Top Banner
lecture 21 1 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003
22

Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

Dec 19, 2015

Download

Documents

Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 1

Impedance and Admittance (7.5)

Prof. Phillips

April 18, 2003

Page 2: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 2

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, )

Page 3: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 3

Impedance

• Resistor: V = I R

– The impedance is ZR = R

• Inductor: V = I jL

– The impedance is ZL = jL

Page 4: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 4

Impedance

• Capacitor:

– The impedance is ZC = 1/jC

Cj1

IV

Page 5: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 5

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.

Page 6: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 6

Impedance Example:Single Loop Circuit

20k+–

1F10V 0 VC

+

= 377

Find VC

Page 7: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 7

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

Page 8: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 8

Impedance Example20k 0

+–

2.65k -9010V 0 VC

+

Page 9: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 9

Impedance Example

Now use the voltage divider to find VC:

0k2090-k65.2

90-k65.20 10VCV

4.82- 1.31VCV

Page 10: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 10

What happens when changes?

20k

1F10V 0 VC

+

= 10

Find VC

+–

Page 11: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 11

Low Pass Filter:A Single Node-pair Circuit

Find v(t) for =2 3000

1k0.1F

5mA 0

+

V

Page 12: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 12

Find Impedances

1k

-j5305mA 0

+

V

Page 13: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 13

Find the Equivalent Impedance

5301000

5301000

j

jeq

Z

5mA 0

+

VZeq

Page 14: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 14

Parallel Impedances

9.271132

90530010

5301000

5301000 3

j

jeqZ

1.622.468eqZ

Page 15: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 15

Computing V

1.622.4680mA5eqIZV

1.62V34.2V

)1.623000t(2cosV34.2)( tv

Page 16: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 16

Change the Frequency

Find v(t) for =2 455000

1k0.1F

5mA 0

+

V

Page 17: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 17

Find Impedances

1k

-j3.55mA 0

+

V

Page 18: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 18

Find an Equivalent Impedance

5.31000

5.31000

j

jeq

Z

5mA 0

+

VZeq

Page 19: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 19

Parallel Impedances

2.01000

905.3010

5.31000

5.31000 3

j

jeqZ

8.895.3eqZ

Page 20: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 20

Computing V

8.895.30mA5eqIZV

8.89mV5.17V

)8.89455000t(2cosmV5.17)( tv

Page 21: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 21

Impedance Summary

Element Impedance

Capacitor ZC = 1 / jC = -1/C 90

Inductor ZL = jL = L 90

Resistor ZR = R = R 0

Page 22: Lecture 211 Impedance and Admittance (7.5) Prof. Phillips April 18, 2003.

lecture 21 22

Class Examples