PHYSICS 3050
Total - 42 marks
Solutions to Problem Set 2
was due October 18, 2011
”There is no reason anyone would want a computer in their home.”
Ken Olson, president, chairman and founder of Digital Equipment Corp., 1977
1. (6 marks) At time t = 0, the capacitor in the circuit below carries no charge and the
switch is moved to position A. After one time constant τ , the switch is then moved to
position B. Sketch the voltage VR across the resistor as a function of time and label
your sketch with quantitative values.
A
B
1 kΩ
1V
2V
0.1 µF
• From KVL VR + VC = Vs where
Vs = 0V t < 0
= 2V 0 ≤ t ≤ τ
= −1V t > τ
and VC = 0 V for t ≤ 0 (so VR = 0 V for t ≤ 0). The time constant is τ = RC =
(103 Ω)(10−7 F) = 10−4 s or 100 µs. During the time interval 0 ≤ t ≤ τ the capacitor
starts to charge up to 2 V with time constant τ . Therefore,
VC (0 ≤ t ≤ τ ) = (2 V)(1 − e−t/τ )
with VC (t = τ ) = (2 V)(1-e−1 ) = 1.26 V. So
VR (0 ≤ t ≤ τ ) = (2 V) − [(2 V)(1 − e−t/τ )] = (2 V)e−t/τ
For t > τ Vs = -1 V and the capacitor will now start to “charge up” to -1 V. But now
it is not starting at 0 V but VC (t = τ ) = 1.26 V so;
VC (t > τ ) = (1.26 V)e−(t−τ )/τ + (−1 V)(1 − e−(t−τ )/τ ) = (2.26 V)e−(t−τ )/τ − (1 V)
Can see that this has the right form since VC (t = τ ) = 1.26 V and VC (t → ∞) → -1
V. And from this we get
VR (t > τ ) = (−1 V) − VC (t > τ ) = (−2.26 V)e−(t−τ )/τ
2. Variation on Text Problem 3, page 86 (2 marks)
Represent the following complex numbers in polar (i.e., exponential) form: 3 + 4j and
4 − 3j.
√
−1
2
2
• For Z=3+4j, |Z| =
(4/3) = 53.13◦ so Z=5exp (j53.13◦ )
√ 3 + 4 = 5 and θZ =tan
−1
2
2
For Z=4-3j, |Z| = 4 + 3 = 5 and θZ =tan (−3/4) = -36.87◦ so Z=5exp (−j36.87◦ )
3. Variation on Text Problem 7, page 86 (2 marks)
If V1 = 7cos (ωt + 20◦ ) and V2 = 5cos (ωt − 30◦ ) then what is V3 = V1 + V2 (in the
form V3 = |V3 | cos (ωt + φ))?
• Treat them as phasors (or use the complex representation) then
V1 = 7ej20 ejωt = (6.58 + j2.39)ejωt
◦
V2 = 5e−j30 ejωt = (4.33 − j2.5)ejωt
◦
V1 + V2 = (10.91 − j0.11)ejωt = 10.91e−j0.56 +jωt
◦
So V1 + V2 = 10.91 cos (ωt − 0.56◦ ).
4. Variation on Text Problem 11, page 86 (2 marks)
What is the impedance of a resistor R in parallel with a series combination of a capacitor
C and an inductor L?
• Using the complex impedances ZR = R, ZC = 1/jωC, and ZL = jωL
1
Z
=
1
1
ZR + Z C + Z L
+
=
ZR ZC + Z L
ZR (ZC + ZL )
1
R jωC
+ jωL
ZR (ZC + ZL )
Z =
=
1
ZR + Z C + Z L
R + jωC + jωL
This can be simplified a number of ways, e.g.,
Z=
(1 − ω 2 LC)
(1 − ω 2 LC) + jωRC
5. (5 marks) What is the resonant frequency of the circuit below? The voltage source
has a magnitude of 1 V. Find the magnitudes of the current and the voltage through
each component at resonance. Why don’t the magnitudes of the voltages across the 3
elements add to 1 V?
100 Ω
0.1 µF
25 mH
•
i =
|i| =
V
V
V
=
=
Z
ZR + Z C + Z L
R + j ωL −
|V |
= r
|Z|
V0
R2 + ωL −
1
ωC
So |i| resonates when ωL − (1/ωC) = 0 or
fr =
1
ωC
2
ωr
1
1
= q
= √
= 3180 Hz = 3.2 kHz
2π
2π LC
2π (.025 H)(10−7 F)
For each component V = iZ so
VR = iZR =
VR
R + j ωL −
1
ωC
√
|VR (ω = 1/ LC)| = |V | = 1 V
!
√
1
V
VC (ω = 1/ LC) = iZC =
R
jωC
√
|VC (ω = 1/ LC)| =
|V |
|V |
=
ωr RC
R
s
L
(1 V)
=
C
(100 Ω)
√
V
jωL
VL (ω = 1/ LC) = iZL =
R
s
√
|V | L
= 5V
|VL (ω = 1/ LC)| =
R C
v
u
u (.025
t
(10−7
H)
= 5V
F)
|VR | + |VC | + |VL | 6= 1 V because the three are out of phase. Specifically, when ωL −
(1/ωC) = 0, VC = −VL so V = VR + VC + VL = VR .
6. (10 marks) Given that V=V0 cos ωt, what are the magnitude of the current i (i.e., |i|),
the magnitude of VA -VB (i.e., |VAB |) and the phase difference between i and V for the
following circuit.
i
o
R
V
L
A
C
B
o
• First transform to the complex representation of the circuit.
i
A
ZR
V
ZL
V
ZC
A
ZR
B
where
Z0 =
Then we have;
ZC ZL
ZL
jωL
=
=
ZC + Z L
1 + ZL /ZC
1 − ω 2 LC
V jωL
Z0
1−ω 2 LC
=
=
V
jωL
ZR + Z 0
R(1 − ω 2 LC) + jωL
R + 1−ω2 LC
V0 ωL
|VAB | = q
i =
R2 (1 − ω 2 LC)2 + ω 2 L2
V
V (1 − ω 2 LC)
V
=
=
jωL
ZR + Z 0
R(1 − ω 2 LC) + jωL
R + 1−ω
2 LC
|i| = q
V0 (1 − ω 2 LC)
R2 (1 − ω 2 LC)2 + ω 2 L2
And from the expressions for V and i we have;
V
= i
/
B
jωL
VAB = V
Z
[R(1 − ω 2 LC) + jωL]
= iT
(1 − ω 2 LC)
θV
= θi + θT
∆θ = tan
−1
"
ωL
R(1 − ω 2 LC)
#
√
a) When ω = 1/ LC (i.e., 1−ω 2 LC = 0) we have |VAB | = V0 , |i| = 0 and ∆θ = tan−1 (∞)
= 90◦ .
b) ω = 0 we have |VAB | = 0, |i| = V0 /R and ∆θ = tan−1 (0) = 0◦ .
c) ω → ∞ we have |VAB | → 0, |i| → V0 /R and ∆θ = − tan−1 (0) = 180◦ .
7. (5 marks) Calculate the gain and phase shift as a function of ω for the following circuit.
What kind of filter is this?
C
V in
V out
L
• First we go to the complex representation of the circuit
ZC
C
V in
V 0e
L
jω t
ZL
with ZC = 1/jωC and ZL = jωL. Therefore, now, using voltage division, we have
Vout = Vin
ZL
ZL + Z C
jωL
Vout
=
Vin
jωL + 1/jωC
2
−ω LC
=
1 − ω 2 LC
T =
Therefore,
gain = AV =
Vout Vin
=
ω 2 LC
1 − ω 2 LC
The gain → 0 as ω → 0 and the gain → 1 as ω → ∞. Further, the gain → ∞
2
when ω√
LC = 1, so it is a resonant + high-pass circuit, with the resonant frequency of
ω = 1/ LC. The transfer function is completely real so the phase shift, given by
tan
−1
I(T )
R(T )
!
is 0.
8. (10 marks) For electrocardiographs (electrical measurement of the heart), a real problem is 60 Hz AC line interference. Design the “notch” filter in the following circuit so
that the gain is a minimum at f = 60 Hz. You also want as narrow a bandwidth as is
practical. The output is taken as the voltage across RL .
C
V AC
L
RL
V signal
• Go to the complex representation of the circuit. Taking Vin = Vsignal + VAC and Vout
= VL and Z 0 as the parallel combination of the capacitor and inductor (i.e., 1/Z 0 =
jωC + 1/jωL so Z 0 = jωL/(1 − ω 2 LC)) then
Vout = Vin
RL
RL + jωL/(1 − ω 2 LC)
RL (1 − ω 2 LC)
=
RL (1 − ω 2 LC) + jωL
RL (1 − ω 2 LC)
= q
RL2 (1 − ω 2 LC)2 + ω 2 L2
= Vin
Vout
Vin
AV
RL
RL + Z 0
√
The “tank” (notch) circuit gain is a minimum (=0) for ω = 2πf = 1/ LC. We want
this to occur at 60 Hz so
LC =
1
= 7.04 × 10−6 HF
(2πf )2
The bandwidth for a tank circuit is ∆ω = R/L so ∆f = R/2πL. You’d like this to be
on the order of 10’s of Hz so that you’re filtering out around 60 Hz and not much else.
Say RL is 50 Ω then L=R/2π∆f ≈ 8/∆f H. For ∆f = 20 Hz (filtering out 60 ± 20 Hz)
then L = 400 mH and C = 18 µF.