ECE 1100: Introduction to Electrical and Computer Engineering Notes 20 Power in AC Circuits and RMS 1 Spring 2011 Wanda Wosik Associate Professor, ECE

ECE 1100: Introduction toECE 1100: Introduction toElectrical and Computer EngineeringElectrical and Computer Engineering

Notes 20

Power in AC Circuits and RMS

1

Spring 2011

Wanda WosikAssociate Professor, ECE Dept.

Notes prepared by Dr. Jackson

AC PowerAC Power

22 cosp

abs

V tv tP t

R R

R []+-v (t)

Goal: Find the average power absorbed by resistor:

cos

cos 2 [V]

p

p

v t V t

V f t

f = frequency [Hz]

Vp = peak voltage

Note: The phase of the voltage wave is assumed to be zero here for convenience.

AVEabsP

2

AC Power (cont.) AC Power (cont.)

2

2cospabs

VP t t

R

T = 1/f [s]c os

(t )

Tp = T / 2 = 0.5 / f [s]

c os2

(

t)

3

AC Power (cont.)AC Power (cont.)

0

22

0

22

0

22

0

1

1cos

1cos

1cos

p

p

p

T

AVEabs abs

p

T

p

p

T

p

p

Tp

P P t dtT

Vt dt

T R

Vt dt

T R

Vt dt

T R

Note: We obtain the same result if we integrate over Tp or T.

4


2

2

0

1cos

TpAVE

abs

VP t dt

T R

Consider the integral that needs to be evaluated:

2

0

cosT

cI t dt

5


2 2

0 0

2

0

0

0

cos cos 2

cos 2 /

1 cos 2*2 /

2

sin 2*2 /

2 2(2*2 / )

2

T T

c

T

T

T

I t dt f t dt

t T dt

t Tdt

t Tt

T

T

“The average value of cos2 is 1/2.”6


21

2pAVE

abs

V TP

T R

Hence 21

2pAVE

abs

VP

R

Ic = T/2

2

2

0

1cos

TpAVE

abs

VP t dt

T R

so

7

R []+-v (t)

cospv t V t

21

2pAVE

abs

VP

R

SummarySummary

8

Effective Voltage Effective Voltage VVeffeff

21

2pAVE

abs

VP

R

2p

eff

VV

2effAVE

abs

VP

R

Define:

Then we have:Note: Veff is used the

same way we use V in a DC power calculation.

9

Effective Voltage Effective Voltage VVeffeff

2effAVE

abs

VP

R

R []+-v (t)R []V

2AVE

abs abs

VP P

R

DC AC

same formula

10

ExampleExampleIn the U. S., 60 Hz line voltage has an effective voltage of 120 [V]. Describe the voltage waveform mathematically.

Veff = 120 [V]

2 120 2 169 71 [V]p effV V .

cos cos 2p pv t V t V ft

169 71cos 2 60v t . t so

11

ExampleExample60 Hz line voltage is connected to a 144 [] resistor.Determine the average power being absorbed.

R = 144 []+-

120 [V] (eff)

22 120100

144effAVE

abs

VP

R

100 [W]AVEabsP

12

RMS (Root Mean Square)RMS (Root Mean Square)

This is a general way to calculate the effective voltage for any periodic waveform (not necessarily sinusoidal).

t

v(t)

T digital pulse waveform

tp

Duty cycle: D = tp / T13

RMS (cont.)RMS (cont.)

2effAVE

abs

VP

R

2 2

0

1 TeffV v t

dtR T R

Hence,

2

0

1 TAVE

abs

v tP dt

T R Also,

By definition,

14


2 2

0

1 TeffV v t

dtR T R

Hence

2

0

1 T

effV v t dtT

15


Define

VRMS is the root (square root) of the mean (average) of the square of the voltage waveform

2

0

1 T

RMSV v t dtT

Veff = VRMS

Comparing with the formula for Veff , we see that

16


For sinusoidal (AC) signals, 2

pRMS

VV

For other periodic signals, there will be a different relationship between VRMS and Vp.

(See the example at the end of these notes.)

17

RMS CurrentRMS Current

R []+-

i (t)

v (t)

The concept of effective (RMS) current works the same as for voltage.

cospi t I t

2eff RMS pI I I / 2AVEabs RMSP R I

2 2 2cosabs pP t i t R I R t

21

2AVE

abs pP I R

Define:

18

RMS Current (cont.)RMS Current (cont.)RMS current can be easily related to RMS voltage.

cos

cos

p

p

Vv ti t t

R RI t

cospv t V t

2

2p pRMS

RMS pp

V / VVR

I II /

R []+-

i (t)

v (t)

pp

VI

R

where

19

ExampleExample60 Hz line voltage is connected to a 144 [] resistor. Determine the RMS current and the average power absorbed (using the current formula).

1200 83333

144RMS

RMS

VI .

R

0 83333 [A]RMSI .

120 [V] (RMS)

R = 144 []+-

IRMS 22 144 0 83333 100AVEabs RMSP R I .

100 [W]AVEabsP

20

RMS Voltage and CurrentRMS Voltage and Current

Power can also be expressed in terms of both RMS voltage and current.

2AVE RMS RMS

abs RMS RMS RMS

V VP V V I

R R

AVEabs RMS RMSP V I

R []+-

IRMS

VRMS

-

+

21

ExampleExample60 Hz line voltage is connected to a 144 [] resistor. Determine the average power (using the voltage-current formula).

0 83333 [A]RMSI .

R = 144 []+-

120 [V] (RMS)

IRMS

100 [W]AVEabsP

120 [V]RMSV

120 0 83333AVEabs RMS RMSP V I .

22

Summary of AC PowerSummary of AC Power

2AVE RMS

abs

VP

R

AVEabs RMS RMSP V I

2AVEabs RMSP R I

RMSRMS

VI

R

2p

RMS

VV

2p

RMS

II

R []+- VRMS

IRMS

-

+

23

Example (non-sinusoidal)Example (non-sinusoidal)Find the RMS voltage of a sawtooth waveform:

for 0pv t V t / T t T

t

v (t)

T

Vp

2

2

0 0

1 1T T

RMS p

tV v t dt V dt

T T T

24

Example (cont.)Example (cont.)

2

0

2 22 3

2 200

23 2

2

1

1 1 1

3

1 1 1

3 3

T

RMS p

TTp p

pp

tV V dt

T T

V Vt dt t

T T T T

VT V

T T

Hence 3RMS pV V /25

Example (sawtooth wave)Example (sawtooth wave)

t

v (t)

T

Vp

R []+-

v (t)

Given: Vp = 10 [V]

R = 100 []

Find the average power absorbed by the resistor.

26

Example (cont.)Example (cont.)

t

v (t)

T

Vp = 10 [V]

100 []+-

v (t)

3RMS pV V /2

AVE RMSabs

VP

R

25 7735

0 33333100

AVEabs

.P .

10 3 5 7735 [V]RMSV / .

0 33333 [W]AVEabsP .

(for sawtooth)

27

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ECE 1100: Introduction to Electrical and Computer Engineering Notes 20 Power in AC Circuits and RMS 1 Spring 2011 Wanda Wosik Associate Professor, ECE