lecture 21 1

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, )

lecture 21 3

Impedance

• Resistor: V = I R

– The impedance is ZR = R

• Inductor: V = I jL

– The impedance is ZL = jL

lecture 21 4

Impedance

• Capacitor:

– The impedance is ZC = 1/jC

Cj1

IV

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.

lecture 21 6

Impedance Example:Single Loop Circuit

20k+–

1F10V 0 VC

+

–

= 377

Find VC

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

lecture 21 8

Impedance Example20k 0

+–

2.65k -9010V 0 VC

+

–

lecture 21 9

Impedance Example

Now use the voltage divider to find VC:

0k2090-k65.2

90-k65.20 10VCV

4.82- 1.31VCV

lecture 21 10

What happens when changes?

20k

1F10V 0 VC

+

–

= 10

Find VC

+–

lecture 21 11

Low Pass Filter:A Single Node-pair Circuit

Find v(t) for =2 3000

1k0.1F

5mA 0

+

–

V

lecture 21 12

Find Impedances

1k

-j5305mA 0

+

–

V

lecture 21 13

Find the Equivalent Impedance

5301000

5301000

j

jeq

Z

5mA 0

+

–

VZeq

lecture 21 14

Parallel Impedances

9.271132

90530010

5301000

5301000 3

j

jeqZ

1.622.468eqZ

lecture 21 15

Computing V

1.622.4680mA5eqIZV

1.62V34.2V

)1.623000t(2cosV34.2)( tv

lecture 21 16

Change the Frequency

Find v(t) for =2 455000

1k0.1F

5mA 0

+

–

V

lecture 21 17

Find Impedances

1k

-j3.55mA 0

+

–

V

lecture 21 18

Find an Equivalent Impedance

5.31000

5.31000

j

jeq

Z

5mA 0

+

–

VZeq

lecture 21 19

Parallel Impedances

2.01000

905.3010

5.31000

5.31000 3

j

jeqZ

8.895.3eqZ

lecture 21 20

Computing V

8.895.30mA5eqIZV

8.89mV5.17V

)8.89455000t(2cosmV5.17)( tv

lecture 21 21

Impedance Summary

Element Impedance

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

Inductor ZL = jL = L 90

Resistor ZR = R = R 0

lecture 21 22

Class Examples