Chapter 10 – Sinusoidal Steady-State Analysis
Exercises
Ex. 10.3-1
(a)
T = 2π / ω = 2 π /4
v leads i by 30 − ( −70) = 100°
(b)
Ex. 10.3-2
v = 3cos 4t + 4sin 4t =
Ex. 10.3-3
i = −5cos 5t +12sin 5t =
(
(3)2 +(4) 2 cos 4t − tan −1 4
3
)=5 cos(4t −53° )
12 

( −5)2 +(12)2 cos  5t −(180+ tan −1 ) =13 cos (5t −112.6° )
−5 

Ex. 10.4-1
KCL: i s = v R + C dv dt
dv
v
=
+
dt RC
16
Im
cosωt
C
Try v f t = Acosωt + Bsinωt & plug into above D.E.
1
6
⇒ − ωAsinωt + ωBcosωt + 1 RC Acosωt +Bsinωt =
equating sinωt & cosωt terms yields A =
16
v 1t6 =
∴ vf t =
f
Im
RI m
1+ ω 2 R 2 C 2
cosωt
C
and B =
ωR 2 C I m
1+ ω 2 R 2 C 2
ωR 2 C I m
RI m
ω
cos
t
+
sinωt
1+ ω 2 R 2 C 2
1+ ω 2 R 2 C 2
RI m
cos ωt − tan −1 (ωRC)
2 2 2
1+ ω R C
Ex. 10.4-2
KVL : − 10 + j3I + 2I = 0
jωL = j 3 × 1 = j 3
10
10∠0,
10
=
=
∠− 56.3,
,
2+ j3
13
13 ∠56.3
10
cos(3t − 56.3, )
∴ i (t) =
13
⇒I=
Ex. 10.5-1
10
2.36e j45
= 4.24e − j45 = 3− j3
255
Ex. 10.5-2
j 32
−3+ j8
=
32e j90
8.54e
j111
=
32
e j (90-111) = 3.75 e − j21
8.54
Ex. 10.6-1
,
jωt − j80°
}
(a) i = 4cos(ωt − 80 ) = Re{4e e
∴ I = 4e − j80° = 4∠− 80,
(b) i = 10cos(ωt + 20° ) = Re{10e jωt e j20° }
∴ I = 10e j20° = 10∠20°
(c)
i = 8sin(ωt − 20, ) = 8cos(ωt − 110, ) = 8Re{e jωt e − j110° }
∴ I = 8e − j110° = 8∠−110°
Ex. 10.6-2
(a)
V = 10∠ − 140° = 10e − j140°
∴ v(t) = Re{10e − j140° e jwt } = 10cos(ωt −140° )
V = 80 + j75 = 109.7∠43.2 ° = 109.7e j43.2°
(b)
∴ v(t) = Re{109.7e j43.2° e jωt } = 109.7 cos(ωt +43.2 ° )
Ex. 10.6-3
0.01
d
v + v = 10 cos 100 t
dt
(0.01)( j 100 )V + V =10
10
=7.071 ∠− 45°
V=
1+ j
v = 7.071 cos 100 t
A
256
Ex. 10.6-4
J
v s = 40cos100t = Re 4e j100t
KVL : i(t) +10 × 10 −3
di(t)
+
dt
L
1
5×10 −3
I
t
−∞
i(t)dt = v S
Assume i(t) = Ae j100t where i s is complex number to be determined
Plugging into D.E. yields
Ae j100t + jAe j100t + ( − j2A)e j100t = 4e j100t ⇒ A =
so β = tan −1
1
4
1− j
= 2 2 e j45°
= 45°
1
J
i(t) = Re2 2e j100t e j45° = Re 2 2 e j(100t -45° )
L=2
2 cos(100t + 45° )
Ex. 10.7-1
(a)
v = Ri = 10(5cos100t) = 50cos100t
(b)
v = L
di
I
= 0.01 5( −100) sin100t = − 5sin100t = 5cos(100t +90° )
I
dt
(c)
v = 1C idt = 103 5cos100tdt = 50sin100t = 50cos(100t − 90° )
Ex. 10.7-2
i=C
dv
−6
°
= 10 × 10 [100( −500)sin(500t+30 )]
dt
= −0.5sin(500t+30° ) = 0.5sin(500t+210° ) = 0.5cos(500t+120° )
Ex. 10.7-3
From Figure Ex. 10.7 - 3 we get
i(t) = I m sinωt ; I = I m ∠ − 90° A
v(t) = Vm cosωt ; V = Vm ∠0° V
i(t) = I m sinωt = I m cos(ωt − 90° )
The voltage leads the current by 90° , ∴ it is an inductor
⇒ Z eq =
V
I
=
Vm ∠0°
I m ∠− 90
°
=
also Z eq = jωL = ωL∠90°
Vm
∠90° Ω
Im
⇒ ωL =
Vm
Im
or
L=
Vm
ωI m
(H)
257
Ex. 10.8-1 ZR = 8 Ω, ZC =
1
1
j5
12
=
2.4 j 2.4
=
= − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω,
j
j× j
ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V.
Ex. 10.8-2 ZR = 8 Ω, ZC =
1
j3
1
12
=
j4
4
=
= − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω,
j j× j
ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V.
Ex 10.9-1
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
Ex 10.9-2
V1 (ω ) =
V 2 (ω ) =
8 ( j 6 ) j15
4 e = 19.2 e j 68
8 + j6
j12 ( − j 4 ) j15
4 e = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
258
Ex. 10.10-1
Va
V − Vb
+ a
=1
− j10
4 − j2
⇒ (4 − j12) Va + ( −4 + j2) Vb =−20 − j40
KCL at Va :
V −V
V
°
KCL at V : b a + b + .5∠ − 90 = 0 ⇒ ( −2− j4)Va + (2− j6)Vb =10+ j20
b
− j10
2+j4
( −20− j40) ( − 4+ j2)
(10+j20)
(2− j6) −200+ j100
=
= 5∠ 296.5°
Using Cramer's rule Va =
(4− j12) ( − 4+ j2)
−80− j60
( − 2− j4) (2-j6)
∴va (t)= 5 cos (100t+296.5° )= 5 cos (100t −63.5° )
Ex. 10.10-2
KVL a I1 : j15I1 + 10(I1 − I 2 ) = 20
⇒ (10+ j15)I1 −10I 2 = 20
(1)
KVL a I 2 : − j5I 2 +10(I 2 − I1 ) = − 30∠− 90°
⇒ − 10I1 + (10 − j5)I 2 = j30
(2)
From Cramer' s rule
20
−10
j30 10 − j5
I1 =
10+ j15 −10
−10
=
200 + j200
= 2.263∠− 81
.°
75+ j100
10 − j5
. ° ) = 24 2 ∠82 °
Now VL = ( j15)I1 = (15∠90° )(2.263∠− 81
∴ v L ( t) = 24 2 cos(ωt +82 ° )V
Ex. 10.10-3
Writing mesh equations:
(10 + j50)I1 −10I 2 = j30
−10I1 + (10 − j20)I 2 + j20I 3 = j50
j20I 2 + (30 − j10)I 3 = 0
Solving these equations gives
. + j1.27,
I1 = − 0.87 − j0.09, I 2 = −132
Then
I 3 = 0.5+ j1.05
Va = 10(I1 − I 2 ) = 14.3∠− 72 ° V
Vb = Va + j50 = 36.6 ∠83° V
259
Ex 10.11-1
V1 =
V2 =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
Zt =
8 ( j10 ) − j 2.4 ( j 20 )
+
= 4.9 + j 1.2
8 + j10 − j 2.4 + j 20
Ex 10.11-2
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
260
V1 (ω ) =
8 ( j 6 ) j15
4 e = 19.2 e j 68
8 + j6
V 2 (ω ) =
j12 ( − j 4 ) j15
4 e = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t - 22° )
Ex. 10.11-3
a) Turn off current source, use phasors with ω = 10 rad/sec
Z eq = − j
10⋅10
= 5(1 − j)
10 − j10
KVL a: −10 + 5I1 + j15I1 + 5(1− j)I1 = 0
⇒ I1 =
10
10 + j10
= 0.707∠− 45°
∴ i1 ( t) = 0.707cos(10t − 45° ) A
b) Turn off voltage source, ω = 0 rad/sec
Current divider I 2 = −
10
3 = −2 A
15
So by superposition i(t) = 0.707cos(10t − 45° ) − 2 A
Ex. 10.12-1 ω 2 =
1
1
=
= 106
LC
(1×10 −3 )(1×10 −3 )
∴ ω = 1000rad sec
Ex. 10.12-2
Diagram drawn with relative magnitudes arbitrarily chosen
261
Ex. 10.12-3
Two possible phasor diagrams for currents
∴ I LC =
425 9−415 9 = 20 = I
2
2
CL
Now if I LC = I L − I C ⇒ I C = 6 − 20 = − 14 ( impossible)
∴ from case (2)
I CL = I C − I L ⇒ I C = 20 + 6 = 26
Ex. 10.14-1
Z1 =
R 1X1 ( X1 − jR 1 )
and
R 12 +X12
∴ Z1 =
(1)(1)(1− j1)
1+1
Z 2 = R 2 = 1kΩ
∴
Vo
Vs
= −
Z2
Z1
=
1
2
=
−j
1
R1 = 1kΩ , X1 =
1
ωC1
=
1
110006410 9
−6
= 1kΩ
kΩ
2
−1
= −1− j
1 − j1
2
2
Problems
Section 10-3: Sinusoidal Sources
P10.3-1
(a)
i(t) = 2 cos(6t +120° ) + 4 sin(6t − 60° )
= 2 (cos6t cos120° − sin6t sin120° )+4 (sin6t cos60° − cos6t sin60° )
= 2.46 cos6t +0.27 sin6t = 2.47 cos(6t − 6.26° )
(b)
v(t) = 5 2 cos8t +10 sin(8t + 45° )
= 5 2 cos8t +10[sin8t cos45° +cos8t sin45° ]
= 10 2 cos8t +5 2 sin8t
v(t) =
250 cos(8t − 26.56° ) = 5 10 sin(8t + 63.4 ° ) V
262
2−π
= 6283 rad sec
ω = 2π f = 2π T =
1×10−3
P10.3-2
v(t) = Vm sin(ω t+φ ) = 100 sin(6283t+φ )
v(0) = 10 = 100 sinφ ⇒ φ = sin −1 (0.1) = 6°
So v(t) = 100 sin(6283t+6° )V
P10.3-3
ω
=
1200π
= 600Hz
2π
2π
i(2 ×10 −3 ) = 300 cos(1200π(2 ×10 −3 ) + 55° ) = 3 cos(2.4 π +55° )
f =
but 2.4 π ×
180 = 432
π °
°
So i(2 × 10 −3 ) = 300 cos(432 ° + 55° ) = 300 cos(127 ° ) = −180.5 mA
P10.3-4
P10.3-5
a)
A = 10
T = 3.9ms− 0.6ms = 3.3ms
2π
ω=
= 1900 rad s
T
10 cos (θ) = 0.87 ⇒ θ = 30°
v s (t) = 10cos (1900t +30° ) V
b)
A = 10
1
(10.9 ms− 0.9 ms) = 5ms
2
2π
ω=
= 1260 rad s
T
10 cos (θ) = 0.87 ⇒ θ = 30°
T=
v(t) = 10cos (1260+30° ) V
263
Section 10-4:
Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function
P10.4-1
L
Try i f
di
+ R i = − v s yields
di
+ 120i = − 400 cos 300t
dt
dt
= A cos 300t + B sin 300t
di f
= − 300 A sin 300t + 300B cos 300t
dt
A = − 0.46
yields − 300A +120B = 0
B = −1.15
and
300B+120A = −400
()
*
so i(t) = − 0.46 cos 300t −115
. sin 300t = 1.24 cos (300t − 68° ) A
P10.4-2
KCL : − i s +
v
+C
dv
= 0⇒
2
dt
Try v f = A cos 1000t + B sin 1000t
dv
+ 500v = 500 cos 1000t
dt
dv f
= −1000A cos 1000t +1000B cos 1000t
dt
yields −1000A + 500B=0
= 0.2
solving A
B = 0.4
and 1000B+500A=500
()
*
so v (t)=0.2 cos 1000t +0.4 sin 1000t=0.447 cos (1000t − 63° ) V
P10.4-3
(j4) (.05) = j(0.2)
I(ω )=
12e j45 ~12e j45
=(2⋅10−3 )e j45 ⇒ i(t)=(2) cos (4t+45° ) mA
6000+j(0.2) 6000
Section 10.5: Complex Exponential Forcing Function
a) Complex Numbers
P10.5-1
(5∠36.9 ° ) (10∠− 531
. ° ) 50∠−16.2 ° 10∠−16.2 °
=
=
= 2 5∠10.36°
(4 + j3)(6 − j8)
10 − j5
5∠− 26.56°
P10.5-2
5∠ + 81.87
3 2∠− 45° 
° 
°
°
 4− j3+
 = 5∠ + 81.87 [4 − j3+ 3 5 ∠ − 36.87 ]
°


5 2∠−8.13 
= 5∠+81.87° (4.48− j3.36) = 5∠+81.87° (5.6∠−36.87° ) = 28∠ +45° = 14 2 + j14 2
264
P10.5-3
A *C *
(3− j7)5e − j2.3
=
= 0.65− j6.31
B
6e j15
P10.5-4
(6∠120° ) ( −4 + j3 + 2e j15 ) = −12.1 − j21.3
so a= −12.1 and b= −213
.
P10.5-5
3− b j tan −1 −4 a) Ae j120 = −4 + j(3 − b) = 4 2 +(3− b) 2 e
120 = tan −1
3− b −4 ⇒ b = 3 + 4 + tan (120° ) = −3.93
A = 4 2 +(3− b) 2 =
b)
c)
4 2 + (3− ( −3.93)) 2 = 8.00
−4 + 8 cosθ + j(b + 8 sinθ) = 3e − j120 = − 15
. − j2.6
2.5
−4 + 8 cosθ = −15
= 72 °
. ⇒ θ = cos−1
8
b + 8 sin (72 ° ) = − 26 ⇒ b = −10.2
−10 + j2a = Ae j60 = A cos 60° − jA sin 60°
A =
−10
−20 sin 60
= −20; a =
= − 8.66
2
cos 60°
b) Response of a circuit
P10.5-6
Z R = 100, Z L = j(107 )(1 × 10 −3 ) = j 10,000, Z c =
Vs
I(ω ) =
=
Z R +Z L +Z C
01
. ∠90°
100 + j 10000 − j 10000
−j
(107 )(10 ×10 −12 )
= − j 10,000
= 0.001∠90°
°
i(t) = 1 cos(ωt +90 ) mA, ω = 107 rad /sec.
P10.5-7
Z L = j(25 × 106 )(160 × 10 −6 ) = j 4000
ZL =
−j
= − j 0.004
(25×106 )(10 ×10 −6 )
Z L / / Z c =− j 0.004
I(ω ) =
V
Z R +Z c
=
20∠45°
200 − j 0.004
= 01
. ∠45°
i(t) = 0.1 cos(ωt + 45° ) A
265
Section 10-6:
The Phasor Concept
P10.6-1 Phasor ckt: Z = R = 300, Z = jω = j(1 × 105 )(1 × 10 −3 ) = j100
R
L
L
Z
c
=
−j
−j
=
= − j100
5
ωC
(1×10 )(0.1×10−6 )
KCL:
V
ZL
+
V
ZR
+
V
= −I
ZC
yields V(ω ) = 1.5∠60°
so v(t) = 1.5 cos (ωt +60° ) V, ω = 105 rad sec
P10.6-2
Z R = R = 680Ω, Z L = jωL = j(1000)(500 × 10 −3 ) = j500
−j
−j
=
= − j303, I s (ω ) = 25×10 −3 ∠−120° A = 25∠−120° mA
ZC =
−6
ωC
(1000)(3.3×10 )
I 2 (ω ) =
Z I
Z C
S
T
Z T = Z1 + Z 2 = (680 + j500) + ( − j303)
I 2 (ω ) =
− j303 25∠−120 =
9
680+ j197 4
7575∠− 210°
°
708∠16
°
= 10.7∠ − 226° = 10.7∠134 °
°
so i(t) = 10.7 cos (1000t +134 ) mA
P10.6-3
Convert to phasor circuit:
Z R = R; Z C =
−j
ωC
=
−j
(500)(0.125×10 −6 )
= − j16000
VS = 2∠ − 90°
voltage divider
− j16000 2∠− 90 = 416000∠− 90 942∠− 90 9
V(ω ) = 9
25612∠− 39
20000− j16000 4
°
°
°
°
= 125
. ∠ − 141°
so v(t) = 1.25 cos (500t −141° ) V
266
Section 10-7:
P10.7-1
Phasor Relationships for R, L, and C Elements
(a)
Rotate 45° ⇒ I = 6 + j8 = 10∠53.1°
subtract 45°
I ′ = 10∠8.1° = 7 2 + j 2
(b)
Rotate 90° ⇒ I = 10∠53.1°
add 90°
I ′ = 10∠143.1° = − 8 + j6
P10.7-2
(a)
V1 = 3∠60° = 15
. + j2.598
V2 = 8∠− 22.5° = 7.391− j3.061
∴ V1 + V2 = 8.891− j0.463 = 8.90∠− 2.98°
⇒ v1 + v 2 = 8.90 cos(2t − 2.98° )
(b)
v1 = 2 2 cos (4t − 90° ) ⇒ V1 = 2 2 ∠ − 90° = − j2 2
V2 = 10∠30° = 5 3 + j5
∴V1 + V2 = 5 3 + j(5− 2 2 ) = 8.93∠14.1°
⇒ v1 + v 2 = 8.93 cos(4t +14.1° )
2A + 5B is pure imaginary and on the '+' imaginary axis
P10.7-3
∴ 2 A sin θ = 5B sin(75° − θ)
4 9
2 5 2 sin θ = 5B sin(75° − θ)
= 20[sin 75° cos θ − cos 75° sin θ)
⇒ tan θ =
sin 75°
°
1
2
+ cos 75
=1
∴θ =45°
so A = 5 2 ∠45° and B = 4∠90° +(75° − 45° ) = 4∠120°
P10.7-4
(a)
°
v = 15 cos(400t+30 )
i = 3 sin(400t+30° ) = 3 cos (400t −60° )
∴v leads i by 90° ⇒ element is an inductor
Now ZL =
(b)
v peak
i peak
i leads v by 90°
Zc =
v peak
i peak
=
=
15
= 5 = ω L = 400L ⇒ L = 0.0125 H = 12.5 mH
3
∴ capacitor
1
1
8
=4=
=
⇒ C=277.77µF
ωC 900C
2
267
v = 20 cos (250t + 60° )
(c)
i = 5 sin (250t +150° )=5 cos (250t +60° )
Since v & i are in phase ⇒ element is a resistor
∴R =
v peak
i peak
=
20
= 4Ω
5
P10.7-5
For algebraic addition, the rectangular form is convenient,
V1 = 150 cos( −30° ) + j150 sin( −30° ) = 130 − j75 V
V2 = 200 cos 60° + j200 sin60° = 100+ j173 V
By the rules for equality and addition
V = V1 + V2 = 230 + j98 = 250∠23.1° V
Thus v(t) = v1 (t)+v 2 (t) = 250 cos (377t +23.1° ) V
Section 10-8:
Impedance and Admittance
ω = 2 πf = 2 π(10 × 10 3 ) = 62830 rad sec
P10.8-1
Z R = R = 36Ω
Z L = jωL = j(62830)(160 ×10 −6 ) = j10Ω
−j
−j
=
= − j 16Ω
ωC
(62830)(1×10 −6 )
ZC =
1
= 0.0278 S
36
1
= − 01
YL =
. j S
ZL
YR =
YC =
Yeq = YP = YR + YL + YC = 0.0278 − j0.00375
Z eq =
1
1
= 0.0625 j S
ZC
S = 0.027 ∠9 °
= 36.5∠ 9 ° = 36 − j5.86 Ω
Yeq
P10.8-2
Z =
−10 ∠ 40°
V
°
=
= − 5000∠ − 155 Ω = 4532 + 2113 j = R + jω L
−I
2×10−3 ∠195°
so R =4532 Ω and L =
2113 2113
=
= 1.06 m H
ω
2×106
268
P10.8-3
j
L R
−j
(R+jω L)
ω
ωC
C
C
Z(ω )=
=
j
1 

−
+ (R+jω L) R+j ω L −

ωC
ωC 

−
2
1   RL − R  ω L − 1 − j R + L  ω L − 1  
 L R 

−
−
−
ω
j
R
j
L








ω C   ω C C 
ω C  
ω C   C ω C 
 C ω C 

=
=
2
2
1 
1 


R 2 + ω L −
R 2 + ω L −


ωC 
ωC 


So Z(ω ) will be purely resistive when
ωL− 1 = 0 ⇒ ω
ωC C
R2
L
R
=
− CL L 1
2
+
ωC
when R=6Ω , C=22µF, and L=27 mH, then ω =1278 rad /s.
2
ω = 2 πf = 6283 rad sec
P10.8-4
R
Zc R
jωC
Z = ZL +
= jωL +
1
R +Z c
R+
jωC
becomes (after manipulation)
R + j(ωL − ωR 2 C+ ω 3 R 2 LC 2 )
1+(ωRC) 2
Set real part equal to 100Ω to get C
Z=
R
= 100 ⇒ C=0.158µF
1+(ωRC) 2
Set imaginary part of numerator equal to 0 to get L
L − R 2 C + ω 2 R 2 LC 2 = 0 ⇒ L=0.1587 H
P10.8-5
ω = 2 πf = 6.28 × 106 rad sec
Z L = jωL= j (6.28 ×106 )(47 ×10 −6 ) = j 300
Z eq = Z c ||( Z R +Z L
1 300+ j300
6
jωC 1
)=
= 590.7
Rearranging 590.7 =
1
+ 300 + j300
jωC
300+300j
1+300jωC − 300ωC
or 590.7 − (590.7)(300ωC)+ j(590.7)(300ωC) = 300+ j300
equating imaginary terms
(590.7)(300ωC) = 300 ⇒ C=0.27 nF
269
Section 10-9: Kirchhoff’s Laws Using Phasors
P10.9-1
(a)
Z1 = 3+ j4 = 5∠53.1°
Z 2 = 8 − j8 = 8 2 ∠− 45°
(b) Total impedance = Z1 + Z 2 = 3+j4 + 8 − j 8 = 11 − j4 = 11.7∠− 20.0°
(c)
I=
100∠0°
Z1 +Z 2
=
100
117
. ∠− 20
°
=
100
∠20.0°
117
.
∴ i(t) = 8.55 cos (1250t +20.0° ) A
P10.9-2
Using voltage divider
V10 = Vs
°
2 ∠− 45 10
°
= 0 2 ∠45°
°
∴v10 ( t) = 10 2 cos (100t +45 ) V
P10.9-3
I =
(a)
160 ∠0°
( −1326) (300 + j37.7)
− j1326 + 300+ j 37.7
=
160 ∠0°
= 0.53 ∠5.9 °
303 ∠− 5.9 °
I=
160∠0°
( − j199)(300+ j251)
− j199 + 300 + j251
∴ i(t) = 0.53 cos (120πt +5.9 ° ) A
(b)
10 = 20∠0 10− j10 10
=
160∠0°
= 0.625∠59.9 °
°
256∠− 59.9
∴ i(t)=0.625 cos (800πt +59.9 ° ) A
P10.9-4
Vs = 2 ∠30°
I =
2 ∠30°
= .185 ∠ − 26.3°
6+ j12+3/ j
∴ i(t) = 0.185 cos (4t − 26.3° ) A
270
P10.9-5
j15 = j(2 π ⋅ 796)(3 ⋅ 10 −3 )
I=
12
= 0.48 ∠ − 37
20+ j15
i(t) = 0.48 cos (2 π ⋅ 796t − 37 ° )
P10.9-6
current divider I =
Z Z + Z I
1
so
I
Is
=
Z1 = R = 8, Z 2 = jωL = j3L
1
B ∠− 5187
. °
I = B ∠− 5187
. ° , I s = 2 ∠−15°
Z1
=
2 ∠−15°
s
2
Z1 + Z 2
=
8
8 + j3L
8 ∠0°
=
82 + (3L) 2 ∠ tan −1
3L 8 Set angles and magnitude equal
angles: +36.87 = + tan −1
magnitude:
8
2
64+9L
3L ⇒ L=2 H
8 =
B
.
⇒ B=16
2
P10.9-7
The voltage V can be calculated using Ohm's Law.
V = (1.72 ∠-69) (4.24∠ 45) = 7.29∠-24
The current I can be calculated using KCL.
I = (3.05 ∠ -77) - (1.72∠ -69) = 1.34∠ -87
Using KVL to calculate the voltage across the inductor and then Ohm's Law gives:
j 2L =
24 - 4(1.34∠-87)
⇒ L=4
3.05∠-77
271
Section 10-10: Node Voltage and Mesh Current Analysis Using Phasors
(a) Node Voltage Analysis
P10.10-1
Draw phasor circuit and use nodal analysis
( VA − VC )
10 +
j5 = 0
⇒ (2 + j)VA − 2 VC = j20
KCL at VA : − 2 +
KCL at VC :
VA
( VC − VA )
+
(1)
VC
− (1+ j) = 0
− j4
j5
⇒ 4VA + VC = 20 − j 20
(2)
Using Cramer' s rule
(2 + j)
Vc =
j20
4
20 − j20
60 − j100
116.6 ∠− 59 °
=
=
= 116
. ∠− 64.7 ° V
(2+ j) −2
10+ j
101 ∠5.7 °
4
1
P10.10-2
KCL at V:
(V −100)
V
V
V
+
+
=0
+
− j125 j80 250
150
⇒ V = 57.6 ∠22.9 °
100 − V
= .667 − .384 ∠22.9 ° = .347 ∠− 255
. °
150
V
= 0.461 ∠112.9 °
IC =
125 ∠− 90°
V
= 0.720 ∠− 67.1°
IL =
80∠90°
V
IR =
= 0.230∠22.9 °
250
∴ IS =
P10.10-3
V1 − Vs V1 V1 − V2
+
+
= 0
− j5 5+ j2
10
⇒ (11 + j2)V1 − (5 + j2)Vs = 10V2 (1)
KCL at V1 :
V2 − V1
V
+I+ 2 = 0
5+ j2
8+ j3
⇒ (8 + j3)V1 = (13+ j5)V2 + (34 + j31)I
KCL at V2 :
°
°
also V2 = 4I = 4(3∠45 ) = 12∠45 = 6 2 + j6 2
(2)
(3)
Plugging I and (3) into (2) yields
(8 + j3)V1 = 74.24 + j 290.62
∴ V1 =
300∠75.7 °
8.54∠20.6
°
= 351
. ∠551
. ° = 201
. + j28.8
272
Now plugging V1 and V2 into (1) yields
(5+ j2)Vs = − 209.4 + j 4731
.
∴ Vs =
517.4∠113.9 °
= 961
. ∠92.1° V
5.38∠218
. °
P10.10-4
Va
(2) into (1) yields Vb = 2.21 ∠ − 144 °
and from (2)
+
Va − Vb
= 0
200
j100
V −V
V
V −12
.
=0
KCL at Vb : b a + b + b
j100 − j50
j80
⇒ Va = (1 / 4) Vb − 3 / 2
KCL at Va :
(1)
(2)
Va = 0.55∠−144 ° −15
. = 197
. ∠−171°
∴ v a (t) = 1.97 cos (5000t −171° ) V
v b (t) = 2.21 cos (4000t −144 ° ) V
P10.10-5
KCL at Vo :
Vo
V
V − j20
+ o + o
= 6∠0°
4+ j3 3− j4
1
⇒ Vo = 16.31∠715
. ° V
v o ( t) = 16.31 cos (105 t + 715
. °) V
P10.10-6
ω = 104 rad s
I s = 20∠53°
1 + 1 + j V + − 1 V = 20∠5313
.
20 40 60 40 1 1 j j
KCL at b: − V + − + V − j80 V = 0
40 40 40 80 −j
1 j
KCL at c:
V + + V = 0
40 80 80
°
KCL at a:
a
b
a
b
b
c
c
(1)
(2)
(3)
Solving (1) - (3) simultaneously for Va
Va = 2 ⋅240∠45° ; thus v a ( t) = 339.4 cos (ωt +45° )
273
P10.10-7
v X = sin (2 π ⋅ 400t), ω = 2 π ⋅ 400
R=100Ω
L R = 40mH
LS = 40mH, door opened
= 60mH, door closed
with the door open →
VA −VB = 0 since bridge circuit is balanced
with the door closed → ZLR = j(800π )(0.04)=j100.5Ω
ZLS = j(800π )(0.06)=j150.8Ω
using nodal analysis
node B:
VB − VC VB
j100.5
+
=0 ⇒VB =
VC
R
ZL R
j100.5+100
VB =0.709∠ 44.86°
for VC = VX =1
node A :
VA − VC VA
+
=0 ⇒ VA =0.833∠33.55° for VC = Vx =1
R
ZLS
∴VA − VB = .833∠33.55° −.709∠ 44.86° = (.694+j.460) −(.503+ j0.500) = .191− j0.040
VA − VB = 0.195∠−11.83°
P10.10-8
node V1 :
node V2 :
V1 − ( −1+ j)
j2
V2 − V1
− j2
also : I C = 2I x = 2
+
V1
2
+
V2
− j2
+
V1 − V2
− j2
=0
−IC = 0
−1+ j "# =−1− j
! -2j #$
Solving (1) through (3) yields V2 =
(1)
(2)
(3)
−3− j
= 2 ∠−135°
1+ j2
∴ v(t) = v 2 (t) = 2 cos (40t −135° ) V
274
P10.10-9
V2 = 0.7571∠66.7 °
V3 = 0.6064∠ − 69.8°
(K
Using: I = I +I K
K
V −V K
I =
) yields
j10 K
KK
V
I =
− j2 K*
1
2
3
3
2
2
3
%K I = 0.3032 ∠20.2
∠−184
.
&K I = 01267
. ∠36
K' I = 0195
°
3
2
°
1
3
so i1 (t)=0.195 cos (2t +36° ) A
(b): Mesh Current Analysis
P10.10-10
KVL loop1: (4 + j6)I1 − j6I 2 = 12 + j12 3
KVL loop2: − j6I 1 + (8 + j2)I 2 = 0
(1)
(2)
Using Cramer' s rule to solve I1
I1 =
(12 + j12 3 )(8 + j2)
= 2.5∠29 ° = 2.2 + j1.2
(4+ j6)(8+ j2) − ( − j6)( − j6)
from (2) I 2 =
A
j6
6∠90°
(2.5∠29 ° ) =
(2.5∠29 ° ) = 182
. ∠105°
°
8+ j2
68∠14
Now VL = j6(I1 − I 2 )=(6∠90° )(2.5∠29 ° −182
. ∠105° )
= (6∠90° )(2.71∠−113
. °)
VL = 16.3∠78.7 ° V
and Vc = − j4I 2 = (4∠ − 90° )(182
. ∠105° ) = 7.28∠15° V
P10.10-11
Using voltage divider
V = 10∠0°
5+-j10 "#
!1+5− j10 $
= 9.59∠ − 4.4 °
Now VL = −100(100V) = −9.59 × 104 ∠ − 4.4 = 9.59 × 104 ∠175.6°
∴ v L ( t) = 9.59 ×10 4 cos (108 t +175.6° ) V
275
°
P10.10-12
Three mesh equations: (1 − j) I1 + (j) I 2 + 0I 3
=10
j I1 − jI 2 + jI3
= 0
0I1 + jI2 + (1− j) I3 = j10
(1− j)
j
0
So I2 =
(1− j)
j
0
∴
10
0
0
j
j 10 (1− j)
j
0
−j
j
j (1− j)
10 − j10
=
=10
1− j
i (t)=10 cos 103t A
P10.10-13
(2+ j4)
!
Using Cramer' s rule
I3 =
2+ j8
−1
−1
(2 +1/ j4)
− j4
−1
"# I "# 10∠30 "#
## I ## = 0 ##
(3+ j4) $ !I $
! 0 #$
− j4
−1
°
1
2
3
10∠30° = 3.225 ∠44 °
12+ j 22.5
V = 2 I3
∴ v(t) = 6.45 cos (105 t + 44 ° ) V
P10.10-14
ω = 400 rad sec
KVL : j 75I1 − j100I 2 = 375
(1)
KVL : − j 100I1 +(100+ j100)I 2 = 0
(2)
Vs = 375∠0°
Solving for I 2 yields I 2 = 4.5 + j1.5 ⇒ i 2 (t) = 4.74 ∠18.4 ° A
276
Section 10-11: Superposition, Thèvenin and Norton Equivalents and Source
Transformations
(a) Superposition Principle
P10.11-1
Use superposition
ω = 3000 rad sec
Z L = j1500
Vs = 12∠45°
ω = 4000 rad sec
Z L = j2000
I1 =
− V2
V1
−5∠0°
12∠45°
=
= 3.3∠11.3° mA; I 2 =
=
= 15
. ∠153° mA
3000 + j1500
ZT
3000+ j2000
ZT
i(t ) = 3.3cos (4000 t +11.3° )+1.5 cos(3000t +153° ) mA
P10.11-2
Use superposition
I ' (ω ) =
−1∠45°
6000 + j0.2
= −0.166 × 10 −3 ∠45°
I" =
3
= 0.5mA
6000
i(t) = i ' (t) + i '' = ( −0.166 cos (4t +45° ) + 0.5) mA
= [(0.166 cos (4t-135° ) +0.5)] mA
P10.11-3
Use superposition
4
v 2 = 5 sin 3t = 5 cos 3t − 90°
9
Note direction choice
'
I (t) =
12∠ 45°
°
= 2∠ 45 mA
6000 + j0.2
'
"
I (t) =
''
°
5∠−90°
°
= 0.833∠ − 90 mA
6000 + j0.15
°
i(t) = i (t) − i (t)=2 cos (4t+45 ) − 0.833 cos (3t − 90 ) mA
277
Section 10-11 (b): Thévenin and Norton Equivalent Circuits
P10.11-4
Find Voc
Using voltage divider
Voc = 5 ∠− 30°
80 + j80 = 5∠− 30 80 2∠− 219. 100∠36.90 80 + j80- j20 °
°
°
= 4 2 ∠− 219
. °V
Find Z T (Kill volatage source)
=
1− j206180 + j806
− j20 + 80 + j80
= 23 ∠ − 81.9 ° Ω
? have the equivalent circuit
P10.11-5
Find Voc
Voc − 2 Voc
V
− 10 + oc = 0
− j5
j 10
⇒ Voc = − j100 3 = 100 3 ∠ − 90° V
KCL at top:
Find I sc
Vx = 0 ⇒ I sc = 10∠0° A
ZT =
So have the equivalent circuit
Using voltage divider
100 ∠− 90°
Voc
3
=
= 10 3 ∠− 90° Ω
I sc
10∠0°
30 300
. ∠ − 83.66
30+ = − j 9 − j = 3313
Zt . cos 420t − 83.66 9 V
So v1 t 6 = 3313
V = Voc
°
V
°
278
P10.11-6
KVL loop I1: 600I1 − j300(I1 − I 2 ) = 9
Find Voc
⇒ (600 − j300)I1 + j300I 2 = 9
(1)
KVL loop I 2 : − 2V+300I 2 − j300(I 2 − I1 ) = 0
also V= − j300(I1 − I 2 )
⇒ j3I1 + (1 − j3)I 2 = 0
(2)
°
Using Cramer’s rule for equations (1) and (2) ⇒ I 2 = .0124∠−16
∴Voc = 300I 2 = 3.71∠−16° V
Find I sc
KVL loop I sc : − 2 V − V = 0
∴ I sc =
So Z T =
⇒ V=0
°
9 ∠0
= .015∠0°
600
Voc 3.71∠−16°
=
= 247∠−16° Ω
I sc
0.015
P10.11-7
+
V = 100 − j100 = 100 2 ∠ − 45°
−
20∠0°
16
= .1 2 ∠45° =.1+.1 j
100 2 ∠− 450°
1
1
+ = j 100C − 140 + 110
Also YIN = j100c+
j40 10
Equating the imaginary terms
⇒ C = 1800 F
100C − 140 = .1
YIN = 1 Z = I V =
IN
4
9
P10.11-8
a) Voc
Node equation :
Voc Voc − (6 + j8) 3 Voc − (6 + j8)
=0
+
−
− j4
j2
2
j2
yields Voc = 3+ j4 = 5∠531
.°
279
b)
Node equation:
I sc
yields V =
Thus I sc =
!
"#
$
V
V
V − (6 + j8) 3 V − (6 + j8)
+
+
−
=0
2 − j4
j2
j2
2
V 3+ j4
=
2 2 − j2
∴Z T =
3 + j4
1− j
Voc
2 − j2
= 3+ j4
= 2 − j2
3+ j4
I sc
P10.11-9
1
+ jωC
R
1
1
R
=
=
Z1 =
Y1 1 + jωC 1+ jωRC
R
Y1 =
Now Z IN = jωL +
R
1+ jωRC
R (1− ω 2 LC)+ jωL
1+ jωRC
C=39pF, ω =10 rad sec
8
with L=97.5nH,
Z IN =
=
R(1−.038) + j9.75 .962R+ j9.75
=
1+ j0.0039R
1+ j0.0039R
Z IN = 258
. ∠16.5° = 24.7 + j7.33Ω
for R=25Ω
for R = 50Ω
Z IN = 48.2∠0.43° = 48.2 + j0.36Ω
P10.11-10
Y = G + YL + YC
Y= G when YL +YC = 0 or
ωO =
1
LC
, fO =
1
=
1
jωL
+ jωC = 0
1
2 π LC 2 π 39.6 ×10 −15
= .07998 ×10 7 Hz = 800 KHz
(80 on the dial of the radio)
280
V1
Z1 = 50; I1 =
P10.11-11
also Z 2 =
I2 =
V2
Z2
1
jωC
=
Z1
=
100
=
25
= 0.5 A
50
1
j(2000)(2.5×10 −6 )
= − j200
= 0.5 A
200
and Z 3 = jωL = j(2000)(50 × 10 −3 ) = j100
I3 =
V3
50
=
= 0.5A
Z 3 100
Since I is the same for all three cases, Z thev and Z n must also be equal.
So Z t +Z1 = Z t +Z 2 = Z t +Z 3
or
(R +50) 2 + X 2 = R 2 + ( X − 200) 2 =R 2 + ( X+100) 2
which requires that (X − 200) 2 = ( X +100) 2 ⇒ X = 50Ω
Using this in (R +50) 2 + (50) 2 = R 2 +( −150) 2 ⇒ R=175Ω
so
Z t =175+ j50Ω and if Vt = Vt ∠0°
1
we get Vt = I1 Z t +R 1 =(0.5)[(175+50) 2 +(50) 2 ] 2 =115.25V
Section 10-11: (c) Source Transformation
P10.11-12
Z1 =
( − j3)(4)
= 2.4∠ − 531
.°
− j3+ 4
= 1.44 − j1.92
°
Z 2 = Z1 + j4 = 1.44 + j2.08 = 2.53∠55.3
Z 3 = 351
. ∠− 37.9 ° = 2.77 − j2.16
281
351
37.9 2.77. −∠−j2.16+2
. ∠− 37.9 9
4351
= 19
I = 2.85∠ − 78.4
. ∠− 92
45.24∠− 24.4 9
°
∴ I = 2.85∠ − 78.4 °
°
°
°
°
A
P10.11-13
ZP =
I=
0.4∠− 44 °
−4 j+100+ j4
(200)( − j4)
= 4∠ − 88.8°
200 − j4
= 4∠ − 44 ° mA; i(t)=4 cos (25000t − 44 ° ) mA
Section 10-12: Phasor Diagrams
P10-12-1
V = −4 + j3
V1 = 3 + j3
− V2 = −4 − j2
=5∠180° − tan −1
= 5∠143°
*
V3 = −3 + j3
3 4
P10.12-2
I =
10∠0°
10+ j1− j10
= 0.74∠42 °
VR = RI = 7.4∠42 °
VL = Z L I = (1∠90° )(0.74∠42 ° )
= 0.74∠132 °
VC = Z C I = (10∠− 90° )(0.74∠42 ° ) = 7.4∠− 48°
VS = 10∠0°
282
P10.12-3
I = 72 3 + 36 3∠(140° − 90° ) + 144∠210° + 25∠φ
= 40.08 − j24.23+ 25∠φ
= 46.83∠ − 3115
. ° + 25∠φ
Clearly for I to be maxima, the above 2 terms must add in same direction (in phase) ⇒ φ = −3115
. °
Section 10-14: Phasor Circuits and the Operational Amplifier
P10.14-1
Vo (ω )
 104 || − j104 
−j
10 − j 225
= −
=
e
H (ω ) =
 = −10
1− j
Vs (ω )
2
 1000 
 10 − j 225 
− j 225
e
Vs (ω ) = 2 ⇒ Vo (ω ) = 
 2 = 10e
 2

vo (t ) = 10 cos (1000t − 225°) V
P10.14-2
I1 =
VS − V1
R1
I1Z C = V1
− ( V1 − 0)
= I2
R2
V1 − V0
= I2
R3
Using equations (1) through (4) yields
(1)
(2)
(3)
(4)
R
1+ 3 R
V0
2
=
VS
1+ jωR 1C1
P10.14-3
The equations are:
VS − V1
(1)
= I1
ZC
V1
(2)
= I1
R1
V0
=
Solving these four equations yields
VS
jωCR 1 (1+
− ( V1 − 0)
= I2
R2
(3)
V1 − V0
= I2
R3
(4)
R3
R2 )
jωCR 1 +1
283
P10.14-4
ω = 62832 rad sec
= 2 πf
VS = 5∠0° mV
Z C = − j1.6
Equations:
VS − V1
175
V1 − 0
− j1.6
= I1
= I1
(1)
− ( V1 − 0)
(2)
1000
V1 − V0
10K
= I2
= I2
(3)
(4)
Using equations (1) through (4) yields V0 = 0.5∠− 89 °
or
v 0 (t) = 0.5cos(ωt − 89 ° ) mV
PSpice Problems
SP 10-1
Circuit:
v S = 10cos(6t + 45° ) V
iS =
2cos(6t + 60° ) A
Input file:
Vs
L2
R3
C4
Is
1
1
2
3
3
0
2
3
0
0
ac
ac
10
1
3
83.3m
2
45
60
.ac lin 1 0.9549 0.9549 ; ω = 6
.print ac Im(L2) Ip(L2)
.END
Output:
FREQ
9.549E-01
IM(L2)
2.339E+00
IP(L2)
-2.743E+01
So i(t) = 2.34 cos(6t − 27 ° ) A
284
SP 10-2
Circuit:
v S = 5cos(5t − 30° ) V
Input file:
Vs
L2
R3
C4
1
1
2
3
0
2
3
0
.ac lin 1 0.7958 0.7958
.print ac Vm(2) VP (2)
.END
Output:
FREQ
7.958E-01
ac
5
4
12
55.6m
-30
; ω = 5
VM(2)
3.082E+00
VP (2)
-1.005E+02
So v 0 (t) = 3.08 cos(5t-100° ) V
SP 10-3
Circuit:
v S = 200cosωt
i S = 8cos(ωt +90° )
ω = 1000 rad sec
Note: The 100m: resistor was added to provide the dc path to ground required by Spice. Since the resistance is so
large, it has little effect on the solution calculated.
Input File:
V1
R2
R3
L4
R5
C6
I7
R8
C9
R9
1
1
2
3
2
4
0
5
6
6
0
2
3
0
4
5
5
6
0
0
ac
ac
200
20
20
40m
40
50u
8
20
50u
100MEG
90
.ac lin 1 159.15 159.15
.print ac Im(R5) Ip(R5)
.END
Output:
FREQ
1.592E+02
IM(R5)
1.335E+00
IP(R5)
-7.018E+01
So i(t) = 1.34 cos(1000t - 70° ) A
285
SP 10-4
v S = 4cosωt
ω = 2 πf
= 2 π(1000) rad sec
Input file:
V1
C2
R3
R4
R5
R6
Vdummy
F7
R8
C9
R10
1
1
2
2
3
3
6
4
4
4
5
0
2
0
6
0
4
3
3
0
5
0
ac
4
10n
20K
1000
100
100K
0
100
10K
20n
5K
Vdummy
0
.ac lin 1 1000 1000
.print ac Vm(5) VP (5)
Output:
VM(5)
4.245E+01
VP (5)
-8.472E+01
So v(t) = 42.5 cos(ωt − 85° ) V
SP 10-5
Circuit:
i S = 2cos(3t +10° ) A
v S = 3cos(2t + 30° ) V
Input file:
Is
C2
R3
L4
Vs
0
1
1
2
0
1
0
2
3
3
ac
ac
2
500m
2
4
3
10
30
.ac lin 1 0.4776 0.4776
.print ac Im(C2) Ip(C2)
.END
Output:
FREQ
4.776E-01
IM(C2)
1.999E+00
IP(C2)
1.706E+01
So i(t) = 2 cos(2t +17 ° ) A
286
SP 10-6
Circuit:
v S = 5cos2t
i S = 5cos2t
Input file:
Vs
R2
C3
C4
L5
L6
R7
Is
1
1
2
2
2
3
3
0
0
2
0
3
3
0
0
3
ac
5
500m
500m
1
500m
250m
1
5
ac
0
0
.ac lin 1 0.3183 0.3183 ; ω = 2
.print ac Vm(2) VP (2)
.print ac Im(R7) Ip(R7)
.END
0utput:
FREQ
3.183E-01
FREQ
3.183E-01
VM(2)
2.236E+00
IM(R7)
4.472E+00
VP (2)
-2.657E+01
IP(R7)
6.344E+01
So v a (t) = 2.24 cos(2t − 27 ° ) V
and i(t) = 4.47 cos(2t +63° ) A
SP 10-7
Circuit:
v S = 4cos 500t V
See note on problem SP 10-3 for explanation.
287
Input file:
Vs
R2
R3
C4
Vdummy
H5
R6
C7
R7
1
5
2
2
1
3
3
4
4
0
2
0
0
5
2
4
0
0
ac
4
500
2K
200n
0
300
2K
200n
100MEG
Vdummy
0
.ac lin 1 795.8 ; w = 5000 rad/sec
.print ac Im(R2)
Ip(R2)
.END
Output:
FREQ
7.958E+02
IM(R2)
4.180E-03
IP(R2)
3.103E+01
So i(t) = 4.18 cos(500t +31° ) mA
SP 10-8
Circuit:
f = 60 Hz
°
Use 1∠0 A test source
V
Then Z =
1∠0°
Input file:
Is
0
L2
1
R3
2
R4
1
R5
1
C6
3
.ac lin 1 60 60
.print ac Vm(1)
1
ac
2
0
0
3
0
400u
; f = 60Hz
VP (1)
1
15m
8
5
6
0
.END
Output:
FREQ
6.000E+01
So
Z =
V
1∠0
°
VM(1)
2.784E+00
VP (1)
-3.831E+00
= 2.78∠ − 4 ° Ω
288
SP 10-9
Circuit:
v S = 120sin(ωt + 30° )
f = 10 kHz
Input file:
Vs
R2
L3
C4
1
1
2
3
0
2
3
0
ac
120
1000
100m
1000n
-60
.ac lin 1 10e3 10e3
.print ac Im(R2) Ip(R2)
.END
Output:
FREQ
IM(R2)
1.000E+04 1.891E-02
So i(t) 18.9 cos(ωt-141°)mA
IP(R2)
-1.409E+02
Verification Problems
VP 10-1 Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this
case, it is more convenient to do the division in rectangular form.
Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40
KCL at node 1: 2 −
V1
10
−
V1 − V 2
j 10
= 2−
− j 20 − j 20 − ( 20 − j 40 )
−
= 2+ j2−2− j2 = 0
j 10
10
KCL at node 2:
V1 − V 2
j 10
−
 V1  − j 20 − ( 20 − j 40 ) 20 − j 40
 − j 20 
+ 3  =
−
+ 3
 = (2 + j 2) − (2 − j 4) − j 6 = 0
j 10
10
10
 10 
 10 
V2
The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are
correct.
289
VP 10-2
I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180°
Generally, it is more convenient to multiply complex numbers in polar form.
Sometimes, as in this case, it is more convenient to do the multiplication in
rectangular form.
Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284
KVL for mesh 1:
8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − ( − j 5) = j 10
Since KVL is not satisfied for mesh 1, the mesh currents are not correct.
Here is a MATLAB file for this problem:
% Impedance and phasors for Figure VP 10-2
Vs = -j*5;
Z1 = 8;
Z2 = j*10;
Z3 = -j*2.4;
Z4 = j*20;
% Mesh equations in matrix form
Z = [ Z1+Z2
0;
0
Z3+Z4 ];
V = [ Vs;
-Vs ];
I = Z\V
abs(I)
angle(I)*180/3.14159
% Verify solution by obtaining the algebraic sum of voltages for
% each mesh. KVL requires that both M1 and M2 be zero.
M1 = -Vs + Z1*I(1) +Z2*I(1)
M2 = Vs + Z3*I(2) + Z4*I(2)
VP 10-3
V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V
KCL at node 1 :
19.2 ∠ 68° 19.2 ∠ 68°
+
− 4∠15 = 0
2
j6
KCL at node 2:
24 ∠105° 24 ∠105°
+
+ 4∠15 = 0
− j4
j12
The currents calculated from V1 and V2 satisfy KCL
at both nodes, so it is very likely that the V1 and V2
are correct.
290
Here is a MATLAB file for this problem:
% Impedance and phasors for Figure VP 10-3
Is = 4*exp(j*15*3.14159/180);
Z1 = 8;
Z2 = j*6;
Z3 = -j*4;
Z4 = j*12;
% Mesh equations in matrix form
Y = [ 1/Z1 + 1/Z2
0;
0
1/Z3 + 1/Z4 ];
I = [ Is;
-Is ];
V = Y\I
abs(V)
angle(V)*180/3.14159
% Verify solution by obtaining the algebraic sum of currents for
% each node. KCL requires that both M1 and M2 be zero.
M1 = -Is + V(1)/Z1 + V(1)/Z2
M2 = Is + V(2)/Z3 + V(2)/Z4
VP 10-4
ZP =
I=
(3000)( − j1000)
= 949 ∠ − 72 ° = 300 − j900 Ω
3000 − j1000
VS
100 ∠0°
=
= 0.2∠53° A
j500+Z P
j500+300 − j900
Now using current divider:
− j1000 40.2 ∠53 9= 63.3 ∠−18.5 mA
3000− j1000 3000 40.2 ∠53 9 = 190∠714. mA
=
3000− j1000 I1 =
I2
°
°
°
291
Design Problems
DP 10-1
R2
R2
1
=
jω C 1 + jω CR 2
R2
R2
R1
1 + jω CR 2
Vo (ω )
=−
=−
R1
1 + jω CR 2
Vi (ω )
R2
R1
Vo (ω )
j (180 − tan −1 ω CR 2 )
=
e
2
Vi (ω )
1 + (ω CR 2 )
In this case the angle of
tan (180 − 76 )
Vo (ω )
= 0.004 and the
is specified to be 104° so CR 2 =
1000
Vi (ω )
Vo (ω )
8
is specified to be
so
2.5
Vi (ω )
R2
R1
=
R2
8
⇒
2.5
= 132 . One set of values
R1
1 + 16
that satisfies these two equations is C = 0.2 µ F, R1 = 1515 Ω, R 2 = 20 kΩ .
magnitude of
DP 10-2
R2
Vo (ω )
=
Vi (ω )
where K =
R2
1
=
jω C 1 + jω CR 2
R2
1 + jω CR 2
K
=−
R2
1 + jω CR p
R1 +
1 + jω CR 2
R1
R1 + R 2
and R p =
R1 R 2
R1 + R 2
Vo (ω )
K
− j tan −1 ω CR p
=
e
2
Vi (ω )
1 + (ω CR p )
292
In this case the angle of
and the magnitude of
R1 R 2
Vo (ω )
tan ( 76 )
=
= 0.004
is specified to be -76° so C R p = C
Vi (ω )
R1 + R 2
1000
Vo (ω )
2.5
is specified to be
so
Vi (ω )
12
R2
K
2.5
=
⇒ 0.859 = K =
. One set of values that satisfies these two equations is
R1 + R 2
1 + 16 12
C = 0.2 µ F, R1 = 23.3 kΩ, R 2 = 142 kΩ .
DP 10-3
jω L R 2
L
R 2 + jω L
R1
Vo (ω )
=
=−
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
jω
R1 R 2
R1 + R 2
Vo (ω )
=
Vi (ω )
ω
L
R1
 L 
1 + ω


 Rp 
2
e

L 

j  90 − tan −1 ω

R p 

Vo (ω )
L L ( R1 + R 2 ) tan (90 − 14 )
=
=
= 0.1
is specified to be 14° so
Vi (ω )
Rp
R1 R 2
40
L
40
R1
Vo (ω )
2.5
L
2.5
=
⇒
= 0.0322 . One set
and the magnitude of
is specified to be
so
Vi (ω )
8
R1
8
1 + 16
In this case the angle of
of values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .
293
jω L R 2
DP 10-4
L
R 2 + jω L
R1
Vo (ω )
=
=−
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
R1 R 2
R1 + R 2
Vo (ω )
=
Vi (ω )
In this case the angle of
jω
Vo (ω )
is specified to be -14°. This requires
Vi (ω )
ω
L
R1
 L 
1 + ω


 Rp 
2
e

L 

j  90 − tan −1 ω

R p 

L L ( R1 + R 2 ) tan (90 + 14 )
=
=
= −0.1
Rp
R1 R 2
40
This condition cannot be satisfied with positive
DP 10-5
Y1 = 1/10
Z1 =10
Z 2 = 1 jωC
Y2 = jωC
Z 3 = R + jωL
Y3 = 1 R + jωL
(K
K)
KK
*
Use the fact that
admittances in parallel
add and V(∑Y) = I
So V(Y1 + Y2 + Y3 ) = I S with v(t) = 80 cos (1000t − θ) ⇒ V = 8∠ − θ
i S ( t ) = 10 cos 100t
So have 80∠− θ
⇒ I s =10∠0°
1 + 1 + jωc"# =10∠0
!10 R + jωL $
°
⇒ R +10 −10ω 2 LC+ j (ωL+10ωRC)=1.25R + j1.25ωL
Equate real part: 40 − 40ω 2 LC=R
(1)
Equate imaginary part: 40 RC=L
(2)
Plugging (2) into (1) yields
Now try R=20Ω
R=40(1− 4 ×10 7 RC 2 ) ω =1000 rad sec
⇒ 1− 2(1− 4 ×107 (20) C 2 )
which yields C=2.5×10 −5 F=25 µF
∴L=40 RC=0.02H=20mH
Now check θ : Y1 = 1/ 10 = 0.1
Y2 = j0.25
Y3 = 1/(20+ j20) = .025− j.025
∴Y=Y1 +Y2 +Y3 =.125 , so V=YI s =(.125∠0° )(10∠0° )=1.25∠0°
∴ θ =0° meets the design spec
294