Lecture 21 1

Phasor Diagrams and Impedance

Lecture 21 2

Set Phasors on Stun

1. Sinusoids-amplitude, frequency and phase (Section 8.1)

2. Phasors-amplitude and phase (Section 8.3 [sort of])

2a. Complex numbers (Appendix B).

Lecture 21 3

Set Phasors on Kill

3. Complex exponentials-amplitude and phase

4. Relationship between phasors, complex exponentials, and sinusoids

5. Phasor relationships for circuit elements (Section 8.4)

5a. Arithmetic with complex numbers (Appendix B).

Lecture 21 4

Set Phasors on Vaporize

6. Fundamentals of impedance and admittance (some of Section 8.5)

7. Phasor diagrams (some of Section 8.6)

Lecture 21 5

Phasor Diagrams

• A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes).

• A phasor diagram helps to visualize the relationships between currents and voltages.

Lecture 21 6

An Example

-

1F VC

+

-

2mA 40

1k VR

+

+

-

V

Lecture 21 7

An Example (cont.)

I = 2mA 40VR = 2V 40

VC = 5.31V -50

V = 5.67V -29.37

Lecture 21 8

Phasor Diagram

Real Axis

Imaginary Axis

VR

VC

V

Lecture 21 9

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.

Lecture 21 10

Impedance

• Resistor:

– The impedance is R

• Inductor:

– The impedance is jL

RIV

LjIV

Lecture 21 11

Impedance

• Capacitor:

– The impedance is 1/jL

Cj1

IV

Lecture 21 12

Some Thoughts on Impedance

• Impedance depends on the frequency .

• 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 13

Impedance Example:Single Loop Circuit

20k+

-1F10V 0 VC

+

-

= 377

Find VC

Lecture 21 14

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 15

Impedance Example20k 0

+

-2.65k -9010V 0 VC

+

-

Lecture 21 16

Impedance Example

Now use the voltage divider to find VC:

0k2090-k65.2

90-k65.20 10VCV

4.82- 1.31VCV

Lecture 21 17

What happens when changes?

20k+

-1F10V 0 VC

+

-

= 10

Find VC

Lecture 21 18

Low Pass Filter:A Single Node-pair Circuit

Find v(t) for =2 3000

1k0.1F

5mA 0

+

-

V

Lecture 21 19

Find Impedances

1k

-j530k5mA 0

+

-

V

Lecture 21 20

Find the Equivalent Impedance

5301000

5301000

j

jeq

Z

5mA 0

+

-

VZeq

Lecture 21 21

Parallel Impedances

9.271132

90530010

5301000

5301000 3

j

jeqZ

1.622.468eqZ

Lecture 21 22

Computing V

1.622.4680mA5eqIZV

1.62V34.2V

)1.623000t(2cosV34.2)( tv

Lecture 21 23

Change the Frequency

Find v(t) for =2 455000

1k0.1F

5mA 0

+

-

V

Lecture 21 24

Find Impedances

1k

-j3.55mA 0

+

-

V

Lecture 21 25

Find an Equivalent Impedance

5.31000

5.31000

j

jeq

Z

5mA 0

+

-

VZeq

Lecture 21 26

Parallel Impedances

2.01000

905.3010

5.31000

5.31000 3

j

jeqZ

8.895.3eqZ

Lecture 21 27

Computing V

8.895.30mA5eqIZV

8.89mV5.17V

)8.89455000t(2cosmV5.17)( tv