09averagepower [compatibility mode]

8/13/2019 09AveragePower [Compatibility Mode]

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AVERAGE POWER


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v(t)= Vmcost

i(t)= Im

cost

p(t) =v(t) i(t)

= i2(t)R

= v2(t)/R

POWER in RESISTOR

v(t)

t2

i(t)

p(t)

RVAC


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AVERAGE POWER

From the previous formula we can

find the instantaneous power, but it

has little practical use.

What we really need is the average

power over one period of the current

or voltage.

The instantaneous power is

p = i2R

The average power P is

P= average (i2R)

= (average i2)R

P

p(t)


4/14

AVERAGE POWER

The expression for the average power brings into the average of the

current squared.

Im

i(t)

t

Im2

i

2

(t)Average i2(t) = Im

2/2


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6/14

AVERAGE POWER

Using the same reason we can write

P= Vrms2/R

where

Vrms2 = Vm

2/2

Vrms = Vm/2


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Root Mean Square

RMS stand for root mean square. For sinusoidal

current we have

Vrms = Vm/2

Irms =Im/2

The average power in resistor is

P =Irms2R

or P = Vrms2/R

When we say the voltage is 220 V we always

mean that 220V is the RMS value


8/14

POWER IN CAPACITOR

v(t) = Vmcos(t +)

)sin()( ++++==== tItim

p(t) = v(t) i(t)

= VmImcos(t + )sin(t + )

v(t)

2

i(t)

+ +

--

p(t)

CVAC

Average Power P= 0


9/14

POWER IN INDUCTOR

v(t) = Vmsin(t +)

v(t)

2

i(t)

)cos()( ++++==== tItim

p(t) = v(t) i(t)

= VmImcos(t + )sin(t + )

+ +

- -

p(t)

v(t)

Average Power P= 0


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Power in Impedance

v(t)

t

i(t)

+ +--

p(t)

i=Imcos t

v= Vmcos (t+)

p = vi

= (VmIm){cos+cos(2t+)}

Average power is then

P = (VmIm)cos

= VrmsIrms cos= VIcos


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Power in Impedance

p = (VmIm){cos+cos(2t+)}

= VIcos+ VI cos(2t+)

+ +

--

VIcosVI cos(2t+)


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Complex Power

Complex power is defined as

S= VI*

The unit of complex power is Volt Ampere (VA)S= VI*=I2Z =I2(R+jX) =I2R+jI2X

=I2Z cos +jI2Z sin

S= VI cos +jVI sin = P + jQ

Sis called apparent power and the unit is va

Pis called active power and the unit is watt and

Qis called reactive power and the unit is var


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Complex Power

Using this concept the calculation of power is easier

E.g.: A generator is to supply heater, welder and motor

GenMotor

60 kw

0.8 pf

Welder

Z=4+j3

Heater

15 kw

What is the power and

current must be supplied

by the generator.


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Complex Power

Gen

120 V

Motor

60 kw

0.8 pf

Welder

Z=4+j3

Heater

15 kw

Anwer:

P Q

Motor 60 45Heater 15 0

Welder 2.3 1.7

Total 77.3 46.7

S=77.3+j46.7=90.231.2 kva

I*

=S/V= 751.67 amp

09averagepower [compatibility mode]

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