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Impedance and Admittance (7.5)
Prof. Phillips
April 18, 2003
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Impedance
• AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:
V = I Z
• Z is called impedance (units of ohms, )
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Impedance
• Resistor: V = I R
– The impedance is ZR = R
• Inductor: V = I jL
– The impedance is ZL = jL
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Impedance
• Capacitor:
– The impedance is ZC = 1/jC
Cj1
IV
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Some Thoughts on Impedance
• Impedance depends on the frequency, f
• Impedance is (often) a complex number.
• Impedance is not a phasor (why?).
• Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.
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Impedance Example:Single Loop Circuit
20k+–
1F10V 0 VC
+
–
= 377
Find VC
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Impedance Example
• How do we find VC?
• First compute impedances for resistor and capacitor:
ZR = 20k= 20k 0
ZC = 1/j (377·1F) = 2.65k -90
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Impedance Example20k 0
+–
2.65k -9010V 0 VC
+
–
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Impedance Example
Now use the voltage divider to find VC:
0k2090-k65.2
90-k65.20 10VCV
4.82- 1.31VCV
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What happens when changes?
20k
1F10V 0 VC
+
–
= 10
Find VC
+–
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Low Pass Filter:A Single Node-pair Circuit
Find v(t) for =2 3000
1k0.1F
5mA 0
+
–
V
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Find Impedances
1k
-j5305mA 0
+
–
V
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Find the Equivalent Impedance
5301000
5301000
j
jeq
Z
5mA 0
+
–
VZeq
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Parallel Impedances
9.271132
90530010
5301000
5301000 3
j
jeqZ
1.622.468eqZ
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Computing V
1.622.4680mA5eqIZV
1.62V34.2V
)1.623000t(2cosV34.2)( tv
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Change the Frequency
Find v(t) for =2 455000
1k0.1F
5mA 0
+
–
V
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Find Impedances
1k
-j3.55mA 0
+
–
V
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Find an Equivalent Impedance
5.31000
5.31000
j
jeq
Z
5mA 0
+
–
VZeq
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Parallel Impedances
2.01000
905.3010
5.31000
5.31000 3
j
jeqZ
8.895.3eqZ
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Computing V
8.895.30mA5eqIZV
8.89mV5.17V
)8.89455000t(2cosmV5.17)( tv
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Impedance Summary
Element Impedance
Capacitor ZC = 1 / jC = -1/C 90
Inductor ZL = jL = L 90
Resistor ZR = R = R 0
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Class Examples