33-228 Electronics Spring 2014
Carnegie Mellon University Department of Physics
Homework 3 SOLUTIONS
Problems 2.17, 2.21, 2.27, 2.28, 2.29 and 2.D42.
2.17
 For a capacitor and an inductor in series, the equivalent impedance is just
 
eq
 =
 +
 Z 
L
. Expanding this , we have that
eq
 = 1
 jω
 +
 jωL
eq
 = 1
ω
2
LC  jω
eq
 = (
 j
)1
ω
2
LC ωC 
The magnitude of this impedance is zero when
 ω
 =
 1
√ 
LC 
.
2.21
 For three components in parallel, the impedances add as1
eq
= 1
R
 + 1
+ 1
L
.
1
eq
= 1
R
 +
 jω
 + 1
 jωL
1
eq
= 1
R
 + 1
ω
2
LC  jωL
1
eq
=
 L
 +
 R
(1
ω
2
LC 
)
 jωLR
Inverting, we get that
eq
 =
 jωLRR
(1
ω
2
LC 
) +
 jωL
eq
 = (
 jωLR
)(
R
(1
ω
2
LC 
)
 jωL
)
R
2
(1
ω
2
LC 
)
2
+
 ω
2
L
2
eq
 =
 ω
2
L
2
RR
2
(1
ω
2
LC 
)
2
+
 ω
2
L
2
+
 j ωLR
2
(1
ω
2
LC 
)
R
2
(1
ω
2
LC 
)
2
+
 ω
2
L
2
As we care looking for the case when
 
eq
 is real, we only need to find the case whenthe imaginary part of 
 1
eq
is zero. Thus, we have1
eq
= 1
R
 +
 jω
 + 1
 jωL
1
eq
= 1
R
 + 1
ω
2
LC  jωL
1
 
The imaginary part is zero when
 ω
2
LC 
 = 0, or
 ω
 =
 1
√ 
LC 
. At this frequency, themagnitude of the impedance is just
 Z 
 =
 R
.
2.27
 Assuming a linear fall-off of attenuation versus frequency on a log-log plot, we wantto show that a 20
dB
 change over a factor of 10 in frequency corresponds to a 6
dB
change over a factor of 2 in frequency. This corrsponds to20
dB
log
10
(10) =
 xdB
log
10
(2)20
dB
1 =
 xdB
(0
.
301)
x
 = 6
dB
2.28
 If we have an attenuation of 60
dB
 per decade, then the power is given as:
 p
 =
 
60
dB
20
dB p
 =
 
3which means that we have
|
G
|
 
3
2.29
 If we have a gain that goes as
 G
(
ω
) =
 A/
√ 
ω
, the constant
 A
 must have units of 
 s
12
.We also see that
|
G
|
 
12
so
 p
 =
12
 and the attenuation per deacde is then
12 =
 
A
20
dBA
 = 10
dB/decade
2.D42
 The basic idea would be to gave the capacitors rapidly discharge through the hotdogs,thus cooking them. The energy stored in a fully charged capacitor would be
 = 12
C
 2
 = 12(0
.
5
)(35
 )
2
 = 306
In order to cook the hot dog, we need to raise its temperature to about 65 C from arefrigerated temperature of about 10 C, or ∆
 = 55 C. The energy needed to do this2
 
is
 =
 m
·
c
·
 = (50
g
)
·
(4
.
2
J/
(
gC 
))
·
(55
)
 = 11
.
5
kJ .
Thus, we need the energy stored in 38 capacitors to be able to cook the hot dog. Todo this, we could put the 38 capacitors in parallel, to yield a total capacitance of 17
.We now need to discharge this (quickly) through the hotdog. The discharge time isgiven by tha characteristic time of the circuit, or
 τ 
RC 
 =
 RC 
. This time is 1700 s orabout half and hour. This would seem far too long to cook the hot dog.In order to cook the hot dog, we need the charateristic time to be on the order of 50s, which for a fixed resistance of 100Ω, yields a capacitance of about 0
.
5 F. To store11
.
5 kJ in this capacitor, we would need the voltage across the capacitor to be
 =
 
2
·
E/C 
 =
√ 
46000
 = 214
V .
To hold 214 V with 35 V capacitors, we need to place about seven of them in series.Unfortunately, the equivalent capacitance of these in series is
 17
 of the individual ca-pacitances. We can recover this by building the circuit shown in Figure 1. Thus, inorder to be able to do this, your friend’s uncle will need 49 of the capacitors and a DCvoltage source that can provide 200 V as well as switching capability for 200 V DC.
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Figure 1:
 The capacitor array for problem D2.31
3
of 3