HW 6 answers
Using the parallel Z formula for two elements in parallel:
1 / ( 1/Z
1
+ 1/Z
2
)
=
Z
1
Z
2
/(Z
1
+Z
2
)
and
Z
1
=
j
ϖ
L
and
Z
2
= 1/
j
ϖ
C = –
j
(1/
ϖ
C)
then:
(
j
ϖ
L)(
j
(–1/
ϖ
C))
(
ϖ
L/
ϖ
C)
j
ϖ
L
(
j
ϖ
L)+
j
(–1/
ϖ
C)
j
(
ϖ
L–1/
ϖ
C)
(1–
ϖ
2
LC)
for
ϖ
=500
Z =
j
(500)(0.1)/(1 – 500
2
(0.1)(10e–6))
=
j
66.67 ohms
for
ϖ
=1000
Z =
j
(1000)(0.1)/(1 – 1000
2
(0.1)(10e–6))
=
j
∞
ohms
(resonant)
For parallel L and C that are “resonant” the overall impedance becomes infinite if the L
and C have no losses.
In practice, the L and C do have some losses, so the impedance is
very high, but not infinite.
The very high impedance of the resonant parallel LC appears
to the rest of the network as an open circuit.
This is a useful property that is used to
block current flow for a specific frequency, i.e. it is a filter that can stop a signal at that
frequency from getting from one part of a circuit to another part of the circuit.
This is
sometimes referred to as a “trap” (like an animal trap contains the animal).
The
electronic “trap” keeps the signal from getting outside of the trap area.
for
ϖ
=2000
Z =
j
(2000)(0.1)/(1 – 2000
2
(0.1)(10e–6))
=
–
j
66.67 ohms
Note that Z for
ϖ
=500 and Z for
ϖ
=2000 differ only in that they are 180
°
apart from each
other.
1
=
=
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5+
j
15

j
15
I
1
20
=

j
15
j
15
j
10
I
2
(10)
Putting this into my calculator gives:
I
1
=
1.644
∠
80.54°
I
2
=
2.977
∠
74.20°
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 Fall '08
 Staff
 Volt, Complex number, KVL loop equations, KCL NODAL EQUATION

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