CHAPTER 3: AC POWER ANALYSIS Instantaneous & Average Power Max. Average Power Transfer RMS Value &...

CHAPTER 3: AC CHAPTER 3: AC POWER ANALYSISPOWER ANALYSIS

Instantaneous & Average Power Max. Average Power Transfer RMS Value & Apparent Power Complex Power & Power Factor

Correction

Instantaneous Power

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

Instantaneous power (in watts) : the power at any instant of time

)tcos(V)t(v vm Where:

)tcos(I)t(i im

)t2cos()cos(2

IV)t(p iviv

mm

Instantaneous Power

The instantaneous power p(t) entering a circuit

Average Power

)cos(2

IVP iv

mm

Average Power (in watts) : the average of the instantaneous power over one period

iv cos2

IVP mm Where:

Average Power cosIVP effeff

2

IIor I m

effrms 2

or meffrms

VVV

Example 1

V)45t377cos(120)t(v 0 A)10t377cos(10)t(i 0

Find the instantaneous power and average power absorbed by the passive linear network.

Given that

Reference : Alexander, Sadiku Chapter 11 - page 461

Exercise 1

Calculate the instantaneous power and average power if

V)20t10cos(80)t(v 0 A)60t10sin(15)t(i 0

W)10t20cos(6007.385)t(p 0

W7.385P

Circuit Elements

In purely resistive circuit, v and i are in phase. θv = θi. Therefore θ = 0.

RIR

VIV

2

IV0cos

2

IVP 2

eff

2eff

effeffmm0mm

(a) Resistors:

The average power is only dissipated in a purely resistive circuit. For a purely inductive and capacitive, the average power is zero.

Circuit Elements

In purely inductive circuit, v leads by 90o, therefore θ = 90o

0)90cos(2

omm IVP

(b) Inductors:

(c) Capacitors:

In purely capacitive circuit, I leads by 90o, therefore θ = - 90o

0)90cos(2

omm IVP

Example 2

Find the average power supplies by the source and the average power absorbed by the resistor.


Exercise 2

Calculate the average power absorbed by the resistor and inductor. Find the average power supplies by the voltage source.

W6.9P

W0P

W6.9P

L

R

Maximum Average Power Transfer

Finding the maximum average power transfer:

a) circuit with a load b) the Thevenin equivalent


For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTh

LLL

ThThTh

jXRZ

jXRZ

In rectangular form, Thevenin impedance and Load impedance:

Th*

ThThLLL ZjXRjXRZ

Th*

L ZZ


Th

2

Thmax R8

VP

In a situation in which the load is purely real or purely resistive load (XL=0), the load impedance (or resistance RL) is equal to the magnitude of the Thevenin impedance.

Th2Th

2ThL ZXRR

Example 3

Determine the load impedance ZL that maximizes the average power drawn from the circuit of figure below. Calculate the maximum average power.


Exercise 3

Find the load impedance ZL that absorbs the maximum average power for the circuit of figure below. Calculate the maximum average power.

7317.0j415.3ZZ Th*

LW429.1Pmax

Exercise 4

In Figure below, the resistor RL is adjusted until it absorbs the maximum average power. Calculate RL and the maximum average power absorbed by it.

W883.9Pmax 30R L

Effective or RMS Value

The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

Finding the effective current:

a) ac circuit b) dc circuit


dtxT

1X

T

0

2rms

The effective value of a periodic signals is its root mean square (rms) value.

For any perodic function x(t) in general, the rms value is given by:


dttcosIT

1I

T

0

22mrms

2

Idt)t2cos1(

2

1

T

II m

T

0

2m

rms

2

VV m

rms


The average power absorbed by resistor R can be written as:

RIR

VP 2

rms

2rms

)cos(IVP ivrmsrms

The average power can be written as:

Example 4

i(t)

t0

10

-10

2 4 6 8 10

Determine the rms value of the current waveform in figure below. If the current is passed through a 2Ω resistor, find the average power absorbed by the resistor.

A165.8Irms

W3.133P


Exercise 5

Find the rms value of the full wave rectified sine wave in figure below. Calculate the average power dissipated in a 6Ω resistor.

V657.5Vrms W334.5P

v(t)

t0

8

2 3

Apparent Power

The apparent power (in VA) is the product of the rms value of voltage and current.

)cos(IVP ivrmsrms

)cos(SP iv

rmsrmsIVS

S is known as the apparent power.

Power Factor

The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance.

)cos(pf iv

iv

Power Factor :

is Power Factor Angle pf is lagging if the current lags voltage (inductive load)

pf is leading if the current leads voltage (capacitive load)For purely resistive circuit, pf=1. With inductors

and capacitors in the circuit, pf may reduced to less than 1.

where

Example 5

A series connected load draw

a current

when the applied voltage is

Find the apparent power and the power factor of the load. Determine the element values that form the series connected load.

A)10t100cos(4)t(i 0

V)20t100cos(120)t(v 0

V240S )leading(866.0pf uF2.212C


Exercise 6

Calculate the power factor of the circuit below as seen by the source. What is the average power supplies by the source?

)lagging(936.0pf W118P

Complex Power

Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q.

ivrmsrms* IVVI

2

1jQPS

*

2rms2

rms Z

VZIS

Complex power :

VA

*

rmsrmsIV

Complex Power

22

rmsrms QPIVSS Apparent power :

Real power :

)cos(S)SRe(P iv WReactive power :

)sin(S)SIm(Q iv VAR

VA

Q = 0 for resistive loads (unity power factor)

Q < 0 for capacitive loads (leading power factor)

Q > 0 for inductive loads (lagging power factor)

Complex Power

Q

P

SX

R

IZI

Power triangle

Impedance triangle

S

Pcospf

Complex Power

Power Triangle

Example 6

The voltage across a load is

and the current through the element in the direction

of the voltage drop is

Find

a)the complex and apparent powers

b)the real and reactive powers

c) the power factor and the load impedance

V)10tcos(60)t(v 0

A)50tcos(5.1)t(i 0


Example 7

A load Z draws 12kVA at a power factor of 0.856

lagging from a 120 Vrms sinusoidal source.

Calculate:

a)the average and reactive powers delivered to the load

b)the peak current

c) the load impedanceReference : Alexander, Sadiku Chapter 11 - page 476

Exercise 7

A sinusoidal source supplies 10kVAR reactive

power to load

Determine:

a)the power factor

b)the apparent power delivered to the load

c) the peak voltage

075250Z

Power Factor CorrectionThe process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.

Most loads are inductive. A load power factor is improved (to make closer to unity, pf=1) by installing a capacitor in parallel with the load.

a) Original inductive load b) inductive load with improved

power factor

Power Factor Correction

Phasor diagram showing the effect of adding a capacitor in parallel with the inductive load


Power triangle illustrating power factor correction


2

rms

21

2

rms

C

V

)tan(tanP

V

QC

)tan(tanPQQQ 2121C

2

rms

C

2

rmsC CV

X

VQ

Value of required shunt capacitance :

Example 8

When connected to a 120 V (rms), 60Hz power line, a load absorbs 4kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.


Exercise 8

Find the value of parallel capacitance needed to correct a load of 140kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110V (rms), 60Hz line.

CHAPTER 3: AC POWER ANALYSIS Instantaneous & Average Power Max. Average Power Transfer RMS Value &...

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