PHYS 343 Homework #7
SOLUTIONS
1. Consider the two circuit elements below. One is a series RC element and the other is
a parallel RC circuit element.
C
C
R
R
(a) Using qualitative arguments, predict what the impedance of each circuit will look
like at (1) very low frequencies, and (2) very high frequencies. That is, explain for
each case whether it would look like a resistor, a capacitor, or what. Also predict
whether the magnitude of the impedance is very high, very low, or the same as
the resistor.
(b) Now find expressions for the impedance Z for each of the two cases. Express Z
in the form |Z|ejφ. Do your expressions confirm your predictions above? Explain
why or why not by making the necessary approximations.
(a) At very low frequencies, the impedance of the series circuit will look like a capacitor:
the capacitor will have very high impedance and will dominate the series combination.
However, the impedance of the parallel circuit will look like a resistor at low frequencies;
the high impedance of the capacitor will not be noticed.
At high frequencies, the situation is reversed: the capacitor’s impedance is very low,
so the dominant element in the series circuit is the resistor. In the parallel circuit,
the low-impedance capacitor is carrying most of the current, so we mainly see the
capacitor.
(b) Series:
Z =R−
Im
j
ωC
so
|Z| =
q
R2 + (1/ωC)2
q
R
or |Z| = R 1 + (ωRC)−2
Re
φ
To find the phase angle,
tan θ =
−1/ωC
1
=−
R
ωRC
In summary,
q
Z = R 1 + (ωRC)−2 e−j arctan (1/ωRC)
−j
ωC
Z
Parallel:
Im
1
Z
1
1
1
=
+
= + jωC
R −j/ωC
R
1 + jωRC
=
R
Z =
−jωRC
R
1 + jωRC
R
|Z| = q
1 + (ωRC)2
To obtain the phase angle, we must rationalize the denominator. Bummer:
R
1 − jωRC
1 + jωRC 1 − jωRC
R(1 − jωRC)
=
1 + (ωRC)2
Z =
Now find the angle:
tan φ =
In summary,
−ωRC
1
or φ = − arctan ωRC
R
e−j arctan ωRC
Z=q
2
1 + (ωRC)
The results verify our predictions:
Frequency:
Series
Parallel
Low
−j/ωC
R
Re
φ
High
R
−j/ωC
Z
2. Consider the two circuit elements below. One is a series RL element and the other
is a parallel RL circuit element. Note: the parallel configuration is not very realistic,
because most inductors have a resistance due to the wire itself. So we really should
draw another resistor in series with the inductor. We will ignore this and assume the
wire resistance is zero.
L
L
R
R
(a) Using qualitative arguments, predict what the impedance of each circuit will look
like at (1) very low frequencies, and (2) very high frequencies. That is, explain
for each case whether it would look like a resistor, an inductor, or whatever. Also
predict whether the magnitude of the impedance is very high, very low, or the
same as the resistor.
(b) Now find expressions for the impedance Z for each of the two cases. Express Z
in the form |Z|ejφ. Do your expressions confirm your predictions above? Explain
why or why not by making the necessary approximations.
(a) At very low frequencies, the impedance of the inductor is low, so the series circuit
looks mainly like a resistor. The parallel combination, however, is dominated by the
inductor since the inductor’s low-impedance path carries most of the current.
At high frequencies, the inductor has a high impedance, so in the series circuit the
inductor dominates. In the parallel circuit, by contrast, the inductor carries little
current compared to the resistor, so the circuit looks mainly like a resistor.
(b) Series:
Im
Z = R + jωL
|Z| =
tan φ =
ωL
R
q
R2 + (ωL)2
soφ = arctan
φ
ωL
R
R
Summary:
Z=
q
Z
j ωL
ωL
R2 + (ωL)2 ej arctan ( R )
Re
Parallel case:
Im
1
Z
1
1
jωL + R
=
+
=
R jωL
jωRL
jωRL
Z =
R + jωL
ωRL
|Z| = q
R2 + (ωL)2
R
|Z| = s
R 2
1+
ωL
To find φ, we rationalize the denominator:
φ
ωL
jωRL R − jωL
R + jωL R − jωL
ω 2 RL2 + jωR2 L
=
R2 + (ωL)2
ωRL
=
[ωL + jR]
2
R + (ωL)2
Z =
So
φ = arctan
In summary,
Z=s
1+
R
Re Z
= arctan
Im Z
ωL
R
R
R
ωL
2
ej arctan ωL
Our predictions are verified:
Frequency:
Series
Parallel
Z
jR
Low High
R
jωL
jωL
R
Re
3. Let the complex number Z be
Z =1+
Show that
Z ∗Z =
A
B + jC
(A + B)2 + C 2
B2 + C 2
First, it’s probably easier to do this if the expression is a single fraction. We note
1+
jA
A + B + jC
=
B + jC
B + jC
Then
!
!
A + B − jC
A + B + jC
Z Z =
B + jC
B − jC
2
(A + B) + jC(A + B) − jC(A + B) + C 2
=
B 2 + jCB − jCB + C 2
2
(A + B) + C 2
=
B2 + C 2
∗
4. Let the complex number Z be
Z =1+
Show that
Z ∗Z =
We note
1+
jA
B + jC
(A + C)2 + B 2
B2 + C 2
B + jA + jC
B + j(A + C)
jA
=
=
B + jC
B + jC
B + jC
Then
B + j(A + C)
Z Z =
B + jC
2
(B + (A + C)2
=
B2 + C 2
((A + C)2 + B 2
=
B2 + C 2
∗
!
B − j(A + C)
B − jC
!
5. Consider the complex number
Z =1+
5j
2+j
Write this number in the form A + Bj. You will have to “rationalize the denominator”
to do this. Now, sketch a graph showing where this number is in the complex plane.
First we convert to a single fraction:
1+
5j
2 + 6j
=
2+j
2+j
!
!
2 + 6j
2−j
Z =
2+j
2−j
4 + 6 + 12j − 2j
10 + 10j
=
=
22 + j(−j)
5
= 2 + 2j
Im
2
1
−2
−1
1
−1
−2
2 Re