INC 112 Basic Circuit Analysis

Week 12

Complex Power

Complex Frequency

Example

AC

+

- 50cos(2t)

i(t)

2.5H5Ω

3Ω

2H

AC

50sin(2t)

-

+

Find i(t)

AC

+

- 50∟90

i(t)

j55

3

j4

AC

-

+

50∟0

ω = 2

AC

+

- 50∟90

i(t)

j55

3

j4

i1(t)Use superposition

2.2335.65.283.5

5)43(

5)43(5

j

jj

jjZ total

8.6687.72.2335.6

90501

totalZ

VI

i(t) can be found from current divider

3.8515.4)43(5

51

I

jj

jI

i(t)

j55

3

j4

AC

-

+

50∟0

i2(t)2.6874.625.65.2

5)43(

5)43(5

j

j

jjZ total

2.6842.72.6874.6

0502

totalZ

VI

i(t) can be found from current divider

3.8515.4)43(5

52

I

jI

3.8530.83.8515.43.8515.4 totalI

)488.12sin(30.8)(

)3.852sin(30.8)(

tti

tti

Power in AC circuits

R

tvRtitp

22 )(

)()(

In AC circuits, voltage and current values oscillate.This makes the power (instantaneous power) oscillate as well.

However, electric power is best represented as one value.Therefore, we will use an average power.

Average power can be computed by integration ofinstantaneous power in a periodic signal divided by time.

)sin()( tAtv Let v(t) in the form Change variable of integration to θ

sinAv We got Then find the instantaneous power

R

A

R

vp

222 sin integrate from 0 to 2π (1 period)

R

A

R

A

dA

dR

A

dR

AP

24

2sin

2

1

2

2

2cos1

2sin

2

sin2

1

22

0

2

2

0

22

0

22

2

0

22

Compare with power from DC voltage source

DC

AC

+

-

Asin(ωt+Ф) R

i(t)

AC

+

-A R

i(t)

R

AP

2

2

R

AP

2

Root Mean Square Value (RMS)

In DC circuits,R

VRIP

22

In AC, we define Vrms and Irms for convenience to calculate power.Vrms and Irms are defined such that,

R

VRIP rms

rms

22 Note: Vrms and Irms are constant all the time

For sine wave,2

_A

valuerms

V (volts)

t (sec)

311V

V peak (Vp) = 311 VV peak-to-peak (Vp-p) = 622VV rms = 220V

3 ways to tell voltage

0

Reactive Power

Capacitors and inductors have average power = 0 because they have voltage and current with 90 degree phase difference.

)sin()( tAtv )90sin()( tBti

Change variable of integration to θ

sinAv cos)90sin( BBi

cossinABvip

Then integrate from 0 to 2π (1 period)

0)0(sin)2(sin

4

24

)(sinsin2

cossin2

1

22

2

0

2

0

2

0

sin

AB

AB

dAB

dABP

Capacitors and Inductors do not have average power although there are voltage and current.

Therefore, reactive power (Q) is defined

Complex Power

Power can be divided into two parts: real and imaginary

Complex power S = P + jQ

P = real power Q = Reactive power

Inductor has no real power P =0But it has complex power, computed by V, I that are perpendicularto each other.

Example

AC

+

-

5sin(3t+π/3)

i(t)

3H

2Ω

+ vR(t) -

+vL(t)

-

Find i(t), vL(t)

AC

+

-

5∟60

i(t)

j9

2

V

IVR

VL47.1754.0 I

53.7288.4 LV

47.1708.1 RV

Phasor Diagram

Resistor consumes power WRIP rms 292.02

254.0 22

Inductor consumes no real power P = 0 but it has reactive power

VARVIQ rmsrms 318.12

88.454.0

θ = 77.47

S

P

Q

θ = 77.47

Complex Power Diagram

Consider the voltage source,

The voltage source supplies 0.292W real power and 1.318VAR reactive power.

Definition: Power factor = cos θ

Power factor = 0.217

cosFactorPower rmsrmsIV

P

S

P

Complex Frequency

ste is a fundamental waveform of electrical engineering

What if s is a complex number?

tjetetjte

eeeeettt

tjttjttjst

sincos)sin(cos

)(

js

Let s be a complex number composed of real and imaginary parts.

AC

+

-

5cos(3t+π/3) 3H

2Ω

AC

+

-

5∟60 j9

2

)sin()cos()( tjte tj

PhasorFrequency domain

Time domain Euler’s Identity

sincos je j

Complex Frequency

tjte tj sincos AC

+

-

sinωt

tjetee tttj sincos)( AC

+

-

eσtsinωt

Define js

s is called “complex frequency”

Summary of Procedures

• Change voltage/current sources in to phasor form

• Change R, L, C value into phasor form

R L C R sL 1/sC

• Use DC circuit analysis techniques normally, but the value of voltage, current, and resistance can be complex numbers

• Change back to the time-domain form if the problem asks.

Example

AC

+

-20e-2tcos(4t+π/2)

i(t) 1H

5Ω 0.1f

ic(t)

Find i(t), ic(t)

AC

+

-20∟90

i(t) s

5 1/0.1s42 j

js

AC

+

-20∟90

i(t) s

5 1/0.1s

ic(t)

105

50105

105

50)

10||5(

2

s

ss

ss

ssZ total

We then substitute s = -2+j4 and got

1.1435.25.12105

50105 2

j

s

ssZ total

1.5381.1435.2

9020

totalZ

VI

1.538 I

)9268.04cos(8)(

)1.534cos(8)(2

2

teti

tetit

t

ic(t) can be computed from current divider

535.26944.8

1.538565.26118.11.538)5.01(

1.538105

51.538

/105

5

j

s

s

sIC

)4631.04cos(944.8)(

)535.264cos(944.8)(2

2

teti

tetit

C

tC

Complex Frequency Characteristics

te t sin

00

01

1

1