Problems Section 10-2: Sinusoidal Sources
P10.2-1
(a)
i (t ) = 2 cos(6 t + 120°) + 4 sin(6 t − 60° )
= 2 (cos 6 t cos120°−sin 6 t sin 120°)+ 4 (sin 6 t cos 60 60°− cos 6 t sin 60 60° ) = −4.46 co cos 6 t + 0.27 si sin 6t = 4.47 cos(6 t − 176.5° ) (b)
°
v(t ) = 5 2 cos 8t + 10 sin(8t + 45 ) ° ° cos 8t + 10[sin 8t cos 45 45 + cos 8t sin 45 45 ] = 5 2 co
= 10 2 cos 8t + 5 2 sin 8t v(t ) =
°
°
250 cos(8t − 26.56 ) = 5 10 sin(8t + 63.4 ) V
P10.2-2
ω = 2π f =
2π
2π
= 6283 rad sec − 3 1×10 v(t) = Vm sin(ω t + φ ) = 100 sin(6283 t + φ ) T
=
−
1 (0) = 10 = 100 sin φ ⇒ φ = sin (0. (0.1) = 6.7 6.74° v(0)
v(t ) = 100 sin(6283t + 5.74°) V P10.2-3
f =
ω
=
1200π
= 600 Hz 2π 2π −3 −3 i (2 × 10 ) = 300 cos(1200π (2 × 10 ) + 55° ) = 3 cos(2.4π + 55° ) ⎛ 180° ⎞ −3 +55°) = 300 cos(127°) = −180.5 mA ⎟ = 432° ⇒ i (2 ×10 ) = 300 cos(432°+5 ⎝ π ⎠
2.4π × ⎜
P10.2-4
P10.2-5
A = 18 V T = 18 − 2 = 16 ms
ω =
2π
2π
=
0.016
T
16 = 18 cos (θ )
= 393 rad/s
⇒ θ = 27°
v ( t ) = 18 cos ( 393 t + 27° ) V
P10.2-6
A = 15 V T = 43 − 11 = 32 ms
ω =
2π
T
=
2π 0.032
8 = 15 cos (θ )
= 196 rad/s
⇒ θ = 58°
v ( t ) = 15 cos (196 t + 58° ) V
Section 10-3: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Fo rcing Function P10.3-1 di L + R= i sv dt di f
Try i f = A cos 30 300 t + B sin 30 300 t then equating coefficients gives
Then
⇒
dt
di dt
+ 120 =i 400cos300 t
= −300 A sin 30 300 t + 300 B cos 30 300 t . Substituting and
−300 A+120 B = 0 ⎫ A = 0.46 ⎬ B = 1.15 300 B +120 A = 400 ⎭
300 t + 1.15 sin 30 300 t = 1.24 cos (3 (300 t − 68 °) A i (t ) = 0.46 cos 30
P10.3-2 v
dv
2
dt
−is + + C
=0 ⇒
Try v f = A cos1000 t + B sin 1000 t then
dv f
dt Substituting and equating coefficients gives
dv
+ 500 v = 500 cos 1000 t
dt
= −1000 Asin 1000 t +1000 B cos1000 t .
−1000 A+500 B =0 ⎫ ⎬ ⇒ 1000 B +500 A=500 ⎭ Then
A = 0.2 B = 0.4
cos 1000 t + 0.4 sin 1000 t = 0.447 co cos (1 (1000 t − 63 °) V v(t ) = 0.2 co
P10.3-3
I(ω ) =
12 e j 45° 6000 + j (0.2)
~
12 e j 45° 6000
= (2 ⋅10−3) e
j 45°
⇒ i(t ) = 2 cos (4 t + 45 °) mA
Section 10.4: Complex Exponential Forcing Function P10.4-1
(5∠36.9°) (10∠−53.1°) (4 + 3j) +(6− 8j)
=
50∠−16.2 ° 10 − 5j
=
50∠−16.2 ° 5 5∠− 26.56°
= 2 5∠10.36°
P10.4-2
⎡
5∠81 81.87° ⎢ 4 − j3+
⎣
3 2∠− 45° ⎤
3
81.87°[4 − j3 + ∠ − 36.87°] ⎥ = 5∠81 5 5 2∠−8.13° ⎦ = 5∠81.87° (4.48 − j3.36) = 5∠81.87° (5.6∠− 36.87°)
= 28∠45°= 14 2 + j14 2 P10.4-3
A*C* B
=
(3− j 7) 5e− 6e
j 2.3°
j15°
= 0.65 − j 6.31
P10.4-4 j (6∠120° ) (−4 + 3j + 2 e15 ) = −12.1 − 2j 1.3 ⇒
P10.4-5 (a)
j120
Ae
= −4 + (j3 − b) = 4 + (3− b) 2
−
2
j tan
e
−1 ⎛ 3− b ⎞ ⎜ ⎟ ⎝ −4 ⎠
⎛ 3−b ⎞ ° ⎟ ⇒ b = 3 + 4 tan (120 ) = −3.93 ⎝ −4 ⎠
120 = tan 1 ⎜ A =
(b)
42 + (3−b) 2 =
4 2 + (3−( −3.93)) 2 = 8.00
−4 + 8 cos θ + (j b+ 8 sin θ ) = 3 −e j120 = − 1.5 − j 2.6 −4 +8 cos θ = −1.5 ⇒
θ
= cos−1
2.5 8
=
72°
sin (72 (72° ) = − 2.6 2.6 ⇒ b = −10.2 10.2 b + 8 sin
(c)
a= −12.1 and b= −21.3
−10 + j2 a = A e j 60° = Acos 60° − j Asin 60° −10 −20 sin 60° A = = −20 and a = = −8.66 cos 60° 2
P10.4-6
⎛ ⎝
5 ⎜ 0. 1
⎞ dt ⎠ d
v ⎟ + v = cos 2 t
d
⇒
v + 2 v = 2 cos 2 t
dt
Replace the real excitation by a complex exponential excitation to get d dt d
j t Let ve = A e 2 so
dt
j 2t
ve = j 2 A e d dt
(
v+2v =2e
j 2t
and
ve + 2 ve = 2 e
j2 t
j t
⇒ j t
j2 + 2 ) A e 2 = 2 e 2
⎛ 1
e−
⇒
ve = ⎜
Finally
v ( t ) = Re {ve } =
j2 t
2 + j 2 1
j2 t
=
⎟e ⎠
+ 2A e 2
A=
j45° ⎞
so
⎝ 2
j2A e
e
2
1 2
=
j2 t
= 2 e j 2t
1 2
∠ − 45°
j ( 2t −45°)
°) V cos ( 2t − 45
P10.4-7
0.45
d dt
v + v + 0.15
d
2
dt
v = 4 cos 5 t
2
d
⇒
2
dt
v +3
2
20 80 v+ v= cos 5 t dt 3 3 d
Replace the real excitation by a complex exponential excitation to get d2 dt 2
Let ve = A e d2 2
dt
j 5 t
v+3
so d dt
d dt
v+
ve = j 5 A e
20 3
v=
80 3
j 5 t
e
v+3
, and
j5 t
d dt
d 2 dt 2
v+
20 3
v=
80 3
ve = −25 A e
j 5t
e
j 5 t
(
)
⇒ − 25 A e j5 t + 3 j 5 A e j5 t +
20
80 5 Ae 5 )= e ( 3 3 j t
j t
80 20 ⎞ 80 j5 t ⎛ j5 t j A e e 2 5 1 5 − + + = ⇒ ⎜ ⎟ 3 ⎠ 3 ⎝
so Finally
(
ve = 1.126 e
− 1j 41°
3
A=
−25 + j15 +
)e
20
=
80
−55 + j 45
3
j 5t −141° ) = 1.126 e (
j5 t
v ( t ) = Re {ve } = 1.126 cos ( 5 t − 14 1° ) V
= 1.126∠ − 141
Section 10-5: The Phasor Concept P10.5-1
Apply KVL 6i+2
d dt
i − 15 cos 4 t = 0
or 2
d dt
i + 6 i = 15 cos 4 t
j ( 4 t +θ ) 4 t } and 15 cos 4 t = 15 Re{e } to write Now use i = I m Re{e
2
d
( dt
(j4 +t θ )
Im Re{ e
) (
} +6
(j4 +t θ )
Im Re{ e
)
} = 15 Re{e j 4 t }
⎧ d ⎫ j t j j t j j t Im e 4 e θ ) + 6 ( Im e 4 e θ ) ⎬ = Re{15 e 4 } ( ⎩ dt ⎭
Re ⎨ 2
{
}
j t j j t j j t Re 2 ( j4 Im e 4 e θ ) + 6 ( Im e 4 e θ ) = Re{15 e 4 }
(
j θ
j8 Im e j θ
I m e
=
15 6 + j8
)+6( I
m
=
e
15 °
10∠53
j θ
) = 15
= 1.5∠ − 53°
i ( t ) = 1.5 cos ( 4 t − 53° ) A
Finally v (t ) = 2
d dt
i (t ) = 2
d
(1.5 cos ( 4 t − 53° ) ) = 3 ( −4 sin (4 t − 53 °) ) dt = −12 ( cos ( 4 t − 143° ) ) = 12 cos ( 4 t + 37° ) V
P10.5-2
Apply KCL at node a:
v − 4 cos 2 t
1
+ 0.25
d dt
v+i = 0
Apply KVL to the right mesh: 4i + 4
d dt
i −v = 0 ⇒
v = 4i + 4
d dt
i
After some algebra: d
2
dt 2
i+5
d
i + 5 i = 4 cos 2 t dt
j ( 2 t +θ ) j 2 t } and 4 cos 2 t = 4 Re{e } to write Now use i = I m Re{e
d2
d (j 2 +t θ ) ⎤ ⎡ ⎡ Im Re{ e (j2 +t θ )}⎤ + 5 ⎡ Im Re{ e (j2 +t θ )}⎤ = 4 Re{e j 2 t } + I R e { e } 5 2 ⎣ m ⎦ ⎦ ⎣ ⎦ dt dt ⎣
⎧ d2 ⎫ j( 2 t+ ) ⎤ + 5 d ⎡ Im e j(2 +t ) ⎤ + 5 ⎡ Im e j( 2 +t ) ⎤ ⎬ = Re{4 e j 2 t } Re ⎨ 2 ⎡ Im e ⎣ ⎦ ⎦ ⎣ ⎦ dt ⎣ ⎩ dt ⎭ j j t j j t j j t j t Re {−4 e I m e 2 + 5 ( j2 e I m e 2 ) + 5 e Im e 2 } = Re{4 e 2 } θ
θ
θ
θ
θ
θ
−4 e Ij m + 5 ( j 2 e Ij m ) + 5 e Ij m = 4 θ
j θ
I m e
=
4
−4 + 5 ( j 2 ) + 5
θ
=
4 1 + j 10
θ
=
4 10.05∠84
= 0.398∠ − 84°
i ( t ) = 0.398 cos ( 2 t − 85°) A
(checked 7/6/05)
P10.5-3
VS = 2∠ − 90° V Z R = R; ZC =
− j
=
ω C
−j −
(500)(0.125×10 6 )
= − j 16000 Ω
(16000∠−90° )( 2∠−90 ° ) − j 16000 ⎞ ∠ − ° = = 1.25∠ − 141° V 2 9 0 ( ) ⎟ j − ∠ − ° 2 0 0 0 0 1 6 0 0 0 2 5 6 1 2 3 9 ⎝ ⎠ ⎛
V (ω ) = ⎜
therefore v ( t ) = 1.25cos ( 500t − 141°) V
P10.5-4
0.01
d dt
v + v = 10 cos 100 t
( 0.01)( j 100 )V + V =10 V=
10 1+ j
=7.071 ∠ − 45°
v = 7.071 cos (100 t − 45° )
V
P10.5-5
{
vs = 40 cos 100t = Re 4 e
KVL:
i (t ) + 10 × 10
−3
di (t ) dt
+
1 5×10−3
}
j 100 t
t
∫
−∞
i (t ) dt = vS
Assume i (t ) = A e j100 t where A is complex number to be determined. Plugging into the differential equation yields j100 t
Ae
+ j A ej100 t + (− j2 A) ej100 t = 4 e1j 00 t ⇒
A=
4 1− j
= 2 2 ej45°
In the time domain:
{
i (t ) = Re 2 2 e
1j 00 t
e
j45°
} = Re {2
2e
t 45° ) (j 100 +
}=2
° 2 cos (100 t + 45 ) A
Section 10-6: Phasor Relationships for R, L, and C Elements P10.6-1
P10.6-2
P10.6.3
P10.6-4
P10.6-5 (a)
v = 15 cos (4 ( 400 t + 30 °) V i = 3 sin(400t+3 0t+30°) = 3 cos (4 (400 t − 60 °) V v leads i by 90°
ZL
(b)
vpeak
15
=
=
vpeak
element element is an inductor inductor =
3
ipeak
i leads v by 90°
Zc
(c)
=
⇒
=
ipeak
8
5
⇒ =
2
4=
= ω
L= 400 400 L ⇒ L = 0.012 .0125 5 H = 12.5 12.5 mH
the element is a capacitor 1
=
ω C
1 900 C
⇒ C =277.77 μ F
v = 20 cos (250 t + 60 °) V i
=
5 sin (2 (250 t +150 °) =5 cos (2 (250 t +60 °) A
Since v & i are in phase ∴ R =
vpeak ipeak
=
20 5
⇒
=4 Ω
element is a resistor
P10.6-6 V1
= 150 cos( −30 °) + j150 sin( −30 °) =
V2 = 200 co cos 60 60°+ j 200 sin 60 60 ° V = V1 + V2
Thus v(t )
=
=
230 + j 98
v1 (t ) +v2 (t )
=
=
= 100 +
130 − j75 V j173 V
250∠23.1 ° V 250 cos (377 t +2 3.1 °) V
P10.6-7 −
1 R j= Z 20 C
=
V I
=
20∠15° 1.49∠63°
=
20 1.49
∠
(15° − 63°) = 13.42∠ − 48° = 8.98 −
Equating real and imaginary parts gives R = 9 Ω and C =
1 20 × 9.97
=5
j 9. 9 7 Ω
mF .
P10.6-8 V=
( 4j ) (15 ) ( I 1 + I 2 ) =
60j (0. 0 .03∠45 °) − 0.04 ∠0 ° = 60j (0. 0 .0212 + 0.j0212 − 0.04 ) = −1.273 − j1.127 = 1.7∠ − 138.5°
V
so v ( t ) = 1.7 cos ( 4t − 138.5° ) V
(checked: LNAP 8/7/04)
Section 10-7: Impedance and Admittance P10.7-1 ω
= 2π f = 2π (10 ×10 3 ) = 62832 rad sec
Z R = R = 36 Ω ZL =
ω j
ZC =
⇔ YR =
1 ZR
=
1 36
−
L = (j62830)(160 × 10 6 ) = 1j 0.053 ≈ 1j 0 Ω
− j ω C
=
−j (62830)(1×10−6 )
= 0.0278 S ⇔ YL =
= − 15j.915 ≈ − 16j Ω ⇔ YC =
1
= − 0.1 j S
ZL
1 ZC
= 0.0j 625 S
Yeq = YR + YL + YC = 0.0278 − j0.0375 = 0.0467 ∠ − 53.4 ° S
Zeq =
1 Yeq
= 21.43∠ 53.4° = 12.75 + j17.22 Ω
P10.7-2 Z=
V
−I
=
10 ∠40° = − 5000∠205°Ω = 4532 + j2113 = 3 − −2×10 ∠−165°
so R = 4532 Ω and L =
2113 ω
=
2113 2×106
R + ω j L
= 1.057 mH
P10.7-3
− Z(ω ) =
−
ωC
j
j ( R+ jω L)
+ ( R+ jω L)
ω C
L
=
C
⎛ ⎝
− j
R
ω C
R + j ⎜ ω L −
1 ⎞ ⎟ ω C ⎠
R ⎞⎛ 1 ⎞⎞ ⎛ L ⎛ ω j R j L − − − ⎜ C ω C ⎟⎜ ⎜ ⎟⎟ ω C ⎠ ⎠ ⎝ ⎠⎝ ⎝ = 2 1 ⎞ 2 ⎛ R +⎜ ω L − ⎟ ω C ⎠ ⎝ 2 RL R⎛ 1 ⎞ ⎛ R L 1 ⎞⎞ ⎛ ω − L − − ⎜j + ⎜ ω −L ⎜ ⎟ ⎟⎟ ωC ⎠ ωC C ⎝ ω C ⎠ ⎠ C ωC ⎝ ⎝ = 2 1 ⎞ 2 ⎛ R +⎜ ω L − ⎟ ω C ⎠ ⎝ Z (ω ) will be purely resistive when
⎛L 1 ⎞ + ⎜ ω L − ⎟= 0 ⇒ ωC C⎝ ω C ⎠ 2
when
=6 R Ω,
C= 22
μ F,
R
and
= 27 mH, then L
2
ω
ω = 1278
2
⎛ ⎞R = −⎜ ⎟ CL ⎝L⎠ 1
rad/s.
P10.7-4 R R+ j (ω L−ω R C +ω R L C ) j ω C = jω L + = Z = ZL + 1 R +Z c 1+ (ω R C ) 2 R + j ω C Set real part equal to 100 Ω to get C 2
Z c R
R
1+ (ω R C ) 2
3
2
2
= 100 ⇒ C =0.158 μF
Set imaginary part of numerator numerator equal to 0 to get L ( ω = 2π f = 6283 rad sec ) L− R2 C + ω 2 R2 LC2 = 0 ⇒ L= 0.1587 H
P10.7-5 ZL =
L=
ω j
−6
j(6.28 × 10 ) (4 (47 × 10 ) = 6
j295 Ω
⎛ 1 ⎞ ⎜ j ω C ⎟( 300 + j 295 ) ⎠ Z eq = Zc || (Z R +ZL ) = ⎝ = 590.7 Ω 1 + 300 + j 295 j ω C
590.7 =
300 +300 j
⇒ 590.7 − (590.7)(295 ω C) + j(590.7)(300 ω C) = 300 + j295
1+ 30 300 jω C−300 ω C
Equating imaginary terms
(
ω=2 π f
= 6.28×10 6 rad sec
)
(590.7) (3 (300ω C ) = 295 ⇒ C = 0.27 nF
P10.7-6 Replace series and parallel capacitors by an equivalent capacitor and series inductors by an equivalent inductor:
Then 100 Z=
ω 4 j+
1
(
jω 5 ×10
100 +
(
jω 5 ×10
)
ω 4 j+ 100
2 ω
)
ω 4 j+
− j
200 ω
1 − j
2
×
1+ j
ω
4 2 ω
=
ω 4 j+ 100
4 − j 2 ω 4 +ω
2
=
400 4+ω
2
2 ω
1+ j
ω
2
− j
1+
−3
⎛ 200 ⎞ ⎟ ω ⎠ ⎝ = ω 4 j+ = ⎛ 200 ⎞ 100 + ⎜ − j ⎟ ω ⎠ ⎝ 100 ⎜ − j
1
4 Z=
−3
200ω ⎞ ⎛ j − + ⎜ 4 ω 2⎟ 4 + ω ⎠ ⎝
2 ω
Section 10-8: Kirchhoff’s Laws Using Phasors P10.8-1
(a) (b) (c)
Z1 =3+ j 4 = 5∠53.1° Ω
and
Z 2 = 8− j 8 = 8 2 ∠ − 45° Ω
Total impedance = Z1 + Z2 = 3 + j 4 + 8 − j 8 = 11 − j 4 = 11.7∠−20.0° Ω I=
100∠0° Z1 +Z 2
=
100 11.7 ∠− 20°
=
100 11.7
∠20.0° ⇒
i (t ) = 8.55 cos (1250 t + 20.0° ) A
P10.8-2 V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠ 47° − 1.59∠125°
= ( 5.23 + j5.62 ) − ( −0.91 + 1.30 ) = ( 5.23 + 0.91) + j (5.62 − 1.30 ) = 6.14 + j 4.32 = 7.51∠35° v1 ( t ) = 7.51 cos ( 2 t + 35° ) V
P10.8-3 I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( − 0.094 + j 0.532)
= ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = − 0.443 − j 0.125 = 0.460∠196° i ( t ) = 460 co c os (2 ( 2 t + 196°) mA mA
P10.8-4 Vs = 2 ∠30° V I=
and
2 ∠30° 6 + j12 + 3 / j
= 0.185 ∠ − 26.3° A
i (t ) = 0.185 cos (4 t − 26.3°) A
P10.8-5 −3
j (2π ⋅ 796) (3 ⋅10 1 0 ) = j15 Ω
I=
12 20 + j15
= 0.48 ∠ − 37° A
i (t ) = 0.48 cos ( 2π ⋅ 796 t − 37 °) A
P10.8-6 Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A I Is
=
B ∠−51.87°
2 ∠−15°
=
Z1 Z1 + Z 2
=
8 8+ j 3L
=
8 ∠0°
⎛ 3 L ⎞ ⎟ ⎝ 8 ⎠
82 + (3 L) 2 ∠ tan −1 ⎜
Equate the magnitudes and the angles.
⎛ 3 L ⎞ ⎟ ⇒ L = 2 H 8 ⎝ ⎠
angles: + 36.87 = + tan −1 ⎜ magnitudes:
8 64 + 9 L2
=
B
2
⇒ B =1.6
P10.8-7
The voltage V can be calculated using Ohm's Law.
V = (1.72 ∠ - 69°) (4.24∠45°) = 7.29 ∠ - 24° V The current I can be calculated using KCL. I = (3.05 ∠ - 77 °) - (1.72 ∠ - 69 °) = 1.34 ∠ - 87 ° A
Using KVL to calculate the voltage across the inductor and then Ohm's Law gives: 2
24 - 4(1.34∠-87° ) j L=
3.05∠-77°
⇒
=4H
L
P10.8-8
⎛
⎞ ⎟ ⎝ 10 − j10 ⎠ 10 ⎛ ⎞ = 20∠0° ⎜ ⎟ ⎝ 10 2∠− 45° ⎠
V10 = Vs ⎜
10
= 10 2∠45° v10 (t ) = 10 2 cos (100 t + 45 °) V
P10.8-9 (a)
I=
160 ∠0° 160 ∠0° = (− j1326) (3 (300 + j37.7) 303 ∠−5.9° − j1326 + 300 + j 37.7 .53 ∠5.9° A = 0.53
i(t ) = 0.53 cos (1 (120π t + 5.9°) A
(b)
I=
160∠0° 160∠0° = (− j199)(300 + j 251) 256∠−59.9° − j199 +300+ j 251
= 0.625∠59.9° A i (t ) = 0.625 cos (8 (800π t + 59.9°) A
P10.8-10 (a) v ( t ) = − i ( t ) =
4 4 + ( 40 10 ) 40
40 + 10
×
× 24 = −8 V
24 4 + ( 40 10 )
=
8 5
= 1.6 A
(b) Represent the circuit in the frequency domain using impedances and phasors.
V=−
j16 j16 + ( 40 − j 25 )
I=
so
× 24∠15° =
40 40 − j 25
(16∠ − 90° ) ( 24∠15° ) = 33.66∠ − 65° V 40 ( − j 25 ) j16 +
×
40 − j 25
24∠15° = 1.78∠57° A 40 ( − j 25 )
j16 +
40 − j 25
v ( t ) = 33.66 cos ( 4t − 65° ) V
and i ( t ) = 1.78 cos ( 4t + 57° ) A
(checked: LNAP 8/1/04)
P10.8-11 I=
5∠30° 30 + j 40
+
5∠30° 20 − j 50
+
5∠30° j 50 − j 20
= 0.100∠ − 23.1° + 0.0923∠98.2° + 0.1667∠ − 60° = 0.186∠ − 29.5° A
so i ( t ) = 0.186 cos (10t − 29.5° ) A
(checked: LNAP 8/1/04)
P10.8-12
45° ⎡⎣( 20 j 20 ) + ( 30 (− j40 ) ) + ( j10 ( − j50 ) ) ⎤⎦ V = 0.01∠45
⎡ 20 ( j 20 ) 30 ( − j 40) j10 ( − j50) ⎤ = 0.01∠45° ⎢ + + ⎥ j j j10 − j50 ⎦ 2 0 2 0 3 0 4 0 + − ⎣ = 0.01∠45° [14.14∠45° + 24∠ − 36.9° + 12.5∠90° ] = 0.01∠45° [10 + j10 + 19.2 − j14.4 + j12.5 ] = 0.303∠60.5° V so v ( t ) = 0.303 cos ( 5t + 60.5° ) V
(checked: LNAP 8/1/04)
P10.8-13 Let
⎛
Z 1 = ⎜ 20 − j ⎜
⎝
⎞ ( 20 − j 25) j10 250 − j 200 = = 12.81∠75.5° Ω ⎟⎟ j10 = 4 ( 0.01) ⎠ 20 − j 25 + j10 20 − j15 1
and
⎛ ⎜ ⎝
Z 2 = j 20 ⎜ − j
⎞ j 20 ( 40 − j 50 ) 1000 + j800 + 40 ⎟⎟ = = = 25.61∠75.5° Ω 4 ( 0.005 ) 2 0 4 0 5 0 4 0 3 0 + − − j j j ⎠ 1
Then V=
Z2 Z1 + Z 2
×10∠60° =
25.61∠75.5° 12.81∠75.5° + 25.6∠75.5°
×10∠60° = 6.67∠60° V
so v ( t ) = 6.67 cos ( 4t + 60° ) V
(checked: LNAP 8/1/04)
P10.8-14 Represent the circuit in the frequency domain using impedances and phasors. Let
⎛
Z 1 = j 50 + ⎜ 40
⎝
40 ( − j50 ) ⎞ j 5 0 = + = 39.0∠51.3° Ω −3 ⎟ j10 × 2 ×10 ⎠ 40 − j50 1
and Z2 = − j
1 10 (5 ×10 −3 )
25 = − j 20 + + j 20 25
j 20 ( 25 )
25 + j 20
= 12.5∠ − 38.7 Ω
Z1 and Z2 are connected in parallel. Current division gives
I1 =
Z1 Z1 + Z 2
× 0.025∠15° = 0.024∠32.7° A
so i1 ( t ) = 0.024 cos (10t + 32.7° ) A
(checked: LNAP 8/1/04)
P10.8-15 (a) i ( t ) = v ( t ) =
80
80 + 80 40 + ( 80 + 80 )
0.024 = 19.2 mA mA 1
80 + 80
× ( 40 ( 80 + 80) ) 0.024 = ( 32 ) ( 0.024) = 0.384 V 2
(b) Represent the circuit in the frequency domain using impedances and phasors.
I=
80 + j80
− j 25 + ( 80 + j80 )
V=
× 0.024∠15° = 0.028∠25.5° A
80
× ⎡− j 25 (80 + j80 ) ⎤⎦ × 0.024∠15° = 0.494∠ − 109.5 ° V 80 + j80 ⎣
so i ( t ) = 28 cos (10t + 25.5° ) mA
and v ( t ) = 0.494 co cos (10t − 109.5° ) V
(checked: LNAP 8/1/04)
P10.8-16 Represent the circuit in the frequency domain using impedances and phasors. Let Z 1 = 25 + j ( 20 ) 2 + Z 2 = 20 +
1 j ( 20 ) ( 0.002 )
1 j ( 20 ) ( 0.005 )
= 25 + j15 = 29 2 9.2∠31° Ω
= 20 − j10 = 22.36∠ − 26.6° Ω
Z 3 = 40 + j ( 20 ) 2 = 40 + j 40 = 56.57 ∠45° Ω
and let Z p = Z 2 Z 3 = 18.86∠ − 8° = 18.67 − j2.67 Ω
Then
I=
16∠75° Z1 + Z p
×
Z2 Z2 + Z3
= 0.118∠6.1° A
so i ( t ) = 0.118 cos ( 20t + 6.1° ) A
(checked: LNAP 8/2/04)
P10.8-17 Represent the circuit in the frequency domain using phasors and impedances. The impedance 1 capacitor is = − j 20,000 . When the switch is closed −6 j (100 ) ( 0.5 ×10 )
17.89∠ − 26.6° = V =
− j 20,000 × 20∠0° R 2 − j 20,000
Equating angels gives
⎛ −20, 000 ⎞ −26.6° = −90° − tan −1 ⎜ ⎜ R 2 ⎟⎟ ⎝ ⎠
⇒
R 2 =
−20, 000 = 10015 Ω tan ( −63.4 )
When the switch is open 14.14∠ − 45° = V =
− j 20,000 × 20∠0° R1 + R 2 − j 20,000
Equating angles gives
⎛ −20, 000 ⎞ −45° = −90° − tan −1 ⎜ ⎜ R1 + R 2 ⎟⎟ ⎝ ⎠
⇒
R1 + R 2 =
−20, 000 = 20, 000 tan ( −45° )
So 000 − 10015 = 9985 Ω R1 = 20, 00
(checked: LNAP 8/2/04) P10.8-18 Represent the circuit in the frequency domain using phasors and impedances. Let Z 1 = ( j 20 20 ) +
⎛
1 j 0.05
Z 2 = j 40 + 40 + ⎜ j10
⎝
I=−
Z1 Z1 + Z 2
= 10 − j10 = 14.14∠ − 45° Ω
⎞ ⎟ + 15 = 55 + j56.67 = 79∠46.3 ° Ω j 0.04 ⎠ 1
× 20∠30° = 3.535∠129.3° mA
so i ( t ) = 3.535 cos ( 5t + 129.3° ) mA
(checked: LNAP 8/2/04)
P10.8-19 (a) Using KCL and then KVL gives 20 = 50 i ( t ) + 40 ( 5 i ( t ) )
20
⇒ i (t ) =
250
= 80 mA
Then v ( t ) = 40 (5i ( t ) ) = 200 ( 0.08) = 16 V
(b) Represent the circuit in the frequency domain using phasors and impedances.
Where Z 1 = 40 + j (10 ) 3 +
1 j (10 ) ( 0.005 )
26.6° Ω = 40 + j10 = 41.23∠26
and Z 2 = j (10 ) 2 10 = 8 + j 4 = 8.944∠26.6° Ω Using KCL and then KVL gives 20∠15° = Z 1I + 5Z 2 I
⇒
I = 0.234∠ − 5.6° A
Then V = Z 2 ( 5I ) = 10.47∠21° A so i ( t ) = 0.234 cos (10t − 5.6° ) A
and v ( t ) = 10.47 cos (10t + 21° ) V
(checked: 8/3/04)
P10.8-20 (a) Using voltage division twice v ( t ) =
40 40 + 80
× 24 −
100 20 + 100
× 24 = −12 V
(b) Represent the circuit in the frequency domain using phasors and impedances.
Where Z 1 = 20 Ω
⎛ ⎞ 1 20 ⎟ = 12.2 + j70.2 = 71.30∠80.2 ° Ω ⎜ j ( 20 ) ( 0.002 ) ⎟ ⎝ ⎠
Z 2 = j ( 20 ) 4 + ⎜ Z 3 = j ( 20 ) 3 + Z4 =
1 j ( 20 ) ( 0.005 )
1 j ( 20 ) ( 0.004 )
6 3.4° Ω + 25 = 25 + j50 = 55.90∠63
+ 15 = 15 − j12.5 = 19.53∠ − 39.8° Ω
Using voltage division twice V=
Z2 Z1 + Z 2
× 24∠45° −
Z4 Z3 + Z4
80° V × 24∠45° = 24.8∠80
so v ( t ) = 24.8 cos ( 20t + 80° ) V
(Checked using LNAP 10/5/04)
P10.8-21 Represent the circuit in the frequency domain using phasors and impedances.
4 j 6 =
4 ( j 6 ) 4 + j 6
=
24∠90° 7.2∠56°
= 3.33∠34° = 2.76 + j1.86 Ω
Using voltage division V=
3.33∠34°
− j 5 + 2.76 + j1.86
× 5∠45° =
3.33∠34° 2.76 − j3.14
× 5∠45° =
3.33∠34° 4.18∠ − 48°
× 5∠45° = 3.98∠127° V
The corresponding voltage in the time domain is v ( t ) = 3.98 cos ( 2t + 127° ) V
P10.8-22 V1 (ω ) =
V 2 (ω ) =
j10
8 + j10
5 e− j 90 = 3.9 e − j 51 V
j 20 j 20 − j 2.4
5 e− j 90 = 5.68 e − j 90 V
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47 V
P10.8-23 V1 (ω ) =
V2 (ω ) =
8 ( j 6 ) 8 + j 6
4 e j15 = 19.2 e j 68 V
j12 ( − j 4 ) j12 − j 4
4 e j15 = 24 e − j 75 V
V (ω ) = V1 (ω ) + V2 (ω ) = 14.4 e−
j 22
V
Section 10-9:
Node Voltage and Mesh Mesh Current Analysis Using Phasors
P10.9-1 Draw frequency domain circuit and write node equations:
−2+
KCL at A KCL at C:
VA
10
VC − VA j 5
+ +
VA − VC j 5 VC
− j4
= 0 ⇒ ( 2 + j ) VA − 2VC = j 20
− (1+ j ) = 0 ⇒
4VA + VC = 20 − j 20
Solve using Cramers rule: (2 + j )
j 20
4 20 − j 20 60 − j100 116.6 ∠−59° = = = 11.6 11.6 ∠− 64.7 64.7° V (2 + j ) −2 10+ j 101 ∠5.7°
Vc =
4
1
P10.9-2
KCL:
(V −100)
IS =
150
100− V 150
+
V
− j125
+
V j80
+
V
250
= 0 ⇒ V = 57.6 ∠22.9° V
= 0.667 − 0.384 ∠22 22.9° = 0.347 ∠−25.5° A
IC =
V
125 ∠−90°
= 0.46 0.461 1 ∠112.9 112.9° A
IL =
V
80∠90°
IR =
V
250
= 0.72 ∠−67.1 0.720 0 ∠− 67.1° A = 0.230∠22.9° A
P10.9-3
KCL at node A: Va
200
+
Va − Vb j 100
=0
(1)
KCL at node B: Vb − Va
1j00
+
Vb
− 5j 0
+
⇒ Va =
Vb −1.2
8j 0 1 4
Vb −
= 0
3 2
Substitute Eqn (2) into Eqn (1) to get 2.21 ∠ − 144 144° V Vb = 2.21 Then Eqn (2) gives
Va = ( 0.55∠−144°) − 1.5 = 1.97∠ − 171° V Finally va (t ) = 1.97 co cos (4 (4000 t −171 °) V and vb (t) = 2.21cos (4 (4000 t −144 °) V
P10.9-4
= 104 rad s I s = 20∠53° A ω
The node equations are:
⎛ 1 1 j ⎞ ⎛ 1⎞ + + ⎟ Va + ⎜ − ⎟ Vb = 20∠53.13° ⎝ 20 40 60 ⎠ ⎝ 40 ⎠ j j⎞ j ⎛ 1 ⎞ ⎛ 1 KCL at b: ⎜ − ⎟ Va + ⎜ − + ⎟ Vb − Vc = 0 80 ⎝ 40 ⎠ ⎝ 40 40 80 ⎠ KCL at a: ⎜
(2)
KCL at c:
− j 80
j ⎞ ⎛ 1 + ⎟ Vc = 0 ⎝ 40 80 ⎠
Vb + ⎜
Solving these equations, e.g. using MATLAB yields
Va = 2 ⋅ 240∠45° V ⇒
va (t ) = 339.4 cos (ω t + 45°) V (checked: LNAP 7/19/04)
P10.9-5 vs = sin ( 2π ⋅ 400 t ) V R = 100 Ω LR = 40 mH
⎧ 40 mH
door opened
⎩ 60 mH
door closed
LS = ⎨
With the door open VA − VB = 0 since the bridge circuit is balanced. With the the door closed sed ZLR = (8j 00π )(0 )(0.04) 04)= 10 j 0.5 0.5 Ω and ZLS = (8j 00π )(0 )(0.06) 06)= 1j50.8 0.8 Ω. The node equations are: KCL at node B:
VB − VC R
+
KCL at node A :
VB Z LR
= 0 ⇒ VB =
VA − VC R
+
VA Z LS
j100.5 j100.5 +100
VC
=0
Since Since VC = Vs =1 V VB = 0.709∠ 44.86° V and VA = 0.833∠33.55 V
Therefore VA − VB = 0.833∠33.55 ° − 0.709∠44.86° = (0.694 + .46j 0) − (0.503 + 0.5j 00) = 0.191 − 0.0j 40
= 0.195∠ − 11.83° V
P10.9-6
The node equations are: V1 −(−1+ j ) j 2
V2 − V1
− j 2
+
+
V1
2
+
V2
− j2
V1 − V2
−j2
=0
− IC = 0
Also, expressing the controlling signal of the dependent source in terms of the node voltages yields ⎡ −1+ j ⎤ −1 + j Ix = ⇒ IC = 2 Ix = 2 ⎢ ⎥ = −1 − j A -2 j -2 j ⎣ ⎦ Solving these equations yields V2 =
−3− j = 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 1 35°) V 1+ j 2 (checked: LNAP 7/19/04
P10.9-7
V2 = 0.7571∠66.7° V V3 = 0.6064∠ − 69.8° V
⎫ I1 =I 2 + I 3 ⎪ ⎪ V − V2 ⎪ I2 = 3 ⎬ yields j 10 ⎪ V3 ⎪ I3 = ⎪ − j 2 ⎭
0.3032 32 ∠20.2 20.2° A ⎧ I 3 =0.30 ⎪ ⎨ I 2 =0.1267∠−184° A ⎪ I =0.195∠36° A ⎩ 1
therefore (2 t + 36°) A i1 (t ) = 0.195 cos (2
(checked: MATLAB 7/18/04) P10.9-8
The mesh equations are
(4 + 6j) I1 − 6j I 2 = 12 + 1j2 3 - j 6 I1 + (8 + j 2) I 2 = 0
Using Cramer’s rule yields I1 =
(12 + j 12 3 ) (8+ j 2) (4 + j6) (8+ j2) −( − j6) ( − j6)
= 2.5∠29° = 2.2 + j 1.2 A
Then I2 =
j 6
8+ j 2
(2.5∠29° ) =
6∠90° 68∠14°
(2.5∠29° ) = 1.82∠105° A
and (2.5 ∠29 ° −1.82 ∠105 °) = (6 ∠90 °) (2 (2.71 ∠ −11.3 °) = 16.3 ∠78.7 ° V VL = j 6(I1 − I 2 ) = (6 ∠90 °) (2
Finally
) (1.82∠105°) = 7.28∠15° V V = − j 4I 2 = (4∠ − 90°)( c
P10.9-9
The mesh equations are: (10 − j ) I1 + ( j ) I 2 + 0 I 3 = 10 jI1 − jI 2 +
jI 3 = 0
0 I1 + jI 2 + (1− )j I 3 = 1j 0 Solving these mesh equations using Cramer’s rule yields:
I2 =
(10 − j )
10
0
j
0
j
j 10 (1 (1− j )
0 (10 − j ) j
0
j
− j j
0
=
90 − j 20 = 8.38∠77.5° A −11 j
⇒ i (t ) = 8.38 cos (10 1 0 3 t + 77.5 ° ) A
j
(1− j ) (checked using LNAPAC on 7/3/03)
P10.9-10
The mesh equations are:
−1 − j 4 ⎤ ⎡ I1 ⎤ ⎡10∠30°⎤ ⎡ ( 2+ j 4) ⎢ −1 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ ( 2 +1 / j 4) 1 − ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ − j 4 (3+ j 4) ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ −1 Using Cramer’s rule yields I3 =
2+ j8 30° ) = 3.225∠44 4 4° A (10∠30 12+ j 22.5
Then V = 2 I 3 = 2 ( 3.225 ∠44 ° ) = 6.45 ∠44 ° V ⇒ v( t) = 6.45 cos (10 5 t + 44 °) V
(checked: LNAP 7/19/04)
P10.9-11
Mesh Equations: j 75 I1 − j 100 I 2 = 375
− j 100 I1 + (100 + j 100) I 2 =0 Solving for I 2 yields I 2 = 4.5 + 1.j5 = 3 ∠53.1° A
⇒
2
(i )3tcos (4 00 t +53.1 °) A (checked: LNAP 7/19/04)
P10.9-12 (a) The node equations are
24 − v a
=
40 24 − v b 25
va − v b
+
20
+
va
15
va − vb
20
=
vb
50
or 1 1 ⎡1 + + ⎢ 40 20 15 ⎢ 1 ⎢ − ⎢⎣ 20
⎤ ⎡ v ⎤ ⎡ 24 ⎤ ⎥ ⎢ a ⎥ ⎢ 40 ⎥ 20 =⎢ ⎥ ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 24 ⎥ ⎢v b ⎥ + + ⎥ 25 20 50 ⎦ ⎣ ⎦ ⎢⎣ 25 ⎥⎦ −
1
Solving using MATLAB gives v a = 8.713 V and v b = 12.69 V
(b) Use phasors and impedances to represent the circuit in the frequency domain as
where Z 1 = 25 + j ( 20 ) 4 = 25 + j 80 = 83.82∠72.7° Ω
⎛
Z 2 = ⎜ 40 ⎜
⎝
⎞ ⎟⎟ + j ( 20 ) 5 = 3.56 + j 88.6 = 88.68∠87.7° Ω j ( 20 ) ( 0.004 ) ⎠ 1
Z 3 = 20 Ω Z 4 = 15 + j ( 20 ) 2 = 15 + j 40 = 42.72∠69.4° Z 5 = j ( 20 ) 3 +
1 j ( 20 ) ( 0.005 )
= j 50 = 50∠90° Ω
The node equations are 24∠45° − Va Z2 24∠45° − Vb Z1 1 1 ⎡ 1 + + ⎢Z Z3 Z4 ⎢ 2 ⎢ 1 − ⎢ Z3 ⎢⎣
= +
Va Z4
+
V a − Vb
V a − Vb Z3
=
Vb Z5
⎤ ⎡ 24∠ 45° ⎤ ⎡ ⎤ V ⎥ a ⎢ Z ⎥ Z3 2 ⎥ ⎥⎢ ⎥ =⎢ ⎢ ⎥ ⎢ 24∠45° ⎥ 1 1 1 ⎥ + + ⎥ ⎢⎣ V b ⎥⎦ ⎢ ⎥ Z 1 Z 3 Z 5 ⎥⎦ ⎢⎣ Z 1 ⎥⎦ −
1
Solving using MATLAB gives V a = 7.89∠44.0° V b = 8.45∠45.1°
so
Z3
v a ( t ) = 7.89 cos ( 20t + 44°) V v a ( t ) = 8.45 cos ( 20t + 45.1) V
(checked: LNAP 8/3/04)
P10.9-13 Represent the circuit in the frequency domain using impedances and phasors
The mesh currents are I and 0.05∠ − 30 ° A . Apply KVL to the top mesh mesh to get 15∠45 45 ° + ( − j 25 ) I + (15 + j32 ) ( I − 0.05 ∠ − 30 °) + 25 I = 0
so I=
−15∠45° + (15 + j 32 ) ( 0.05∠ − 30° ) = 0.3266∠ − 143.6° = −0.2629 − j 0.1939 A 25 − j 25 + 15 + j 32
Then V = ( − j 25 ) I = 8.166 ∠126.4 ° = −4.8475 + j6.5715 V
so v ( t ) = 8.166 cos (8t + 126.4 ° ) V
(checked: LNAP 8/3/04)
P10.9-14 Represent the circuit in the frequency domain using impedances and phasors.
The mesh currents are I and 10I. Apply KVL to the supermesh corresponding to the dependant current source to get 10I ) + 40 (10 I ) − 25 ∠ −15 ° = 0 ( j 500 ) I + ( − j5 ) (10 so I=
25∠ − 15° 400 + j 450
= 0.04152∠ − 63.37° A
The output voltage is V = 40 (10I ) = 16.61∠ − 63.37 ° V
so v ( t ) = 16.61 cos (100 t − 63.37 ° ) V
(checked: LNAP 8/3/04)
P10.9-15 Represent the circuit in the frequency domain domain using phasors and impedances. Apply KVL to the center mesh to get
0.8394∠138.5° = I = so
8∠210° − 30∠ − 15° R + j10 L R = 35 Ω
⇒
R+ j10 L = 35 + j25 = 35 + j(10 ) 2.5
and L = 2.5 H
(checked: LNAP 8/3/04)
P10.9-16 Represent the circuit in the frequency domain using phasors and impedances. Apply KCL at the top node of R and L to get
( 50∠ − 75° ) − V 4j 0
+
35∠100° − V 40
50∠ − 75°
⇒
V
=
40∠90°
R ω j L
+
35∠110° 40
⎛ 1 1 1 1 ⎞ =⎜ + + − j ⎟V 40 R 20 L⎠ ⎝ 4j0 40
Using the given equation for v(t ) we get 21.25∠ −168.8° = V = then
1 R
− j
1 20 L
=
1.587∠161.7° 21.25∠ − 168.8°
1.587∠161.7° 1 1 0.025 (1 − j ) + − j 20 L R
− 0.025 (1 − j ) = 0.04 − j 0.01176
finally R =
1 0. 0 4
= 25 Ω and L =
1 20 ( 0.01176 )
= 4.25 H (checked: LNAP 8/3/04)
P10.9-17 Represent the circuit in the frequency domain using phasors and impedances.
The node equations are
50∠0° − Va 15 50∠0° − V b
− 2j0
+
=
Vb − V a j100
V b − Va
1j00
=
+
Va
25 Vb
5j0
or 1 1 ⎡1 + + ⎢15 j100 25 ⎢ 1 ⎢ − ⎢ 1j00 ⎣
⎤ ⎡ 50∠0° ⎤ ⎡ ⎤ V a ⎥ ⎢ ⎥ ⎢ 15 ⎥ j100 ⎥ ⎥ =⎢ 1 1 1 ⎥ ⎢ ⎥ ⎢ 50∠0° ⎥ ⎢ Vb ⎥ + + ⎥ 5j0 1j00 − 2j0 ⎦ ⎣ ⎦ ⎢⎣ − j 20 ⎥⎦ −
⎡ 0.1067 − j 0.010 ⎢ j 0.010 ⎣
1
j 0.010⎤ ⎡ V a ⎤
⎡ 3.333⎤ ⎢V ⎥ = ⎢ ⎥ j 0.020 ⎦ ⎣ b ⎦ ⎣ j 2.5 ⎥⎦
Solving, eg using MATLAB, gives
Va = 33.05∠ − 12.6° V and Vb = 108.9 ∠1.9 ° V Then I=
Va
25
= 1.322∠ − 12.6° A
so i ( t ) = 1.322 cos ( 25t − 12.6 ° ) A
(checked: LNAP 8/3/04)
P10.9-18 Represent the circuit in the frequency domain using phasors and impedances. Label the node voltages.
The node equations are
24∠15° − Va 25 24∠15° − V b
− j 6.25
=
+
Va j 40
+
Va − Vb
V a − Vb
10
10
=
Vb
45
or
⎤ ⎡ V ⎤ ⎡ 24∠15° ⎤ ⎥⎢ a⎥ ⎢ ⎥ 10 25 ⎥⎢ ⎥ = ⎢ ⎥ ∠ ° 1 1 1⎥ 2 4 1 5 ⎥ ⎢Vb ⎥ ⎢ + + j ⎣ ⎦ ⎥ ⎢ 6.25 45 10 10 ⎦ ⎣ 6.25∠ − 90° ⎥⎦
1 1 ⎡1 j − + ⎢ 25 40 10 ⎢ 1 ⎢ − ⎢⎣ 10
−
⎡0.140 − j 0.025 ⎢ −0.10 ⎣
1
15° ⎤ − 0.10 ⎤ ⎡ Va ⎤ ⎡ 0.960∠15 ⎢V ⎥ = ⎢ ⎥ ⎥ 0.1222 + j 0.160⎦ ⎣ b ⎦ ⎣ 3.840∠105° ⎦
Solving gives
Va = 24.67∠32.6° V and Vb = 25.59 ∠25.2 ° V Then I=
Va − Vb
10
= 0.3347∠134.9° A
so i ( t ) = 0.3347 co cos (10 t + 134.9 ° ) A
(checked: LANP 8/4/04)
P10.9-19 Represent the circuit in the frequency domain using phasors and impedances.
The node equations are
20∠0° − V j 40
5V − Vo 10 1 1 ⎡1 − − j j ⎢ 25 5 40 ⎢ 1 ⎢ − ⎢⎣ 2
⎡ 0.04 − j 0.225 ⎢ −0.50 ⎣
=
V
+
V − 5V
25
=
− j 20
Vo
− j10 ⎤ ⎡ V ⎤ ⎡ − j 0.5⎤ ⎥⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎥ 1 1 + j ⎥ ⎢⎣ Vo ⎥⎦ ⎢⎣ 0 ⎥⎦ 10 10 ⎥⎦ 0
⎤ ⎡ V ⎤ ⎡ − j 0 .5 ⎤ ⎢ ⎥= 0.10 + j 0.10⎥⎦ ⎣ V o ⎦ ⎢⎣ 0 ⎥⎦ 0
Solving gives
V = 2.188∠ − 10.1° V and Vo = 7.736∠ − 55.1 ° V so v o ( t ) = 7.736 cos ( 5t − 55.1 ° ) V
(checked: LNAP 8/4/04)
P10.9-20 (a) Use KVL to see that the voltage across the 8 Ω resistor is 20i ( t ) − 4i ( t ) = 16 i ( t ) .
Apply KCL to the supernode corresponding to the dependent voltage source to get 0.036 = i ( t ) +
16i ( t ) 8
= 3i ( t )
so i ( t ) = 12 mA
(b) Represent the circuit in the frequency domain using phasors and impedances.
Where
Z 1 = 20 +
1 j ( 25) ( 0.002 )
⎛ ⎜ ⎝
Z 2 = j 50 + ⎜15
= 20 − j 20 Ω
⎞ ⎟ = 43.3∠83.9° Ω j ( 25 ) ( 0.004 ) ⎟⎠ 1
Use KVL to get
V = Z 1I − 4I = ( Z 1 − 4 ) I Then apply KCL to the supernode corresponding to the dependent source to get 0.036∠0° = I +
(Z
1
− 4) I
Z2
⎛ Z + Z2 − 4 ⎞ =⎜ 1 ⎟⎟ I ⎜ Z 2 ⎝ ⎠
so I=
Z 2 ( 0.036∠0° ) Z1 + Z 2 − 4
= 50.4∠35.7° mA
so i ( t ) = 50.4 cos ( 25t + 35.7 ° ) mA
(checked: LNAP 8/4/04)
P10.9-21
KCL at Va: Va
4− j 2
+
Va − Vb
− j10
=1
(4 − 12j) Va + (−4 + 2)j Vb = −20 − 40j
KCL at Vb:
Vb − Va
− j10
+
Vb
2+ j 4
+ 0.5∠ − 90° = 0 ⇒ (−2 − 4j) Va + (2 − 6)j Vb = 10 + 2j0
Cramer’s rule yields: (−20− j 40) V = a
( −4 + j 2)
(10 + j 20) (2 − j 6) −200 + j100 = = 5∠296.5° V (4 − j12) ( −4 + j 2) −80− j 60 (−2− j 4)
( 2- j 6)
Therefore
°
°
v (t ) = 5 co cos (100 t + 296.5 ) = 5 co cos (100 t − 63.5 ) V a
P10.9-22
The mesh equations are: j15 I1 + 10 (I1 − I 2 ) = 20 ⇒
(10 + j15) I1 − 10 I 2 = 20
− 5 I2 + j 10(I 2 − I1 ) = −30∠− 90° ⇒ −10I1 + (10 − 5) I 2j = 30 j Cramer’s rule yields:
−10 j 30 10 − j 5 200 + j 200 I1 = = = 2.263∠ − 8.1° A −10 10 + j15 75+ j100 −10 10 − j5 20
Next
VL = ( j15) I1 = (15∠90°) (2.263∠−8.1°) = 24 2∠82° V Therefore vL (t) = 24 2 cos (ω t + 82°) V
P10.9-23
The mesh equations are: (10 − j50) I1 − 10 I 2 = j30
−10 I1 +(10 − 2j 0) I 2 + 2j0 I 3 = 5j0 j 20 I 2 + (30− j10) I 3 = 0
Solving the mesh equations, e.g. using MATLAB, gives: I1 = −0.87 − j 0.09 A,
I 2 = −1.32+ j 1.27 A,
I 3 = 0.5+ j 1.05 A
Then Va = 10 (I1 − I 2 ) = 14.3∠ − 72° V
and
36.6 ∠83° V Vb = Va + j50 = 36 (checked: LNAP 7/18/04)
Section 10-10: Superposition, Thèvenin and Norton Equivalents and Source Transformations P10.10-1 Use superposition
I1 =
12∠45° 3000 + j 2000 i (t )
= 3.3∠11.3° mA
I2 =
−5∠0° = 1.5∠153° mA 3000 + j1500
(4000 t + 11.3 °) +1.5 co cos (3 (3000 t +153 °) mA = 3.3 cos (4
P10.10-2 Use superposition
I1 =
3 6000
= 0.5 mA
i(t ) = i 2 (t) + i1 (t)
I 2 (ω ) =
−1∠45° = −0.166 ×10 −3 ∠45° A 6000 + j 0.2
= − 0.166 cos (4 t + 45°) + 0.5 mA = 0.166 co cos (4 (4 t − 135°) + 0.5 mA
P10.10-3 Use superposition
I1 (ω ) =
12∠ 45° 6
+ j 2
= 19∠26.6° mA
I 2 (ω ) =
5∠−90° 6
+ j1.5
= 0.808∠ − 104° mA
i (t ) = i1 (t) − i 2 (t) = 19 cos (4 t + 26.6° ) − 0.808 cos (3t − 1 04° ) mA
P10.10-4 Find Voc :
⎛ 80 + j80 ⎞ ⎟ ⎝ 80 + j80 − j 20 ⎠ ⎛ 80 2∠− 45° ⎞ = ( 5 ∠−30° ) ⎜ ⎟ ⎝ 100∠36.90° ⎠ = 4 2∠ − 21.9° V
Voc = ( 5 ∠−30° ) ⎜
Find Z t :
Zt =
The Thevenin equivalent is
( − j 20 ) (80 +
j80 )
− j 20 + 80 + j80
= 23 ∠ − 81.9° Ω
P10.10-5 First, determine Voc :
The mesh equations are
600 I1 − 30j0 (I1 − I 2 ) = 9
⇒ (600 − 30j0) I1 + 30j0 I 2 = 9∠0°
and V = 300j (I 1 − I 2 ) ⇒ −2 V + 300 I 2 − 300j (I1 − I 2 ) = 0 an
3 I1j+ (1− j 3) I 2 = 0
Using Cramer’s rule:
I 2 = 0.0124∠ − 16° A Then
Voc = 300 I 2 = 3.71∠ − 16° V Next, determine I sc :
−2 V − V = 0 ⇒ V = 0 ⇒ I sc =
9∠0° 600
= 0.015∠0° A
The Thevenin impedance is ZT = The Thevenin equivalent is
Voc I sc
=
3.545∠−16° 0.015∠0°
= 247∠ − 16° Ω
P10.10-6 First, determine Voc : The node equation is:
Voc Voc − (6+ j8) 3 ⎛ Voc −(6 + j8) ⎞ + − ⎜ ⎟=0 − 4j 2j 2⎝ 2j ⎠ Voc =3+ j 4=5∠53.1° V
Vs = 10∠53° = 6 + j 8 V Next, determine I sc : The node equation is:
V 2
+
V
+
− 4j
V − (6 + j8) 2j
V=
I sc =
3 ⎡ V − (6 + j8) ⎤ − ⎢ ⎥=0 2⎣ 2j ⎦
3 + j 4 1 − j
V 3+ j 4 = 2 2 − j 2
Vs = 10∠53° = 6 + j 8 V
The Thevenin impedance is
ZT =
⎛ 2− j 2 ⎞ Voc = 3 + j 4 ⎜ ⎟ = 2 − j2 Ω + I sc 3 j 4 ⎝ ⎠
The Thevenin equivalent is
(checked: LNAP 7/18/04)
P10.10-7 Y = G + YL + YC Y = G when YL + YC = 0 or ω O
=
1
, f O = LC 2π
1
1 jω L
+ jω C = 0 1
=
− LC π 2 39.6×10 15
= 0.07998×107 Hz =800 kHz (80 on the dial of the radio)
P10.10-8 In general: I=
Voc
and V =
Z t +Z L
ZL Z t +Z L
Voc
In the three given cases, we have Z1 = 50 Ω ⇒
Z2 =
1 jω C
=
1 −6
j(2000)(2.5×10 )
I1 =
V1 Z1
=
25 50
200 Ω ⇒ = − j 200
−3 j L= (j 2000)(50 × 10 ) = 1j 00 Ω Z3 = ω
Since |I| is the same in all three cases, Z t +Z1 = Zt +Z 2
0.5 A = 0.5
I2 =
⇒
I3 =
V2 Z2
V3 Z3
=
=
100 200
50 100
This requires ( X− 200) 2 = ( X+ 100) 2 ⇒
X= 50 Ω
Then 2
= 0. 5 A
= Zt +Z3 . Let Z t = R +
( R+ 50) 2 + X2 = R2 + ( X − 200) 2 = R2 + ( X + 1 00) 2
( +R50) 2 + (50) 2 =
0.5 A = 0.5
R+ (−150)
2
⇒
=R175 Ω
so
Z t =175+ j50 Ω and
Voc = I1 Z t + R1 =(0.5) (175+50)2 + (50)2 =115.25 V
j X . Then
P10.10-9
Z1 =
( − j 3)( 4)
− j3+ 4
= 2.4∠ − 53.1° Ω =1.44 − j1.92 Ω
Z 2 = Z1 + j 4
= 1.44 + j 2.08 = 2.53∠55.3° Ω Z3 = 3.51∠ − 37.9° Ω
= 2.77 − j 2.16 Ω ⎛ 3.51∠− 37.9° ⎞ ( 3.51∠− 37.9° ) I = ( 2.85∠− 78.4° ) ⎜ = ( 2.85∠− 78.4° ) = 1.9∠ − 92° A ⎟ 2 . 7 7 j 2 . 1 6 2 5 . 2 4 2 4 . 4 − + ∠ − ° ( ) ⎝ ⎠ (checked: LNAP 7/18/04)
P10.10-10
Z2 =
I=
(200)( − j 4) = 4∠ − 88.8° Ω 200 − j 4
0.4∠− 44°
−4 j +100+ j 4
= 4∠ − 44° mA
i (t ) = 4 cos ( 25000 t − 44 °) mA
P10.10-11 Use superposition in the time domain. Let i s1 ( t ) = 36 cos ( 25 t ) mA and i s 2 (t ) = 48 cos ( 50t + 45 ° ) mA
We will find the response to each of these inputs separately. Let ii(t ) denote the response to isi(t ) for i = 1,2. The sum of the two responses will be i(t ), ), i.e. i ( t ) = i1 (t ) + i 2 ( t )
Represent the circuit in the frequency domain as
Use KVL to get Vi = Z i I i − 4I i
Apply KCL to the supernode corresponding to the dependent voltage source.
I si = I i +
Vi Z2
=
Z1 + Z 2 − 4 Z2
Ii
or
Ii =
Z 2I s i Z1 + Z 2 − 4
Consider the case i = 1 : is1(t ) = 26cos(25t ) mA. Here ω = 25 rad/s and
I si = 36∠0° mA Z 1 = 20 +
1 j ( 25 ) ( 0.002 )
⎛ ⎜ ⎝
Z 2 = j 50 + ⎜15
= 20 − j 20 Ω
⎞ ⎟ = 43.3∠83.9° Ω j ( 25 ) ( 0.004 ) ⎟⎠ 1
and I 1 = 50.4∠35.7 ° mA
so i ( t ) = 50.4 cos ( 25t + 35.7° ) mA
Next consider i = 2 : is2 = 48cos(50t + 45°) mA. Here ω = 50 rad/s and
I s2 = 48∠45° mA Z 1 = 20 +
1 j ( 50 ) ( 0.002 )
⎛ ⎜ ⎝
Z 2 = j100 + ⎜15
= 20 − j10 Ω
⎞ ⎟ = 95.5∠89.1° Ω j ( 50 ) ( 0.004 ) ⎟⎠ 1
(Notice that Z1 and Z2 change when ω changes.) I 2 = 52.5∠55.7 ° mA
so i 2 ( t ) = 52.5 cos ( 50t + 55.7° ) mA
Finally, using superposition in the time domain gives i ( t ) = 50.4 cos ( 25t + 35.7° ) + 52.5 cos ( 50t + 55.7 ° ) mA
(checked: LNAP 8/7/04)
P10.10-12 Use superposition in the time domain. Let i1(t ) be the part of i(t ) due to vs1(t ) and i2(t ) be the part of i(t ) due to vs2(t ). ). To determine i1(t ), ), set vs2(t ) = 0. Represent the resulting circuit in the frequency domain to get
where
Z 1 = 20 + j80 = 82.46∠76° Ω Z 2 = 10 + ( j 40 15 ) = 23.15 + j4.93 = 23.67 ∠12° Ω Z 3 = 20 +
1 j ( 20 ) ( 0.005 )
= 20 − j10 = 22.36∠ − 26.6° Ω
Next, using Ohm’s law and current division gives
I1 =
30∠70°
Z1 + (Z 2 Z 3 )
Z3
×
Z2 + Z3
=
Z 3 ( 30∠70° ) Z 1Z 2 + Z 2 Z 3 + Z 1Z 3
= 0.182∠ − 17.6° A
so i ( t ) = 0.182 cos ( 20t − 17.6° ) A
To determine i2(t ), ), set vs1(t ) = 0. Represent the resulting circuit in the frequency domain to get
where
Z 4 = 20 + j 40 = 44.72∠63.4° Ω Z 5 = 10 + ( j 20 15 ) = 19.6 + j7.2 = 20.88∠20.2° Ω Z 6 = 20 +
1 j (10 ) ( 0.005 )
= 20 − j 20 = 28 2 8.28∠ − 45° Ω
Next, using Ohm’s law and current division gives I2 =
18∠ − 15°
Z 6 + (Z 4 Z 5 )
×
Z4 Z4 + Z5
=
Z 1 (18∠ − 15° ) Z 1Z 2 + Z 2 Z 3 + Z 1Z 3
= 0.377∠18° A
so i 2 ( t ) = 0.377 cos (10t + 18° ) A
Using superposition, i ( t ) = i1 (t ) + i 2 (t ) = 0.182 cos ( 20t − 17.6° ) + 0.377 cos (10t + 1 8° ) A
(checked: LNAP 8/8/04)
P10.10-13 Represent the circuit in the frequency domain as
and V 2 = 4.88 ∠ − 95.8 ° V. where i =1,2 and R1 = 20 Ω, V1 = 3.0∠ − 100.9 ° V, R 2 = 40 Ω an
Using voltage division gives R i
Vi =
Ri
+ Rt + j20 Lt
× Voc
so
⎛ R1 + Rt + j20 Lt ⎞ ⎛ R + Rt + j20 Lt ⎞ = V oc = V 2 ⎜ 2 ⎟ ⎟⎟ ⎜ ⎟ ⎜ R R 1 2 ⎝ ⎠ ⎝ ⎠
V1 ⎜ Solving gives
R j0 L t + 2 t =
So Rt = 52 Ω and L t =
35 20
(
R1 R 2 V1 − V2
V2 R1 − V1 R 2
) = 52 +
3j5 Ω
= 1.75 H . Next Voc =
R1 + Rt
+ j20 Lt
R1
× V1 = 12∠ − 75° V
So A = 12 V and θ = −75°. (checked using LNAP 10/4/04)
P10.10-14 Represent the circuit in the frequency domain as
a nd I 2 = 0.848 ∠ −100.7 ° A . Using where R1 = 20 Ω and I 1 = 1.025∠ − 108.5 ° A, R 2 = 40 Ω an
current division gives
In =
+ j10 Lt × I sc Rn + Rt + j10 Lt Rt
so
⎛ R1 + Rt + j10 Lt ⎞ ⎛ R2 + Rt + j10 Lt ⎞ = = I I ⎟⎟ sc 2 ⎜⎜ ⎟⎟ ⎜ + + R 10 j L R j L t t t t ⎝ ⎠ ⎝ ⎠
I1 ⎜ Solving gives
Rt + 1j0 Lt =
R 2I 2 − R1I 1
I1 − I 2
= 40 + 5j 0
So Rt = 40 Ω and Lt = 5 H. Next
I sc = so
R1 + Rt Rt
+ j10 Lt
+ j10 Lt
× I 1 = 1.25∠ − 120° A
B = 1.25 1.25 A and
θ = −120° .
(checked: LNAP 8/8/04)
P10.10-15 Represent the circuit in the frequency domain as
Three cases are mentioned, so we consider i =1,2,3; with
Z1 = 10 + j75 Ω and V1 = 7.063∠50.2 ° V for experiment 1 and
Z2 = 25 + j250 Ω and V 2 = 8.282∠47.8 ° V for experiment 2. Using voltage division
⎛ Z1 + Z t ⎞ ⎛ Z 2 + Z t ⎞ = = V V oc 2⎜ ⎜ Z 1 ⎟⎟ ⎜ Z 2 ⎟⎟ ⎝ ⎠ ⎝ ⎠
V1 ⎜ Solving gives
Zt =
Z 1Z 2 ( V1 − V 2 ) V 2 Z 1 − V1Z 2
= 20 + j50
and
⎛ Z1 + Z t ⎞ ⎜ Z 1 ⎟⎟ = 10∠45° V ⎝ ⎠
Voc = V1 ⎜ Now when Z3 = 10 + j200
V3 =
Z3 Z3 + Z t
Voc =
10 + j 200
(10 + j 200 ) + ( 20 +
j50 )
×10∠45° = 7.95∠49° V
so v ( t ) = 7.95 cos ( 25t + 49° ) V
(checked: LNAP 8/8/04)
P10.10-16 Represent the circuit in the frequency domain as
Three cases are mentioned, so we consider i =1,2,3; with Z 1 = 25 Ω, V1 = 9.77 ∠31.6 ° V , Z 2 = j 60 Ω, V 2 = 18.9 ∠90.0 ° V
and Z 3 = − j
1 15C
, V3
= B∠ − 45°
Using voltage division
Vi =
Zi Zi + Zt
Voc
so
⎛ Z1 + Z t ⎞ ⎛ Z + Z t ⎞ = Voc = V2 ⎜ 2 ⎟ ⎜ Z1 ⎟ ⎜ Z t ⎟⎟ ⎝ ⎠ ⎝ ⎠
V1 ⎜ Solving gives
Zt =
Z 1Z 2 ( V1 − V 2 ) V2 Z 1 − V1Z 2
= 65 + j 48.75 Ω
and
⎛ Z1 + Z t ⎞ V oc = V1 ⎜ ⎜ Z 1 ⎟⎟ = 40∠60° V ⎝ ⎠ Now 1 B∠ − 45° = V3
=
Z3 Z3 + Zt
j15C
× Voc =
1 j15C
× 40∠60° =
+ 65 + j 48.75
40∠60° 1 − 731.25C + j 975C
Equating angles gives
⎛ 975C ⎞ −45° = 60° − tan −1 ⎜ ⎟ ⎝ 1 − 731.25C ⎠ Then Z 3 = − j
1 15C
C =
tan (105° ) 975 + 731.25 tan (10 105° )
= 2.1277 mF
= 31.33∠ − 90° Ω and V3 =
so
⇒
Z3 Z3 + Zt
Voc = 18.625∠ − 45° V
B = 18.625 V .
(checked: LNAP 8/8/04)
P10.10-17 Use superposition in the time domain. Let vs1(t ) = 5 V and vs2(t ) = 30cos(100t ) V.
Find the steady state response to vs1(t ). ). When the input is constant and the circuit is at steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. So 5 i1 ( t ) = = 1 A 5 Find the steady state response to vs2(t ). ). Represent the circuit in the frequency domain using impedances and phasors.
I2 =
30∠0° 5 + j 5
= 4.243∠ − 45 A
So i 2 ( t ) = 4.243 cos (100t − 45° ) A
Using superposition i ( t ) = i1 (t ) + i 2 (t ) = 1 + 4.243 cos (100t − 45° )
P10.10-18 Use superposition. First, find the response to the voltage source acting alone: Zeq =
− j1010 ⋅ = 5 (1 − j ) Ω 10 − j10
Replacing the parallel elements by the equivalent impedance. The write a mesh equation :
−10 + 5 I1 + j15 I1 + 5(1 − j) I1 = 0 ⇒ I1 =
10 10 + j10
Therefore: i1 (t )
= 0.707 co cos(10 t − 45° ) A
Next, find the response to the dc current source acting alone:
= 0.707∠ − 45° A
Current division:
Using superposition: i (t )
I 2
=−
cos(10 t − 45 °) − 2 A = 0.707 co
10 15
× 3 = −2 A
Section 10-11: Phasor Diagrams P10.11-1
*
*
V = V1 − V2 + V3 = ( 3+ 3j) − ( 4 + 2j) + ( −3 − 2j)
= −4 +
j3
P10.11-2
I=
VR
=
R I = 7.4∠42° V
VL
=
Z L I = (1∠90°)(0.74∠42°) = 0.74∠132 ° V
VC
=
Z C I = (10∠−90°)(0.74∠42 °) = 7.4 ∠−48 ° V
VS
=
10∠0° V
10∠0° 10 + j1− j10
= 0.74∠42°
A
P10.11-3
I = 72 3 + 36 3∠(140 ° − 90 °) +144 ∠210 ° + 25 ∠φ
= 40.08 − j24.23 + 25 ∠φ =
46.83∠ − 31.15° + 25∠φ
To maximize I , require that the 2 terms on the right side have the same angle
⇒
φ = − 31.15° .
P10.11-4 Two possible phasor diagrams for currents:
In both cases: I CL
=
I LC
IL
−
=
( 25 )−(15 ) 2
2
=
20 A
6 − 20
= −14
In the first case: I LC
=
IC
⇒
IC
=
A
Section 10-12: 10-12: Phasor Circuits Circuits and the Operational Operational Amplifier P10.12-1
⎛ 104 || − j10 104 ⎞ 10 − j 225 −j = −⎜ = e H (ω ) = ⎟ = −10 1 0 0 0 1 − j Vs (ω ) 2 ⎝ ⎠ Vo (ω )
⎛ 10 − j 225 ⎞ − j 225 ⇒ Vo (ω ) = ⎜ e ⎟ 2 = 10e ⎝ 2 ⎠ cos (1000t − 225°) V vo ( t ) = 10 co Vs (ω ) =
2
P10.12-2
Node equations: V1 − VS
+ j ω C 1 V1 = 0 ⇒ V1 =
R 1 V1
+
R
2
VS 1+
j ω
C R 1 1
⎛ R 3 ⎞ = 0 ⇒ V0 = ⎜ 1 + ⎟⎟ V1 ⎜ R 3R 2 ⎝ ⎠
V1 − V0
Solving: V0 VS
1+
=
R 3 R 2
1 + jω C1 R1
P10.12-3
Node equations: V1
+ j ω C 1 ( V1 − VS ) = 0 ⇒
V1 =
R
1
V1
+
R 2
V1 − V0 R 3
=0
⎛ ⎜ ⎝
⇒
⎛
VS
j ω C1 R1 ⎜1 + ⎜
=
1+
V0 = ⎜1 +
Solving: V0
jω C1 R1 VS
⎝
R 3 ⎞
⎟ R 2 ⎟⎠
1 + j ω C1 R1
j ω
R 3 ⎞
C R 1 1
⎟ V1
R⎟ 2 ⎠
P10.12-4
Node equations: V1 − VS
175 V1
1000
+
+
V1
− j1.6
= 0 ⇒ V1 =
V1 − V0
10000
VS
1 + j 109
= 0 ⇒ V0 = 11 V1
Solving: V0 =
11 1 + j 109
VS =
11 110∠89.5°
( 0.005∠0° )
= 0.5∠ − 89.5° mV Therefore v0 (t ) = 0.5 cos (ω t − 89.5°) mV
P10.12-5 Label the nodes:
The ideal op amps force Va = 0 and Vc = 0. Apply KCL at node a to get
Vb =
Apply KCL at node c to get
Vo =
Z2 Z1 + Z 2 Z4 Z3 + Z4
Vs
Vb
Therefore Vo Vs
=
Z4 Z3 +Z4
×
Z2 Z1 +Z2
P10.12-6 Label a node voltage as Va in each of the circuits.
In both circuits, we can apply KCL at the node between Z3 and Z4 to get Vo =
Z4 Z3 + Z4
Va
In (a) Va =
=
(
Z 2 || Z 3 + Z 4
)
(
Z 1 + Z 2 || Z 3 + Z 4
)
Vs
(
Z2 Z3 + Z4
(
)
)
(
Z1 Z 2 + Z3 + Z 4 + Z 2 Z 3 + Z 4
)
Vs
so Va Vs
=
Z2 Z4
(
)
(
Z1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4
In (b) Va =
Z2 Z1 + Z 2
Vs
so Vo Vs
=
Z4 Z3 +Z4
×
Z2 Z1 + Z2
)
P10.12-7 Label the node voltages Va and Vb as shown:
Apply KCL at the node between Z1 and Z2 to get Z2 Va = Vs Z1 + Z 2 Apply KCL at the node between Z1 and Z2 to get Z3 + Z4 Vb = Va Z3 Apply KCL at the node between Z5 and Z6 to get Z6 Vo = Vb Z5 + Z6 so Vo Vs
=
Z6 Z5 + Z6
×
Z3 + Z4 Z3
×
Z2 Z1 + Z 2
P10.12-8 The network function of the circuit is
1 R 2 ⎞ ⎛ j ω C = ⎜1 + ⎟ 1 V s ⎝ 1000 ⎠ R1 +
Vo
=
j ω C
1+
R 2
Vs
=
5 ∠71.6° 2
Consequently 5 ∠71.6° 2 Equating angles gives
=
R2
1000 = 1000 1 + j ω C R1 1 + j 10−3 R1
Converting the given input and output sinusoids to phasors gives Vo
1+
1+
R 2
1000 1 + j 10−3 R1
71.6° = − tan Equating magnitudes gives R 2 1+ 5 1000 2
=
1 + 10−3 R1
(
)
2
−1
(10− R ) 3
⇒ R1 = tan (71.6 °) ×10 3 = 3006 Ω
1
1+
=
R2
1000
1 + 10−3 × 3006
(
)
2
⎛5 ⎞ ⇒ R 2 = ⎜ 10 − 1⎟ × 10 3 = 6906 Ω ⎝2 ⎠
P10.12-9 Represent the circuit in the frequency domain as
Apply KCL at the top node of the impedance of the capacitor to get Vs − V 1 V V 5 = + ⇒ Vs = 1 + j ( 5 ×10 ) C V 4 4 1 10 10 2
(
)
j100C Apply KCL at the inverting node of the op amp to get V
104
+
Vo
=0
R
⇒
Vo = −
so Vo
−
R
4 2 10 × = Vs 1 + j ( 5 ×105 ) C
Converting the input and output sinusoids to phasors gives Vo Vs
so
=
8∠135° 4∠0°
= 2∠135°
R
104
V
−
2∠135° =
R
R
2 ×10 = 1 + j ( 5 × 105 ) C
2 ×10 4
4
1 + ⎡( 5 ×105 ) C ⎤
⎣
∠180° − tan −1 ( (5 ×10 5 ) C )
2
⎦
Equating angles gives
(
135° = 180° − tan −1 (5 ×10 5 ) C
)
⇒
C =
tan ( 45° ) 5 ×10
5
= 2 ×10 −6 = 2 μ F
Next, equating magnitudes gives R
R
2 × 10
2=
1 + ( 5 ×10 10
5
4
10 ) ) ( 2 ×10 −6
4 = 2 × 10
⇒
2
R = 10 4 = 10 kΩ
P10.12-10
Represent the circuit in the frequency domain using phasors and impedances. To calculate the input impedance Z, add a current source as shown. The input impedance will be given by Z=
Vt It
Label the node voltages as shown. Apply KCL at the noninverting input of the lower op amp to get Vb =
R 4 + R 5 R 5
Vt
Apply KCL at the output of the upper op amp to get R 4 + R 5 I2 =
Apply Ohm’s law twice to get
Vb − Vt 3R
=
R 5
Vt − Vt 3R
=
R4 3R R 5
Vt
V t − V a = R1 I t and V t − V a =
R 4 j ω C 2 R 3 R5
Vt
so R1 I t =
ω j eq L=
Z=
Vt It
=
R 4 j ω C 2 R 3 R5
jω R1 C 2 R3 R5 R 4
Vt
⇒
eq L=
R1 C 2 R3 R5 R4
Section 10.15 How Can We Check…?
P10.15-1
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
V2
= 20 − j 40
KCL at node 1: 2−
V1
10
−
V1 − V 2
j 10
= 2−
− j 20 − j 20 − ( 20 − j 40 ) − = 2 + j 2 − 2 − j 2 = 0 10
j 10
KCL at node 2: V1 − V2
j 10
−
V2
10
⎛ V ⎞ − j 20 − ( 20 − j 40 ) 20 − j 40 ⎛ − j 20 ⎞ + 3⎜ 1 ⎟ = − + 3⎜ ⎟ = (2 + j 1 0 1 0 1 0 1 0 ⎝ ⎠ ⎝ ⎠
The currents calculated from and V2 are correct.
V1
j ) − ( 2 − 4j ) − 6 j =0 2
and V2 satisfy KCL at both nodes, so it is very likely that the
V1
P10.15-2 I1
= 0.390 ∠ 39° and
I2
= 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:
I1
= 0.305 + j 0.244 and
I2
= −0.284
KVL for mesh 1: j 0 ( 0.305 + j0.244) − (− j5) = j10 ≠ 0 8 ( 0.305 + j0.244 ) + 1
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)
P10.15-3 V1
= 19.2 ∠ 68° and
V2
= 24 ∠ 105° V
KCL at node 1 : 19.2 ∠ 68° 8
+
19.2 ∠ 68° j6
− 4∠15 = 0
KCL at node 2: 24 ∠105°
− j4
+
24 ∠105° j 12
+ 4∠15 = 0
The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. 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 P10.15-4
First, replace the parallel resistor and capacitor by an equivalent impedance ZP
=
(3000)( − j 1000) 3000− j 1000
= 949 ∠ − 72° = 300 − j900 Ω
The current is given by I
=
VS
5j00 + Z P
=
100 ∠0° 5j00 +300 − 9j00
= 0.2∠53° A
Current division yields
⎛ − j 1000 ⎞ =⎜ 5 3° ) = 63.3 ∠ − 18.5° mA ⎟ ( 0.2 ∠53 − j 3 0 0 0 1 0 0 0 ⎝ ⎠ ⎛ ⎞ 3000 I2 = ⎜ ⎟ ( 0.2 ∠53° ) = 190∠71.4° mA − j 3 0 0 0 1 0 0 0 ⎝ ⎠ I1
The reported value of I1 is off by an order of magnitude.
P10.15-5
Represent the circuit in the frequency domain dom ain using phasors and impedances. Use voltage division to get 1 j 200C 1
18.3∠ − 24° =
R +
× 20∠0°
j 200C
so 0.915∠ − 24° =
1
1
=
1 + 2j 00 CR
1 + ( 200CR )
2
∠ − tan −1 ( 200CR )
Equating angles gives
−24° = − tan −1 ( 200CR )
⇒
200CR = tan ( 24° ) = 0.4452
The nominal component values cause 200 CR = 0.5. So we expect that the actual component values are smaller than the nominal values. Try C = 5 (1 − 0.10 ) × 10
−6
= 4.5 μ F
Then R =
0.4452 200 × 4.5 × 10−6
= 494.67 Ω
500 − 494.67
= 0.01066 = 1.066% this resistance is within 2% of 500 Ω. We conclude 500 that the measured angle could have been caused by a capacitance that is within 10% of 5 μF and the resistance is within 2% of 500 Ω. Let’s check the amplitude. We require
Since
1 1 + ( 0.4452 )
2
= 0.9136
0.915
So the measured amplitude could also have been caused by the given circuit with C = 4.5 μF and R = 494.67 Ω. We conclude that he measured capacitor voltage could indeed have been produced by the given circuit with a resistance that is within 2 % of 500 Ω and a capacitance that is within 10% of 5 μF.
