. 8 + j14 Ω; 180 W 19. 96 W 21. 52 W; 15 W, 31 W 23. 289 W, 145 W; 90.3 W, 181 W 25. 54 W, 1.6 W, 0, 0 27. 1.4, 1.4, 1.4, 1.4 29. 4.04 A 31. 12.6 V; 12; 10 33. 8.5;12.4 35. 42.7 W; 25 W; 7.32 W; 55.2 W; 80.2 W 37. 30 V, 30 V; 34.6 V, 34.2 V 39. 9.88 41. 655 W; 320 W; 335 W; 800 VA; 320 VA; 568 VA; 0.6 lagging 43. 1230 VA, 774 VA, 86.5 VA, 865 VA, 3020 VA 45. 4.79 Arms; 0.91 lagging 47. 7.5 μF; 40 μF o 49. 211 + j442 VA, 289 + j0 VA, 0 + j192 VA, 562 + j0 VA, 640 – j390 VA, 142∠-90 VA 51. 1600 + j1800 VA; 0.66 lagging; 0.95 lagging 53. 70 kW; 81.4 kVA; 0.86 lagging P ( tan θ old - tan θ new ) 55. C = 2 ω Vrms 57.5.1 Arms; 1200 W; -1200 W; 1200 VA; 1200 VA; j1200 VA o o 59. 520∠3 kVA, 38 kVA, -j49.6 kVA, j77 kVA, 480 + j0 kVA; 520∠3 kVA; NO!; 520 kW; 28 kVAR
CHAPTER 12
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1. -9.3 V; -0.7 V; 9.3 V 3.
Van
o
Vdn
Van
o
Vdn
= |Vp| ∠ 0 o Vbn = |Vp| ∠ -60 o Vcn = |Vp| ∠ -120 = |Vp| ∠ 0 o Vbn = |Vp| ∠ 60 o Vcn = |Vp| ∠ 120 o
o
= |Vp| ∠ -180 o Ven = |Vp| ∠ -240 o Vfn = |Vp| ∠ -300 o
= |Vp| ∠ 180 o Ven = |Vp| ∠ 240 o Vfn = |Vp| ∠ 300 o
5. 56.7 ∠ -11.5 V; 190 ∠ 35.0 V 7.
The phase sequence is negative, since sequence is acbacb…. A positive sequence would be abcabc…
9. The temptation is to extend the procedure for voltages, but without the specific circuit topology, we do not have sufficient information to determine I31. o
o
o
o
11. 22.8 ∠ -18.5 A; 34.4 ∠ -12.1 A, 7.60 ∠ -109 A, 36 ∠ 180 A 91.5μ F ; 6.68 kVA 13. 91.5 o
15. 15. 1040 1040 W; 81.3 81.3 ∠ 144 V 17. 1.18 Ω ; 282 ∠ 20.8 ° V ; 450 ∠173° V ; 15.8 – j6.00 kVA 19. 6.80∠ − 96.1° A rms 21. 2.97∠17° A ; 52.8 W; 1990 W; 0.96 leading
CHAPTER 12
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
23. 0.894; 22 μF; 541 VAR 25. 346∠ − 30° V ; 48 − j 24 Ω ; 11.2∠86.6° A rms 27. 5.48 A rms; 3.16 A rms; 240 V rms rms; 60.5 60.5∠ − 170 170 ° A rms; rms; 36∠ − 30° A rms; rms; 4320 4320 + j4320 4320 VA 29. 40.2 ∠45° A rms;
31. 243 ∠30 °V ; 24 ∠ − 1° A rms ; 41.7 ∠ −31° A rms 33. 33.9 ∠45° ; 25.2 ∠ − 7.6° A ; 53 ∠ − 157° A rms ; 6100 + j 3300 VA 35. 1.54 kW; 2.16 kW; 615 W. 37. 186 W. 39. 862 W 41. We assume that the wire resistance cannot be separated from the load, so we measure from the source connection.
CHAPTER 13
ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1. M 21 = 663 μ H 3. 1 and 3, 2 and 4; 1 and 4, 2 and 3; 3 and 1, 2 and 4 5. 60.8 sin 800t pV; 36 sin 800t pV 7. −2300e − t + 3400e −3t A/s ; −1700e − t + 4600e −3t A/s ; i2 = 1700e − t + 4600e −3t A 9. −10.4 W; 4.8 W; 0 each; 0 11. 106 + j 76 Ω ; 25 W 13. ic (t ) =
30t (t + 0.01) 2 2
μ A,
t >0
15. (6 + j5ω) I – j2ω I – 6 I = 100, -j2ω I + (4 + j5ω) I – j4ω I = 0, -6 I - j4ω I + 1
(11 + j6ω) I = 0 ; 3
2
I 3
3
o
1
2
= 4.32 ∠ -54 A
⎡ 0.02ω2 ⎤ + jω ⎢ 0.1 − 17. ; 2.8 + j1.2 Ω 2 ⎥ + ω 25 + 0.25ω2 2 5 0 . 2 5 ⎣ ⎦ 0.2ω2
19. 19. 27.3 27.3co coss(10t + 69° ) V; 23.6 23.6co cos( s(10 10t + 66° ) V; 9.6 W, 5.76 W 21. 1.3 ∠ − 60° A 23.
2.16K 2 K 4 − 1.82K 2 + 1.1881
W
25. 0.84 W; 0.26 W; 1.1 W 27. 1.7 ∠42° ; 0.39 ∠ -80 ; 2.2 ∠0.05° o
29.
V2
=−
j1.7 k L1L 2 + 1
31. 4.56 − j 4 nΩ ; 10 + j 63 Ω 33. M = 5 H, L 1 = 9 H, L 2 = 11 11 H 35. 600 mH; 880 mH; 750 mH
3
1
2
CHAPTER 13
ENGINEERING CIRCUIT ANALYSIS
37. OC
SC
T×A Z oc
= jω4 M Ω
T ×B Z oc T×A Z in
= jω4 M Ω = − jω4 − j ω10 + j ω8 M Ω
T ×B
Z in
39.
SELECTED ANSWERS
;
T×A
Z SS
= Z STS× B = − jω4 M Ω + jω8 j ω10 M Ω ;
= jω26 j ω12 − jω8 M Ω j 4.9ω
1 + j 0.5ω
Ω
41. 25 + j 0.62 Ω ; j 24 Ω ; − j 25 Ω 43. 20 + j 31 Ω ; 20 + j 28 Ω ; 20 + j 25 Ω ; 21 + j 24 Ω 45. 192 W, 73 W, 61 W, 550 W 47. 8 W; 2.1 W; 5 kW 49. 0.89, 5 51. −9.2 V 53. 4.8 A 1 ⎛
2
⎞ ⎜⎜ ⎟ × 576 × Age ; half a century 57. IQ = 3 1000 ⎝ 28.8 × 10 + 576 × Age ⎟⎠ 120
59. You need to purchase (and wire in) a three-phase transformer rated at 3 (208 )(10 ) = 3.6 kVA.
)
CHAPTER 14 ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1. s = 0; s = ± j9 s ; s = -8 s ; s = -1000 ± j1000 s ; s = 0, s = ± 2 s -1
-1
-1
-1
–t 3. 8e ; 19; 11∠38o ; 1∠0° ; 1∠0° ; 1∠0° ; 88∠9°
5. 6.6 μ C ; 9 μ C ; No. 7. 8.1e
−3t
cos(15t − 60°) ; 8.1e
−3t
cos(15t − 60°) ; −4.1; −4.1
−1 1 − 9. 30∠230° V ; 36e 2 t cos (50t − 56°) V ; -19 V; s = −2 + j 50 s ; 2 − j50 s −
11. impedance R; an impedance Z L = sL = (σ + jω ) L , an impedance 1 1 = (0.002) = 20 ∠101 101 ° Ω ; Yes. Zc= ; Z R = 100 Ω. Z L = ( −2 + j10)(0. (σ + jω )C sC − t 13. 0.35∠ − 105° A ; 350e 2 cos(10t − 105°) mA −3t 15. 185∠ − 48° V ; 185e cos(4t − 48°) V
17.
19.
21.
23.
K s
5 s
;
3 s +8
5
K
;0;
1 − e− ) ; ( s 2s
s
5
1 − e− ) ( s 2s
8
8
8
⎡⎣1 − e−6−3s ⎤⎦ ; ⎡⎣1 − e6e −3s ⎤⎦ ; ⎡⎣1 − e −6−3s ⎤⎦ 2+s s−2 s+2
25. e −3t u(t ); ); δ (t ) ; t u(t ) ; 275 δ (t ) ; u(t ) − 27. 0.047 + j 0.11 ; −0.18 + j 0.20 ; (0.47 + j 6.5) × 10 10 3
29.
31.
33.
1 s
2 s
(e −2s − e −5s ) ;
4
(e −
2
s
− e − 3s ) ;
s
s
e −2 s ; F(s) =
e
−4s
;
3 s+2
4 s+3
e
− 4 s −8
e −2 s −6 ; 4e −2 s ; 2.9
− ; 3e −5 s ; −4e s
1 t ⎤ ⎡1 t − t t 90δ (t ) − 4.5u( t ) ; 11δ (t ) + 2u (t ) ; te u (t ) ; ⎢ e − − e −2 + e−3 ⎥ u(t ) 2 ⎣2 ⎦
CHAPTER 14 ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
35. 2.5 mA t 1 37. δ (t ) + u (t ) + 2e − u (t ) ; δ (t − 2) + 2δ (t − 1) + δ (t ) ; 2e − δ (t − 1) ; δ (t − 1) + δ (t − 5)
4 t 4 t 4 t t t t t 39. 5e − u (t ) ; (5e − − 2e − )u (t ) ; 6(e − − e − )u (t ) ; 6(4e − − e − )u (t ) ; t 4 t 18δ (t ) + 6(e − − 16e − )u (t )
3t 2t 6 t t 41. 2 u (t ) − 3 e − u (t ) ; 2δ (t ) + 4e − u (t ) ; 3δ (t − 0.8) ; 3(e − − e − )u (t ) ; 2 t 2 t 6t (3te − − 0.75e − + 0.75e − )u (t )
43. f (t ) = δ (t ) +
2 3
2
u (t ) −
3
cos(2 t +1 35 °) e −3 u (t ) ; f (t ) = 0.5tu(t ) + 0.25u( t) + 0.35 co t
t t t 45. h(t ) = δ (t ) − e −2 u (t ) ; h(t ) = 2e − − e −2 u (t ) ; 1 − t 9 −t 81 −3t d h(t ) = 2 δ (t ) + 6δ (t ) − te u (t ) + e u (t ) − e u (t ) 2 4 4 dt
47. f (t ) = (1.9 − 5.59e
−4t / 3
)u (t )
⎛ 20 130 −3t ⎞ + e ⎟u (t ) V 3 ⎝ 3 ⎠
49. 50 V; 0.1v c' + 0.2v c + 0.1(vc − 20) = 0 ; vc (t ) = ⎜ 51. (4 − 2e −0.15 )u (t ) t
t t 53. y (t ) = (2 + 6e − )u (t ) ; (6e − − 1)u (t )
∞
55. -600 mA; 40 = 100ic + 50
∫ i dt + 100 ;
57. R = 250 mΩ, C = 1 F, L =
1
0
−
3
59. ic (t ) + 0.4δ (t ) − 1.6e −2t u (t ) A 61. STABLE; UNSTABLE. 63. STABLE; STABLE. 65. 7 V ; 7 V 67. 2, 3, 0, 0, ∞ 69.
n
s+b s+a
; b − a, b − a
c
− 0.6e −0.5t u (t )
H ; v (t ) = (75e
−3t
− 12.5e −t − 62.5e −5t )u (t ) V
CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
1.
= 0.032s Ω 12
0.032s Ω
s
= 384 μV
3.
5.
7.
20s 2 + 11, 00 000s + 200, 00 000 s
+ 700s + 100 100,000 ,000
2
16s 2 + 50s + 4000 s
+ 80s
2
1000
;
s
2
+ 700s + 100,000 ,000
Ω ; 6.85∠ −114° Ω ; 910 mΩ ; 1 Ω
2
u(t )
μA –6.67t
v1(t ) = –5.6e + 3.6 V, t ≥ 0 –6.67t v2(t ) = –3.73e + 4.4 V, t ≥ 0
2
13. i1 (t ) =
3
−
1 6
e
− t 4
A, t ≥ 0
⎡( s + 2 ) + 100 ⎤ ⎡( s + 6 )2 + 100 ⎤ ⎣ ⎦⎣ ⎦ -6t
-2t
-0.61t
-0.55t
+ 0.79 e
2
3
2s + 17s + 90s + 185s + 250
1 12
e
− t 4
A, t ≥ 0
[0.092 [0.092 cos10 cos10t - 0.34 sin 10t ] A o
-3t
; [185 e
o
o
-1.25t
cos cos (4t - 48 ) + 86e
21. I1 = 271.7∠-96.5 A and I2 = 272∠-96.5 A o
+
cos(0.34t + 99 )] u(t )
200 200s(s 2 + 9s + 12) 12) 3
2
;
[0.092cos 2t - 1.5 sin 2t ] - e
17. [0.63e
4
i2 (t ) =
and
35s - 131 2
e
19.
; 0.16 − j 4.7
;
20s
Z L rπ R B Cπ C μ s + (g m Z L rπ R B C μ + rπ R B Cπ + rπ R B Cμ + Z L rπ Cμ +Z L RB Cμ )s + rπ + RB -0.28t
15.
−11 Ω ; 8.1∠54° Ω ;
20s 2 + 11,00 ,000s + 200,00 ,000
rπ R B (1 + Z L C μ s )
9. 4.5 e 11.
;
s + 200 s + 500
mA
o
cos (1.9t + 107 )] u(t )
CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS -142.8
SELECTED ANSWERS
o
o
-5
23.
t 2[1.301 e cos (742.3t + 12.54 ) + 0.00202 cos (2t – 6.538 ) – 6.601×10 -142.8t o o 2 - 1.564 e cos (742.3t – 33.56 ) - 2.998 cos (2t + 179.9 )] W
25.
(a)
δ(t )
(b) 2500s + 0.5 0.00 0.001 1s 2 + 5s + 500 500 -2.5×106t
(c)
[-3 e
-0.2t
+3e
7.5 7.5 × 106s + 1500 1500
Ω,
(
5000s + 5 ×10 s s + 5000 2
5
)
-3
V o
+ 3×10 + 21 cos(711t + 89.9 )] u(t )
27. 70 60s 2 + 19s + 3
V;
420s 4 + 133s3 + 21s 2
+ 60s + 9
70
Ω;
60s 2 + 19s + 3
420s 4 + 133s3 + 21s 2
+ 60s + 9
29. V1 = V2 =
3030 30303( 3(0. 0.22 2239 39 ×1013 + 0.16 0.1613 13 ×1013s + 9870 98700 0s 2) 10 s + 0.7 0.7732 ×10 s + 0.5691×10 s + 0.1 0.1936 ×10 ) s(0.4639 ×10 10 3
15 2
18
18
7609 7609(7 (705 0500 000 0s3 + 0.11 0.1175 75 ×1012s 2 + 0.63 0.6359 59 ×1014s + 0.88 0.8897 97 ×10 14) s(0.4639 639 ×10 s + 0.773 7732 ×10 s + 0.5691×10 s + 0.1936 ×10 ) 10 3
15 2
-
-165928t
v1(t ) = [3.504 + 0.3805×10 2 e -2
-165928t
v2(t ) = [3.496 – 0.1365×10 e
18
18
-739t
– 0.8618 e
-739t
+ 0.309 e
] u(t ) V,
-0.3404t
– 2.647 e
] u(t ) V
31. 9870 8700s3 + 1.645 ×1010s 2 + 1.21× 10 1013s + 2.059 059× 1012
;
-0.3404t
– 2.646 e
4.639 639 × 109s3 + 7.732 × 1014s 2 + 5.691× 1017 s + 1.93 .936× 1017
,
Ω,
2.05 2.059 9 ×1012 (5s - 3) s(4.639 ×109s3 + 7.732 ×1014s 2 + 5.691 ×10 10 17s + 1.936 ×10 17)
,
201 μA
A
CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS
5000s
33.
(s
2
+ 10
6
) (s + 6385 )
SELECTED ANSWERS
; [-0.76 e-6385t + 0.77 cos (103t – 8.9o)] u(t ) A;
3
o
0.77 cos (10 t – 8.9 ) A. 35. Poles at
−1 ±
1
Poles at s = ±1, − ± ± j 2, 1 ; zeroes at s = 0, ∞ .; Poles 4
j
3 4
, double at s = 0 , Zeroes at
∞
j 2,
37. (a.) -1 zeros at s = -25 -25 and and -12.5 s , and -1 poles at s = 0 and s = -1.7 -1.7 s . (b.) 5 -1 zeros at s = -9. 1 and -10 s , and 5 -1 poles at s = -1.55×10 and s = -3.2 s . Z in
39.
41. 0,
10 π
=
5(s + 1)(s + 4) 6(s + 1.5)
(1 − cos π t ) ,
20 π
− 1.5, ∞ ; − 1, − 4 s −1
;
, (10/ π ) (1 + cos π t t ), ), 0
43. 8t − 8 V, 16 V, zero 45. 15 s
- 15e -2s , 15 u(t ) – 15 u(t - 2),
u(t ) – 30 u (t - 2) + 15 u(t ), ),
15 s
15 2
s +9
2
-
−
15
e − 2 s , 15 t u(t ) – 15 u (t - 2),
s 15s 2
s +9
15 s
2
30 s
e −2 s
+ 15e − 4 s , 15 t
e-2s , 5 sin 3t u(t ) – 15 cos [3(t – 2)] u(t - 2),
47. h(t ) = δ (t ); ); 8e − u (t ) V t
49. (a)
−
(b)
(c) 5
CHAPTER 15 ENGINEERING CIRCUIT ANALYSIS
SELECTED ANSWERS
51.
; 4.7, 10; 15 cm 53.
10s3 + 30s 2 − 10s − 30 s
2
+ 6s + 13
; -2.3,
∞ ; 2.2 ∠117° , 2.2∠63°, 3.6∠34° , 5 ∠53° , 3∠0°
55.
;
100( 2 + jω ) (5 − ω 2 ) + j 2ω
; 100
57.
t t − 5e −6t A (all t ); ); [−5e −6 + e −2 (5 cos 4t + 3 sin 4t ]u (t ) A
59.
−1.7 and − 24 s -1 ;
-24t
iin(t ) = [10 – 2.1e
+4 ; graph; 2 rad/s, 69 − 6ω 2 + 25 ω 2
4
ω
-1.7t
– 0.885 e
] u(t ) A
61. 2.5 s
2
+ 6.75s + 2.5
-6.4t
; [1 + 0.066 e R + 10 5
-0.39t
– 1.1e
] u(t ) V
10s + 10 5
63.
− 5s H (s) = 5 s + 10
65.
0 F, 400 Ω ; 5 nF, nF, R 1 = 200 k Ω ; 50 nF, 200 k Ω ; R fA = 1 k Ω , C fA = 10 nF,
;
5s
;
s + 10
5
R fB = 100Ω , R 1B = R 1 A = 10k Ω 67.
One possible design: If we use a 1-μF capacitor, then R = 159 Ω. To complete the design, select R f f = 2 k Ω and R1 = 1 k Ω.
69.
One possible design: If we use 100-nF capacitors, then R = 3.167 k Ω.
CHAPTER 17
ENGINEERING CIRCUIT ANALYSIS
⎡ 4 −8 9 ⎤ ⎡ I1 ⎤ ⎡12 ⎤ ⎢5 0 −7 ⎥ ⎢I ⎥ = ⎢ 4 ⎥ ; (b) 651; (c) 21; (d) 600 mA; (e) -141 mA ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣7 3 1 ⎥⎦ ⎢⎣I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
1.
(a)
3.
(a) 390 W; (b) 250 W; (c) 180 W
5.
7.
SELECTED ANSWERS
2s 2 + 15s + 20 2s + 5 15s + 25 s(s + 4)
R1R 3 R 4
; (b) jω 0.8 × 103 Ω ( Lin
= 0.8 mH)
9.
(a) jω C
11.
142 mS, -77 mS
13.
40 mS, -40 mS, 40 mS, -30 mS
15.
(a) 50 Ω; (b) 60 Ω
17.
200 mS, -300 mS, -400 mS, 150 mS
19.
(a) Input is applied between g-s and output taken from d-s; 1 (b) jω ( Cgs + C gd ) , − jω C gd , g m − jω C gd , + jω ( C gs + C gd ) ; r d
R2
−3 1.4ω pS , 4.7 × 10 − − j1.4 − − 10 4 + jω (0.4 + 1.4) ×10 12 S
4.8ω pS , (c) j 4.8
21.
784 Ω, 1.72 k Ω, 367 Ω
23.
9.9 Ω
25.
(a) 56; (b) -9.6; (c) 530; (d) 3.4
27.
(a)
jω (1.4) ×10
Ω; (e) 35 Ω (b)
−12
S,
CHAPTER 17
ENGINEERING CIRCUIT ANALYSIS
29.
(a) 1.55 V; (b) -17.5 μA
31.
⎡ 7.5 ⎢−4.5 ⎣
33.
(a) -2; (b) 4; (c) 8; (d) 1 Ω; (e) 1.3 Ω
35.
SELECTED ANSWERS
1.1⎤
⎥ (Ω )
11 ⎦
= 133∠ − 48° Ω z12 = 94∠ − 2.6° Ω
= 9400 9400∠87° Ω z22 = 565∠ − 3.6° Ω
z11
⎡10Ω ⎢ 20 ⎣
z21
−2 ⎤ ⎡ 42Ω ; (b) ⎥ ⎢ 17 0.2 S⎦ ⎣
−1.7 ⎤ ⎥ 0.17 S⎦
37.
(a)
39.
(a) 1.2; (b) 9.6 Ω; (c) 240 mS
41.
(a)
⎡1000Ω ⎢ 10 ⎣
43.
(a)
jω r C ( jωC ) (1 + jω r C ) ( g − jω C ) r 1 +g + ; (b) 1 + jω r ( C + C ) 1 + jω r ( C + C ) r 1 + jω r ( C + C )
⎤ ⎥ ; (b) 8.6 k Ω 2 × 10−4 S⎦ 0.01
μ
π
π
m
μ
μ
π
m
π
(c) r x +
π
μ
π
r π
1 + jω rπ ( Cπ
(a)
⎡ 0.61 ⎢0.05 ⎣0.0533 S
47.
(a)
⎡ 1.4 ⎢0.2 S ⎣
49.
(a)
⎡ 1 ⎢1 / R ⎣
45.
π
+C
)
; (d)
μ
π
μ
d
π
π
jω Cμ r π
1 + jω rπ ( Cπ
+C
)
μ
3.3Ω ⎤
⎥ ; (b) 11 Ω
0.81 0.81 ⎦
⎡ 1. 5 ⎢ ⎥, 1 1 ⎦ ⎢ S ⎣6
2Ω ⎤
0⎤
⎡1 , ⎢ ⎥ 1 ⎦ ⎣0
R⎤
3Ω ⎤ 1
⎥, ⎥ ⎦
⎡1/ a , ⎢ ⎥ 1⎦ ⎣ 0
⎡ 11 / 7 ⎢1 / 7 S ⎣ 0⎤
4Ω ⎤ 1
⎥ ; (b) ⎦
⎡ 0.58 ; (b) ⎢ ⎥ a⎦ .115 S ⎣0.11
⎡ 4.71 ⎢0.96 ⎣ .962 S 14 Ω ⎤
⎥
4.5 ⎦
15.9 Ω ⎤ 3.47 .47
⎥ ⎦
μ
