Department of Electrical & Computer Engineering
ECE3413 — Introduction to Electronic Circuits
Solution
Homework #5
4.2. We have a capacitor with capacitance C = 100µF. For a capacitor, the current is
iC (t) = C
d
vC (t)
dt
(a) For vC (t) = 40 cos(20t − π/2) V,
d
40 cos(20t − π/2)
dt = 10−4 · 40 · −20 sin(20t − π/2)
= −80 sin(20t − π/2) mA
ic (t) = 100µ ·
(b) For vC (t) = 20 sin(100t) V,
d
20 sin(100t)
dt = 10−4 · 20 · 100 cos(100t)
ic (t) = 100µ ·
= 0.2 cos(100t) A
(c) For vC (t) = −60 sin(80t + π/6) V,
d
−60 sin(80t + π/6)
dt
= 10−4 · (−60) · 80 cos(80t + π/6)
= −0.48 cos(80t + π/6) A
ic (t) = 100µ ·
(d) For vC (t) = 30 cos(100t + π/4) V,
d
30 cos(100t + π/4)
dt = 10−4 · 30 · −100 sin(100t + π/4)
= −0.3 sin(100t + π/4) A
ic (t) = 100µ ·
4.4. The current in the 2-H inductor is
Fall 2012


0 A,
i(t) = t A,


10 A,
−∞ < t < 0,
0 ≤ t < 10,
10 ≤ t < ∞
The energy in the inductor, as a function of time, is
1 2
Li (t)
2
1
= · 2 · i2 (t)
2


−∞ < t < 0,
0 J,
2
= t J,
0 ≤ t < 10,


100 J, 10 ≤ t < ∞
w(t) =
4.7. The voltage on the 0.1-F capacitor is


0 V, −∞ < t < 0
v(t) = t V,
0 ≤ t < 10


10 V, 10 ≤ t < ∞
The energy in the capacitor, as a function of time, is
1 2
Cv (t)
2
1
= · 0.1 · v 2 (t)
2


−∞ < t < 0
0 J,
2
= 0.05t J, 0 ≤ t < 10


5 J,
10 ≤ t < ∞
w(t) =
4.10. Under steady-state conditions, all voltages and currents in the circuit are DC. This means that all capacitors are equivalent
to open circuits and all inductors are equivalent to short circuits. The circuit becomes:
+ vC2 -
iL
2Ω
+
− 6V
+
vC1
-
+
vC3
-
4Ω
8Ω
6Ω
i1
iL
We note that, since the 3-F capacitor is an open circuit, no current flows in the 6-Ω resistor, and its voltage is zero.
We use mesh analysis to find the currents i1 and iL .
KVL around i1 :
2i1 + 4(i1 − iL ) − 6 = 0
6i1 − 4iL = 6
(1)
KVL around iL :
8iL + 4(iL − i1 ) = 0
−4i1 + 12iL = 0
Solving (1) and (2) simultaneously, we arrive at
i1 ≈ 1.286 A
iL ≈ 0.429 A
The capacitor voltages are then:
vC1 = 4(i1 − iL )
≈ 4(1.286 − 0.429)
≈ 3.429 V
vC2 = 0 V
vC3 = 8iL
≈ 8 · 0.429
≈ 3.429 V
(2)
Thus, the stored energies are:
1
2
C1 vC1
2
1
≈ · 2 · 3.4292
2
≈ 11.76 J
wC1 =
1
2
C2 vC2
2
= 0J
wC2 =
1
2
C3 vC3
2
1
≈ · 3 · 3.4292
2
≈ 17.63 J
1
wL = Li2L
2
1
≈ · 2 · 0.4292
2
≈ 0.184 J
WC3 =
4.29. We have a signal
x(t) = 2 cos(ωt) + 2.5
The period of the signal is
2π
ω
T =
The average value is
1
x(t) =
T
=
1
T
2
=
T
T
Z
x(t) dt
0
Z
0
Z
=0+
T
T
2 cos(ωt) + 2.5 dt
cos(ωt) dt +
0
1
T
1
· 2.5T
T
Z
T
2.5 dt
0
= 2.5
due to the fact that the average of the cosine is zero over an integer number of periods.
The rms value is
s
1
T
T
=
s
Z
1
T
1
T
=
s
Z
T
=
s
Z
1
T
Z
xrms =
x2 (t) dt
0
0
0
0
2
2 cos(ωt) + 2.5 dt
T
T
4 cos2 (ωt) + 10 cos(ωt) + 6.25 dt
4 cos2 (ωt) dt +
1
T
Z
T
10 cos(ωt) dt +
0
Note that the rms value of
y(t) = 2 cos(ωt)
1
T
Z
0
T
6.25 dt
is
2
yrms = √
2
√
since the rms value of a sinusoid of amplitude A is A/ 2. Thus,
1
T
Z
0
T
Z
1 T 2
y (t) dt
T 0
2
= yrms
2
2
= √
2
=2
4 cos2 (ωt) dt =
and then
√
xrms = 2 + 6.25
≈ 2.872