elec 250: linear circuits i chapter 11 5/4/08 chapter 11 ...amiralib/courses/elec250/chapter...

Elec 250: Linear Circuits I Chapter 11 5/4/08

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where the circuits are

through knowing pha-

e same amount of aver-

riodic current

S.W. Neville

Chapter 11AC Power

Sections 10.1 - 10.7 (Text)

11.1 Introduction

• In Chapter 1 we studied power consumption in the DC case.

• This chapter studies power consumption in the case of ac conditions (i.e.excited by sinusoidal power sources)

• The quantity of interest is the average power and how it can be obtainedsor voltages and currents in the circuit.

11.2 RMS Value

• Assume a periodic current passes through a resistor

- We want to know the equivalent dc current ( ) that delivers thage power in the resistor

- This equivalent dc current is called the effective value of the pe

i t( ) Im ωt φ+( )cos=

Irms


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r an interval is

wer consumed by the

a b,[ ]

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- From calculus we know that the average value of a function ovegiven by

- Mathematically, we can write can therefore write the average poresistor over one period of the sinusiod as

- Thus the effective current is found from the equation

- in the subscript is an abbreviation for “root mean square”

favg1

b a–------------ f x( ) xd

a

b

∫=

P RIrms2 1

T--- Ri2 t( ) td

0

T

∫= =

Irms

Irms1T--- i2 t( ) td

0

T

∫=

rms


07

alue of a periodic func-following steps

re”)

is chapter we will

S.W. Neville

- An easy way to remember the formula for finding an effective vtion is from the abbreviation which implies we must do the

1. Find the square of the given function or waveform (“squa

2. Find the average value of the squared waveform (“mean”)

3. Find the square root of the average value (“root”)

- Hence,

- RMS values can be obtained for any periodic waveform

- In this course we mainly deal with sinusoidal waveforms so in thfocus on how to compute values for sinusoidal functions,

Ex. 11.1 Find the if the current waveform is

rms

RMS root of the mean of the square≡

rms

Irms

i Im ωt φ+( )cos=


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given above we have inusiod,

ift the current by and

a current

rage power, , not the

φ

P

S.W. Neville

- By substituting our function for into the equation for that and performing the integration over one period of the s

- Since we are concerned with one complete period we can shdrop from the above equation,

- Therefore the average power consumed by a resistor when flows through it is given by

- Keep in mind that for ac circuits we are interested in the aveinstantaneous power, .

i Irms

Irms1T--- Im ωt φ+( )cos[ ]2 td

φ

T φ+( )

∫=

φ

IrmsωIm

2

2π--------- ωt( )2cos td

0

2πω------

∫Im

2-------= =

i Im ωt φ+( )cos=

PIm2

2-----R Irms

2 R= =

p


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roduce the same power ion over one period of

12---ωt+ 1

ω----

0

πω----

S.W. Neville

- values give us the equivalent dc current which would pconsumption (i.e. they give us the average power consumptthe sinusoidal waveform).

Ex. 11.2 Find the rms value of the current waveform

Solution:

- Note that the period is for the above signal

- apply the formula from above

rms

i t( ) 10 ωt( )sin=

T πω----=

I2rms

1T--- 10 ωt( )sin( )2 td

0

T∫=

I2rms

ωπ---- 100 ωt( )2sin td

0

πω----

∫100ω

π------------- 1

2--- ωt( ) ωt( )cossin–= =

I2rms

100ωπ

------------- π2---⎝ ⎠⎛ ⎞ 1

ω----⎝ ⎠⎛ ⎞ 50 A2= =


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xpress its rms value in

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- If our power source is expressed as a phasor then we can easily ephasor notation as follows,

I Im φi∠=

IrmsIm

2------- φi∠=

V Vm φv∠=

VrmsVm

2------- φv∠=


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corresponding time-.d/s

S.W. Neville

Ex. 11.3 What is the rms value of the sinusoidal voltage waveform

Solution:

Ex. 11.4 Given an rms phasor voltage , write itsdomain waveform if the angular frequency is

Solution:

v t( ) 10 100t 60°+( )cos=

Vrms 100ej30°V=

ω 100 ra=


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elow

t

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Ex. 11.5 What is the rms value of the sawtooth waveform given b

Solution:

i

3

0 0.5 1 1.5


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e across it is .

y

n the voltage and cur-

rs

v

v) ωt φi+( )cos

S.W. Neville

11.3 Average Power

• Consider an impedance in which the current through it is and the voltag

- The instantaneous power consumed by the impedance is given b

- We are interested in the power consumed by the impedance wherent are sinusoidal.

- Assume we have

- From Chapter 7 we know that these correspond to the two phaso

- The instantaneous power in this case is given by

i

p vi=

v Vm ωt φv+( )cos=

i Im ωt φi+( )cos=

V Vmejφv=

I Imejφi=

p

p Vm ωt φv+( )cos[ ] Im ωt φi+( )cos[ ] VmIm ωt φ+(cos= =


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.

then the AC term will

ωt φi+[ ])]

]

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- But we have the trigonometric identity that

- Thus we have

- Therefore, the instantaneous power consists of two terms

1. a DC term:

2. an AC term:

- Since average power is defined over one period of the waveformbe zero,

α βcoscos 12--- α β+( )cos α β–( )cos+[ ]=

p 12---VmIm ωt φv+[ ] ωt φi+[ ]+( )cos ωt φv+[ ] –(cos+[=

p 12---VmIm 2ωt φv+ φi+( )cos φv φi–( )cos+[=

12---VmIm φv φi–( )cos

12---VmIm 2ωt φv+ φi+( )cos

1T--- 1

2---VmIm 2ωt φv+ φi+( )cos td

0

T

∫ 0=


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the instantaneous

f the sinusiod) is there-

ively. (Note that these , but only the magni-

s.

s

S.W. Neville

- So the average power is only a function of the DC component ofpower.

- The average power consumed by the element (over one period ofore given by,

- where

- are the rms values of the voltage and current magnitudes respectare NOT the rms phasors, which would be denoted and tudes of the respective phasors)

- and the phase angle is given by

- is the phase difference between the voltage and current source

P 12---VmIm φv φi–( )cos VrmsIrms φ( )cos= =

VrmsVm

2-------=

IrmsIm

2-------=

Vrms Irm

φ

φ φv φi–=

φ


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lements are . Calculate the aver-V

S.W. Neville

- The rms voltage and current phasors can therefore be written as

Ex. 11.6 The current and voltage time-domain waveforms for an e and

age power consumption in this element.

Solution:

Vrms Vrmsejφv=

Irms Irmsejφi=

i t( ) 15 100t 80°+( ) mAsin= v t( ) 5 100t 60°+( ) cos=


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pedance is given by

bbreviated pf.

.

(i.e. that the element

s) and will be in the

a leading power fac-

lagging power factor,

0° φv φi–( ) 90°≤ ≤

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11.4 The Power Factor

• In the section above we found that the average power consumed by the im

- The term is called the power factor and is a

- For passive loads the phase angle difference is always

- This assures that the real part of the power consumed is positiveconsuming the power is a passive element)

- Hence, the power factor is always positive (for passive elementrange of

- In capacitive circuits the current leads the voltage and we havetor, .

- In inductive circuits the current lags the voltage and we have a

P VrmsIrms φv φi–( )cos VrmsIrms φ( )cos= =

φcos φv φi–( )cos=

φv φi– 9–

0 pf 1≤≤

φv φi– 0<

φv φi– 0>


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as follows.

asor (A)

°

°

S.W. Neville

Ex. 11.7 Assume three elements which have currents and voltages

Find the power factor in each element and the power consumed.

Solution:

Element RMS Phasor Voltage (V)

RMS PhCurrent

Element A

Element B

Element C

15e15° 5e15°

10e45° 2e105

2e20° 15e 70–


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lement as

er.

S.W. Neville

11.5 Complex (Apparent) Power

• In Section 11.3 we derived an expression for the power consumed in an e

- where

- Note that the average power is a real number.

• We can write the average power as the real part of a complex number

- where is interpreted as the complex power, or apparent pow

- where is the real or average power defined as

- and is the imaginary or reactive power defined as

P VrmsIrms φv φi–( )cos VrmsIrms φ( )cos= =

φ φv φi–=

P VrmsIrmsRe φ j φsin+cos[ ] Re S[ ]= =

S

S VrmsIrms φ( )cos jVrmsIrms φ( )sin+=

S P jQ+=

P

P VrmsIrms φ Wattscos=

Q

Q VrmsIrms φ VARsin=


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pressed in Euler’s form

positive.

e negative.

S.W. Neville

- where is measured volt-ampere reactive (VAR).

• The complex or apparent power is a complex number which may be ex

- where is measured in volt-amperes (VA)

• For lagging power factor (inductive circuits), is positive and will be

• For leading power factor (capacitive circuits), is negative and will b

• Based on the above,

- When we are given the rms voltage and current phasors

- the apparent power is given by

Q

S

S VrmsIrmsejφ=

S

φ Q

φ Q

Vrms Vrmsejφv=

Irms Irmsejφi=

S VrmsIrmsej φv φi–( )

=


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rent magnitudes and

s.

ent phasors we must

.

e angles .

power used by an ele-

ctor , and a capacitor each element.

ejφi–

φ φv φi–=

L

S.W. Neville

- Note that this equation is an equation of the rms voltage and curtheir respective phase angles.

- It is NOT written in terms of the rms voltage and current phasor

- To write the apparent power in terms of the rms voltage and currwrite

- were is the conjugate of the rms current phasor

- The conjugation is required in order to get the difference in phas

- The above equation gives us a means of calculating the average ment in phasor notation.

Ex. 11.8 Determine the power consumed by a resistor , and indu when an rms voltage is applied across

Solution:

S VrmsIrms* VrmsIrmse

j φv φi–( )= =

Irms* Irms

* Irms=

RC Vrms V ωtcos=


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plied rms voltage ircuit.

s the power triangle,

S.W. Neville

Ex. 11.9 Given a circuit with impedance and an ap, find the complex power consumed by the c

Solution:

11.6 The Power Triangle

• The different components of the complex power form what is known awhich is the representation of the apparent power in the complex plane.

Z 10 j10 Ω+=V 20 30°∠=

S


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it ( so the circuit

s voltage pedance.

Q 0>

S.W. Neville

- The figure below shows the power triangle for an inductive circumust be inductive)

Ex. 11.10 Assume an impedance of and an applied rm. Determine the power consumed by this im

Solution:

Q (VAR)

P (W)

S (VA)

φ

Z 3 4j Ω+=V 120 30°∠=


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leading. The applied ent values of the power

to a rms current such

7

I

S.W. Neville

Ex. 11.11 An impedance consumes at a power factor of voltage V. Determine the componand the impedance.

Solution:

11.7 Impedance and Power

• Assume an impedance where a rms voltage is applied and gives rise that

500 W 0.70v 170 314t 15°+( )sin=

Z V

Vrms Vejφv=

Irms Iejφi=


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lane where

S.W. Neville

- We can then write

- The figure below shows the representation of in the complex p

- The power consumed by this impedance is given by

Z VI----=

Z2Vrms

2Irms-------------------

VrmsIrms----------- V

I---e

j φv φi–( ) VI---ejφ= = = =

Z

R Z φcos Z φcos= =

X Z φsin Z φsin= =

Z

R

Xφ

S VI* VrmsIrmsejφ= =


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lex plane as

n see that they are very

S.W. Neville

- The power triangle for this impedance can be drawn in the comp

- where

- Comparing the impedance triangle and the power triangle we casimilar.

- For an inductive impedance is positive and is positive.

- For a capacitive impedance is negative and is negative.

Q (VAR)

P (W)

S (VA)

φ

P S φcos S φcos= =

Q S φsin S φsin= =

X Q

X Q


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pedance using the fol-

ys be quickly derived

rms2R

rms2X

S.W. Neville

• Using this realization, we can now express the power consumed by an imlowing relations

- The above equations need not be memorized since they can alwafrom

S VrmsIrms* Vrms

2

Z*---------------- Irms

2Z= = =

P Re S[ ] Vrms Irms* φcos

Vrms2

Z 2----------------R I= = = =

Q Im S[ ] Vrms Irms* φsin

Vrms2

Z 2----------------X I= = = =

V IZ=

S VrmsIrms*=


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and a current er triangle.

S.W. Neville

Ex. 11.12 A circuit has an applied voltage . Determine the impedance and pow

Solution:

v t( ) 100 20t 15°+( )sin=i t( ) 2 20t 20°–( )sin=


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an impedance such tor of lead-

ity power factor.

stor.

e ohmic losses in the

Zpf 0.707=

S.W. Neville

Ex. 11.13 A voltage of is applied across that the power consumed in with a power facing. Find the value of and the value of it elements.

Solution:

11.8 Power Factor Correction

• It is desirable in industrial application to supply power to a load with a un

- In this case, the load appears to the power utility as a simple resi

- Therefore, the current supplied to the load will be reduced and thtransmission lines will also be reduced.

v t( ) 99 6000t 30°+( )cos=P 940 W=

Z


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d produces some rms

, which produces some

t

consumption.

rs that have loads with

I1

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• Assume an rms voltage is applied to a pure restive load (unity pf) ancurrent in that load.

- The power supplied to the load is given by

• Assume the same rms voltage is applied to a load without a unity pfcurrent in the load.

- The power supplied to the load in this case is given by

• Comparing the two cases when the power consumed is equal we have tha

- More current is drawn by the non-unity pf load for equal power

- This is the reason why power utilities penalize industrial customenon-unity power factors.

Vrms

I1

P VrmsI1 φv φi–( )cos VrmsI1 1( ) Vrms= = =

Vrms

I2

P VrmsI2 φv φi–( )cos=

VrmsI1 VrmsI2 φv φi–( )cos=

I2I1φv φi–( )cos

----------------------------- I1>=


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er factors

cing a capacitor in par- the load becomes

an inductor in parallel

oltage at wer factor to a unity pf factor correction.

V 100 30°∠=

S.W. Neville

• How do we correct for non-unity power factors?

- Industry usually has inductive loads which produce lagging pow

- A lagging power factor can be made a unity power factor by plaallel with the load, where the value of the capacities is such thatpurely resistive (i.e. is made equal to 0).

- A leading power factor can be made purely resistive by placing with the load.

Ex. 11.14 An industrial load is supplied by an rms v60 Hz. Find the capacitor value required to change the poand find the current supplied before and after the power

Solution:

Im Z[ ]

Z 5 10° Ω∠=


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r factor of 0.8 lagging. mine the current flow-

cuits operating in DC cuits operating in AC

S.W. Neville

Ex. 11.15 The power consumed in an impedance is 100 W at a poweIf the applied voltage is , detering in the impedance and the value of the impedance.

Solution:

11.9 Maximum Power Transfer

• In Chapter 3, we studied maximum power transfer for purely resistive circonditions, we would like to perform a similar analysis for impedance circonditions.

v t( ) 25 300t 45°+( ) Vsin=


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e can replace any lin-t the load).

um power out of the

S.W. Neville

- Consider the following Thevenin equivalent circuit (remember wear circuit about its load with it Thevenin equivalent circuit abou

- Assume that we can vary the load impedance to get the maximcircuit.

- Applying KVL to the circuit we get

- Thus,

+-

VT

ZT

Z

I+

-

V

Z

VT I ZT Z+( )=

IVT

ZT Z+( )--------------------=


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S.W. Neville
- Applying voltage division across the load

- From the above two equations the power delivered to the load is

- The real power consumed by the load is

- But the source and load impedances can be written as

- Therefore the real power can be written as

VZT

ZT Z+---------------VT=

S VI* VT2Z

ZT Z+ 2---------------------= =

PVT

2RZT Z+ 2---------------------=

ZT RT jXT+=

Z R jX+=

P VT2 R

RT R+( )2 X XT+( )2+------------------------------------------------------=


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e 0.

quation with respect to

maximum power trans-

S.W. Neville

- The value of which maximizes is

- since this causes the term in the denominator to becom

- and the real power becomes

- The maximum value for is found by differentiating the above e

- The maximum occurs when which occurs when

- We now have the real and imaginary parts of required for the fer.

X P

X X– T=

X XT+( )2

P VT2 R

RT R+( )2-----------------------=

PR

dPdR------- VT

2 RT R–

RT R+( )3-----------------------=

dPdR------- 0=

R RT=

Z


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alent circuit is that the impedance.

S.W. Neville

- The maximum power transfer to the load in a Thevenin equivload impedance must be the complex conjugate of the Thevenin

- At this value the output voltage and current will be

- The maximum complex power delivered to the load will be

- And the maximum power available from the circuit is

Z

Z ZT* RT jXT–= =

V' VTZT

*

2RT---------=

I'VT

2RT---------=

S' V'I'*VT

2

4RT2

-----------ZT*= =

P'VT

2

4RT-----------=


7

ent, voltage, and maxi-

nce of is the value of which f that power?

Z 3 30° Ω∠=Z

S.W. Neville

- In terms of the Norton equivalent circuit and , the load currmum power are given by

Ex. 11.16 A voltage source of has an internal impedaand drives a serially connected load impedance . What will maximize the power transfer and what is the value o

Solution:

IN ZT

V'IN ZT

2

2RT-----------------=

I'INZT2RT-----------=

P'IN

2 ZT2

4RT----------------------=

V 10 V rms=Z


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S.W. Neville
Assignment #11

Refer to Elec 250 course web site for assigned problems.

• Due 1 week from today @ 5pm in the Elec 250 Assignment Drop box.

elec 250: linear circuits i chapter 11 5/4/08 chapter 11 ...amiralib/courses/elec250/chapter...

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