ECE201 Lect-5 1

Phasor Relationships for Circuit Elements (8.4); Impedance and

Admittance (8.5)

Dr. Holbert

February 1, 2006

ECE201 Lect-5 2

Phasor Relationships for Circuit Elements

• Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.

• A complex exponential is the mathematical tool needed to obtain this relationship.

ECE201 Lect-5 3

I-V Relationship for a Resistor

Suppose that i(t) is a sinusoid:

i(t) = IM ej(t+

Find v(t)

R v(t)

+

–

i(t)

)()( tiRtv

ECE201 Lect-5 4

Computing the Voltage

jtjM eIRtiRtv )()(

jtjM eVtv )(

RIV

ECE201 Lect-5 5

Class Example

• Learning Extension E8.5

ECE201 Lect-5 6

I-V Relationship for a Capacitor

Suppose that v(t) is a sinusoid:

v(t) = VM ej(t+

Find i(t)

C v(t)

+

–

i(t)

dt

tdvCti

)()(

ECE201 Lect-5 7

Computing the Current

dt

edVC

dt

tdvCti

jtjM

)(

)(

)()( tCvjeCVjti jtjM

ECE201 Lect-5 8

Phasor Relationship

• Represent v(t) and i(t) as phasors:

V = VM

I = jC V

• The derivative in the relationship between v(t) and i(t) becomes a multiplication by j in the relationship between V and I.

ECE201 Lect-5 9

Example

v(t) = 120V cos(377t + 30) C = 2F

• What is V?

• What is I?

• What is i(t)?

ECE201 Lect-5 10

Class Example

• Learning Extension E8.7

ECE201 Lect-5 11

I-V Relationship for an Inductor

V = jL I

L v(t)

+

–

i(t)

dt

tdiLtv

)()(

ECE201 Lect-5 12

Example

i(t) = 1A cos(2 9.15•107t + 30) L = 1H

• What is I?

• What is V?

• What is v(t)?

ECE201 Lect-5 13

Class Example

• Learning Extension E8.6

ECE201 Lect-5 14

Circuit Element Phasor Relations(ELI and ICE man)

Element V/I Relation Phasor Relation Phase

Capacitor I = C dV/dt I = j ω C V = ωCV 90°

I leads Vby 90º

Inductor V = L dI/dt V = j ω L I = ωLI 90°

V leads Iby 90º

Resistor V = I R V = R I = R I 0°

In-phase

ECE201 Lect-5 15

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, )

ECE201 Lect-5 16

Impedance

• Resistor: V = I R

– The impedance is ZR = R

• Inductor: V = I jL

– The impedance is ZL = jL

ECE201 Lect-5 17

Impedance

• Capacitor:

– The impedance is ZC = 1/jC

Cj1

IV

ECE201 Lect-5 18

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.

ECE201 Lect-5 19

Impedance Example:Single Loop Circuit

20k+–

1F10V 0 VC

+

–

= 377

Find VC

ECE201 Lect-5 20

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

ECE201 Lect-5 21

Impedance Example20k 0

+–

2.65k -9010V 0 VC

+

–

ECE201 Lect-5 22

Impedance Example

Now use the voltage divider to find VC:

0k2090-k65.2

90-k65.20 10VCV

4.82- 1.31VCV

ECE201 Lect-5 23

Low Pass Filter:A Single Node-pair Circuit

Find v(t) for =2 3000

1k0.1F

5mA 0

+

–

V

ECE201 Lect-5 24

Find Impedances

1k

-j5305mA 0

+

–

V

ECE201 Lect-5 25

Find the Equivalent Impedance

5301000

5301000

j

jeq

Z

5mA 0

+

–

VZeq

ECE201 Lect-5 26

Parallel Impedances

9.271132

90530010

5301000

5301000 3

j

jeqZ

1.622.468eqZ

ECE201 Lect-5 27

Computing V

1.622.4680mA5eqIZV

1.62V34.2V

)1.623000t(2cosV34.2)( tv

ECE201 Lect-5 28

Impedance Summary

Element Impedance

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

Inductor ZL = jL = L 90

Resistor ZR = R = R 0

ECE201 Lect-5 29

Class Examples

• Learning Extension E8.8

• Learning Extension E8.9