Department of Electrical & Computer Engineering
ECE3413 — Introduction to Electronic Circuits
Solution
1.
Fall 2012
Homework #6
(a)
time domain:
phasor domain (ω = 377):
v(t) = 155 cos(377t − 25◦ ) V
V = 155∠−25◦
(b)
time domain:
phasor domain (ω = 1000):
v(t) = 5 sin(1000t − 40◦ ) V = 5 cos(1000t − 130◦ ) V
V = 5∠−130◦
(c)
time domain:
phasor domain (ω = 10):
i(t) = 10 cos(10t + 63◦ ) + 15 cos(10t − 42◦ ) A
I = 10∠63◦ + 15∠−42◦ ≈ 15.73∠−4.11◦
(d)
time domain:
phasor domain (ω = 500π):
2.
i(t) = 460 cos(500πt − 25◦ ) − 220 sin(500πt + 15◦ ) A
= 460 cos(500πt − 25◦ ) + 220 cos(500πt + 105◦ ) A
I = 460∠−25◦ + 220∠105◦ ≈ 360.4∠2.88◦
We have
i(t) = 17 cos(ωt − π/12) mA
v(t) = 3.5 cos(ωt + 1.309) V
where ω = 628.3 rad/s. In the phasor domain (ω = 628.3), these become
I = 0.017∠−15◦
V = 3.5∠75◦
where we have converted the phase angles to degrees. The impedance of the component is
Z=
V
I
3.5∠75◦
0.017∠−15◦
≈ 205.9∠90◦
=
(a) An inductor has impedance of the form
ZL = jωL = ωL∠90◦
Consequently, the component is an inductor due to the 90◦ phase in Z.
(b) The inductance is thus
|Z|
ω
205.9
≈
628.3
≈ 0.328 H
L=
3.
(a) The source current is
is (t) = 13 cos(1000t + π/6) mA.
In the phasor domain (ω = 1000), this is
Is = 0.013∠30◦ .
(b) The impedance of the capacitor is
ZC =
1
jωC
1
j · 1000 · 0.5µ
= −j2000
=
(c) In the phasor domain (ω = 1000):
+
VC
0.013∠30º
-j2000
-
The capacitor voltage is
VC = Is ZC
= 0.013∠30◦ · (−j2000)
= 26∠−60◦
In the time domain, we have
vC (t) = 26 cos(1000t − 60◦ ) V
4.
Time-domain circuit:
3Ω
3H
i(t)
12cos(3t) V +
−
1/3 F
In the phasor domain:
Inductor:
ZL = jωL = j · 3 · 3 = j9
Capacitor:
ZC =
1
1
=
jωC
j·3·
1
3
= −j
Phasor-domain circuit (ω = 3):
3
12 +
−
j9
I
-j
KVL around the single mesh yields:
3I + j9I + (−j)I − 12 = 0
(3 + j8)I = 12
12
I=
3 + j8
≈ 1.405∠−69.4◦
Thus,
i(t) ≈ 1.405 cos(3t − 69.4◦ ) A
5.
Time-domain circuit:
+
3Ω
10cos(2t) A
3H
1/3 F
v(t)
-
In the phasor domain:
Inductor:
ZL = jωL = j · 2 · 3 = j6
Capacitor:
ZC =
1
1
=
jωC
j·2·
1
3
= −j
3
2
Phasor-domain circuit (ω = 2):
+
3
10
j6
-j 3/2
V
-
KCL at the single node yields:
10 +
0−V 0−V 0−V
+
+
=0
3
j6
−j 23
1
1
2
− −
V = −10
+
3 j6 j3
(0.6∠−123.7◦ ) V = −10
−10
0.6∠−123.7◦
≈ 16.6∠−56.3◦
V≈
Thus,
v(t) ≈ 16.6 cos(2t − 56.3◦ ) V
6.
We have the circuit:
1/4 H
1/8 F
2Ω
In the phasor domain:
Inductor:
ZL = jωL = j · 4 ·
Capacitor:
ZC =
1
1
=
jωC
j·4·
1
=j
4
1
8
= −j2
Phasor-domain circuit (ω = 4):
j
-j 2
The impedance is
Z = j + (−j2) k 2
1
=j+ 1
1
−j2 + 2
= 1Ω
Note that the impedance is purely resistive at this frequency.
2
7.
The time-domain circuit is
0.5 H
5Ω
10 H
1H
6cos(2t) A
0.5 F
i(t)
In the phasor domain:
0.5-H inductor:
jωL = j · 2 ·
1
=j
2
10-H inductor:
jωL = j · 2 · 10 = j20
1-H inductor:
jωL = j · 2 · 1 = j2
Capacitor:
1
1
=
jωC
j·2·
1
2
= −j
Phasor-domain circuit (ω = 2):
5
j
j 20
j2
6
-j
I
6
I
KVL around I:
5I + (j2)I + (−j)(I − 6) + (j20)(I − 6) = 0
(5 + j21)I = j114
j114
I=
5 + j21
≈ 5.28∠13.4◦
Thus,
i(t) ≈ 5.28 cos(2t + 13.4◦ ) A
8.
The time-domain circuit is
75 Ω
+
15cos(1500t) V −
0.5 H
1 µF
i1(t)
100 Ω
i2(t)
In the phasor domain:
Inductor:
ZL = jωL = j · 1500 ·
Capacitor:
ZC =
1
= j750
2
1
1
=
≈ −j666.7
jωC
j · 1500 · 1µ
Phasor-domain circuit (ω = 1500):
75
j 750
+
15 −
-j 666.7
I2
I1
100
KVL around I1 :
75I1 + (−j666.7)(I1 − I2 ) − 15 = 0
(75 − j666.7)I1 + (j666.7)I2 = 15
(1)
KVL around I2 :
(j750)I2 + 100I2 + (−j666.7)(I2 − I1 ) = 0
j666.7I1 + (100 + j83.3)I2 = 0
Solving (1) and (2) simultaneously yields:
75 − j666.7
j666.7
j666.7
100 + j83.3
I1
15
=
I2
0
I1 = 0.00382∠46.6◦
I2 = 0.0196∠−83.2◦
Thus,
i1 (t) = 3.82 cos(1500t + 46.6◦ ) mA
i2 (t) = 19.6 cos(1500t − 83.2◦ ) mA
(2)