ene 104 electric circuit theory - web page for...

ENE 104

Electric Circuit Theory

Lecture 10:

AC Power Circuit Analysis

http://webstaff.kmutt.ac.th/~dejwoot.kha/

(ENE) Mon, 19 Mar 2012 / (EIE) Wed, 28 Mar 2012

Week #11 : Dejwoot KHAWPARISUTH



Objectives : Ch11

Week #11

ENE 104

• the instantaneous power

• the average power

• the rms value of a time-varying waveform

• complex power: average and reactive power

• the power factor, how to improve.

AC Circuit Power Analysis

Instantaneous Power:

Week #11

ENE 104 AC Circuit Power Analysis

𝑝 𝑡 = 𝑣 𝑡 𝑖(𝑡)

the device is a resistor:

𝑝 𝑡 = 𝑖2 𝑡 𝑅 =𝑣2 𝑡

𝑅

the device is entirely inductive:

𝑝 𝑡 = 𝐿𝑖 𝑡𝑑𝑖 𝑡

𝑑𝑡=

1

𝐿𝑣(𝑡) 𝑣 𝑡′ 𝑑

𝑡

−∞𝑡′

the device is entirely capacitive:

𝑝 𝑡 = 𝐶𝑣 𝑡𝑑𝑣 𝑡

𝑑𝑡=

1

𝐶𝑖(𝑡) 𝑖 𝑡′ 𝑑

𝑡

−∞𝑡′


Week #11


Consider the series RL circuit,

𝑖 𝑡 =𝑉0𝑅

1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)

The total power delivered by the source

𝑝 𝑡 = 𝑣 𝑡 𝑖 𝑡 =𝑉0

2

𝑅1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)


Week #11





The power delivered to the resistor is

𝑝𝑅 𝑡 = 𝑖2 𝑡 𝑅 =𝑉0

2

𝑅(1 − 𝑒−𝑅𝑡 𝐿 )2𝑢(𝑡)


Week #11





to determine the power absorbed by the inductor, first

𝑣𝐿(𝑡) = 𝐿𝑑𝑖(𝑡)

𝑑𝑡

= 𝑉0𝑒−𝑅𝑡 𝐿 𝑢 𝑡 +

𝐿𝑉0

𝑅1 − 𝑒−𝑅𝑡 𝐿 𝑑𝑢(𝑡)

𝑑𝑡

= 𝑉0𝑒−𝑅𝑡 𝐿 𝑢 𝑡

Then,

𝑝𝐿(𝑡) = 𝑣𝐿 𝑡 𝑖(𝑡) =𝑉0

2

𝑅𝑒−𝑅𝑡 𝐿 (1 − 𝑒−𝑅𝑡 𝐿 )2𝑢(𝑡)


Week #11



𝑝 𝑡 = 𝑝𝑅 𝑡 + 𝑝𝐿(𝑡)


Week #11


Power Due to Sinusoidal Excitation:

the response is Where

)cos()( tIti m

222 LR

VI m

m

R

L 1tan

Practice: 11.1

Week #11


A current source of 12cos(2000t) A., a 200-ohm

resistor, and a 0.2-H inductor are in parallel.

Assume steady-state conditions exist. At t = 1ms.,

find the power being absorbed by the

(a) resistor

(b) inductor

(c) sinusoidal source

Practice: 11.1

Week #11


Average Power:

Week #11


For Periodic Waveforms:

2

1

)(1

12

t

t

dttptt

P

)()( Ttftf

Tt

t

dttpT

P1

1

)(1

1

Tt

t

x

x

x

dttpT

P )(1

Average Power:

Week #11


For Periodic Waveforms:

2

2

)(1

nT

nT

dttpnT

P

,...3,2,1)(1

ndttpnT

P

nTt

t

x

x

2

2

)(1

lim

nT

nT

dttpnT

Pn

Example:

Week #11


find the average power delivered to a resistor R

by the periodic sawtooth current waveform

TtTTtT

Iti m 2),()(

TttT

Iti m 0,)(

R

tvRtititvtp

)()()()()(

22

Example:

Week #11


TtTTtT

Iti m 2),()(

TttT

Iti m 0,)(

TtRtIT

tp m 0,1

)( 22

2

RIdttT

RI

TP m

T

m 2

0

2

2

2

3

11

TtTTtRIT

tp m 2,)(1

)( 22

2

Average Power:

Week #11


In the Sinusoidal Steady State

the instantaneous power is

Or

By inspection:

)cos()( tIti m

)cos()( tVtv m

)cos()cos()( ttIVtp mm

)cos(2

1 mmIVP

)2cos(2

1)cos(

2

1)( tIVIVtp mmmm

Example:

Week #11


Given the time-domain voltage V.,

find both the average power and an expression for the

instantaneous power that result when the corresponding

phasor voltage V = 4 0o V. is applied across an

impedance Z = 2 60o ohm.

Sol:

6/cos4 tv

Example:

Week #11


W

IVP mm

2)60cos()2)(4(2

1

)cos(2

1

V. V = 4 0o V.

Z = 2 60o ohm.

Sol:

The phasor current is V/Z = 2 -60o

So the average power is:

6/cos4 tv

606/cos2 ti

Example:

Week #11


)603

cos(42)606

cos()6

cos(8

)2cos(2

1)cos(

2

1

)cos()cos()(

ttt

tIVIV

ttIVtp

mmmm

mm

the instantaneous power is:

6/cos4 tv 606/cos2 ti

Example:

Week #11


)603

cos(42)( t

tp

6/cos4 tv 606/cos2 ti

Practice: 11.2

Week #11


Given the phasor voltage V = V. across

an impedance Z = 16.26 19.3o ohm., obtain an

expression for the instantaneous power, and compute

the average power if = 50 rad/s.

Sol:

452115

)cos()cos()( ttIVtp mm

)2cos(2

1)cos(

2

1)( tIVIVtp mmmm

Practice: 11.2

Week #11


Average Power:

Week #11


Absorbed by an Ideal Resistor:

Or

Absorbed by Purely Reactive Elements:

mmmmR IVIVP2

1)0cos(

2

1

R

VRIP m

mR22

1 22

0xP

Example:

Week #11


Find the average power being delivered to an

impedance ZL = 8-j11 ohm. by a current I = 5 20o A.

Sol:

.100852

1

2

1 22 WRIP mR

Practice: 11.3

Week #11


Calculate the average power delivered to the

impedance 6∠25°Ω by the current 𝐈 = 2 + 𝑗5 A.

Practice: 11.4

Week #11


Compute the average power delivered to each of the

passive elements.

Sol:

Practice: 11.4

Week #11


For the circuit in the figure below, compute the

average power delivered to each of the passive

elements. Verify your answer by computing the

power delivered by the two sources.

Practice: 11.4

Week #11


Maximum Power Transfer:

Week #11


An independent voltage source in series with an

impedance Zth delivers a maximum average power

to that load impedance ZL which is the conjugate of

Zth, or ZL=Z*th

Example 11.5:

Week #11


If we are assured that the voltage source is delivering

maximum average power to the unknown impedance,

what is its value?

Sol:

3500*

? jZZ th

Practice: 11.5

Week #11


If the 30-mH inductor of Example 11.5 is

replaced with a 10-𝜇𝐹 capacitor, what is the

value of the inductive component of the

unknown impedance 𝐙? if it is known that 𝐙? is

absorbing maximum power?

Practice: 11.5

Week #11


Average Power:

Week #11


For Nonperiodic Function: for example

the average power delivered to a 1 ohm resistor:

ttti sinsin)(

Watt

dtttttP

12

1

2

1

)sinsin2sin(sin1

lim2

2

22

Average Power:

Week #11


For

the average power delivered to a resistor R:

tItItIti NmNmm cos...coscos)( 2211

RIIIP mNmm

22

2

2

1 ...2

1

Practice: 11.6

Week #11


A voltage source 𝑣𝑠 is connected across a 4-Ω

resistor. Find the average power absorbed by

the resistor if 𝑣𝑠 equals (a) 8𝑠𝑖𝑛200𝑡 (b)

8𝑠𝑖𝑛200𝑡 − 6cos(200𝑡 − 45°) V. (c) 8𝑠𝑖𝑛200𝑡 −4𝑠𝑖𝑛100𝑡 V. (d) 8𝑠𝑖𝑛200𝑡 − 6 cos 200𝑡 − 45° −5𝑠𝑖𝑛100𝑡 + 4 V.

Practice: 11.6

Week #11


Effective Values:

Week #11


If the resistor received the

same average power in part

(a) and (b), then the effective

value of i(t) is equal to Ieff,

and the effective value of v(t)

is equal to Veff

Effective Values:

Week #11


of a Periodic Waveform:

the average power delivered

to the resistor,

the power delivered by the

direct current is:

RIP eff

2

TT

dtiT

Rdti

TP

0

2

0

21

Effective Values:

Week #11


of a Periodic Waveform:

we get

note: the effective value is

often called the root-mean-

square value, or simply the

rms value.

RIdtiT

Reff

T2

0

2

T

eff dtiT

I0

21

Effective Values:

Week #11


of a sinusoidal Waveform:

which has a period

to obtain the effective value

)cos()( tIti m

T

eff dtiT

I0

21

2T

2

0 21

21

0

22

)]22cos([2

)(cos1

dttI

dttIT

I

m

T

meff

Effective Values:

Week #11


of a sinusoidal Waveform

)cos()( tIti m

T

eff dtiT

I0

21

2

][4

)]22cos([2

)(cos1

2

2

0

0 21

21

0

22

meff

m

m

T

meff

II

tI

dttI

dttIT

I

Effective Values:

Week #11


Use of RMS Values to

Compute Average Power

R

VRIRIP

eff

effm

2

22

2

1

2

meff

II

)cos()cos(2

1 effeffmm IVIVP

Effective Values:

Week #11


with Multiple-Frequency

Circuits

RIIIP Neffeffeff

22

2

2

1 ...

2

meff

II

22

2

2

1 ... Neffeffeffeff IIII

Effective Values: Example

Week #11


Ex22 Page 384: find the effective value of:

T

eff dtiT

I0

21

Practice: 11.7

Week #11


Calculate the effective value of each of the

periodic voltages: (a) 6𝑐𝑜𝑠25𝑡; (b) 6𝑐𝑜𝑠25𝑡 +4sin(25𝑡 + 30°) V. (c) 6𝑐𝑜𝑠25𝑡 + 5𝑐𝑜𝑠2(25𝑡) V.

(d) 6𝑐𝑜𝑠25𝑡 + 5𝑠𝑖𝑛30𝑡 + 4 V.

Practice: 11.7

Week #11


Apparent power and Power factor:

Week #11


The sinusoidal voltage

is applied to the network and the resultant

sinusoidal current is

the average power delivered to the network

the apparent power:

the power factor:

)cos( tVv m

)cos( tIi m

effeff IV

)cos()cos(2

1 effeffmm IVIVP

effeff IV

PPF

Example:

Week #11


Calculate values for the average power delivered

to each of the two loads, the apparent power

supplied by the source, and the power factor of

the combined load.

Example:

Week #11


Calculate values

for the average

power delivered

to each of the two

loads, the

apparent power

supplied by the

source, and the

power factor of

the combined

load.

)cos( angIangVIVP effeff

effeff IV

effeff IV

PPF

Example:

Week #11


.432

)]1.53(0cos[)12()60(

)cos(

W

angIangVIVP effeffS

rmsAjj

IS .13.5312)51()12(

060

Example:

Week #11


.720)12()60( VAIV effeff

the apparent power

supplied by the source:

the power factor of the

combined load.

6.0)12()60(

432

effeff IV

PPF

2)12( 2 P

1)12( 2 P

Practice: 11.8

Week #11


For the circuit of the figure below, determine the

power factor of the combined loads if 𝑍𝐿 = 10Ω.

Practice: 11.8

Week #11


Complex Power:

Week #11


the average power, express as,

the complex power,

)cos( effeff IVP

*)( ReReRe effeff

j

eff

j

eff

j

effeff IVeIeVeIVP

)sin( effeff IVQ

jQPeIVIVS j

effeffeffeff )(*

Power Measurement:

Week #11


The complex power

delivered to several

interconnected loads is

the sum of the complex

power delivered to

each of the individual

loads.

S = VI* = V(I1+I2)

* = V(I1*+I2

*) = VI1*+VI2

*

Example:

Week #11


An industrial consumer

is operating a 50-kW

induction motor at a

lagging PF of 0.8. The

source voltage is 230

Vrms. In order to obtain

lower electrical rates,

the customer wishes to

raise the PF to 0.95

lagging. Specify a

suitable solution.

Example:

Week #11


A purely reactive load must be added to the

system, in parallel, since the supply voltage must

not change. The complex power supplied to the motor must

have a real part of 50-kW and an angle of

cos-1(0.8)=36.9o

Hence, S1

kVAj 5.3750

8.0

9.3650

Example:

Week #11


S1 =

To achieve a PF of 0.95, the total complex power

must become:

S

Thus the corrective load is

S2 =

kVAj 5.37508.0

9.3650

kVAj 07.21

kVAj 43.1650

)95.0(cos95.0

50 1

Example:

Week #11


S2 =

We select a phase angle of 0o for the voltage

source, and the current drawn by Z2 is

Or

Therefore,

.6.91230

210702*

2 Ajj

V

SI

kVAj 07.21

.6.912 AjI

51.26.91

230

2

2 jjI

VZ

Practice: 11.9

Week #11


Find the complex power

absorbed by the

(a) 1-ohm R

(b) -j10 C

(c) 5+j10 Z

(d) source

Practice: 11.9

Week #11


Practice: 11.10

Week #11


A 440-Vrms source supplies power to a load

𝑍𝐿 = 10 + 𝑗2Ω through a transmission line

having a total resistance of 1.5Ω. Find (a) the

average and apparent power supplied to the

load; (b) the average and apparent power lost in

the transmission line; (c) the average and

apparent power supplied by the source; (d) the

power factor at which the source operates.

Practice: 11.10

Week #11


Example: Final 2/47

Week #11


จากวงจรตามรูป ใหห้า 1) the average power ท่ีจ่ายโดย แหล่งจ่าย (source) 2) ค่า power factor ท่ีแหล่งจ่าย (source) 3) ขนาดของ ตวัเกบ็ประจุ (capacitor) ท่ีเม่ือน าไปต่อ

ขนานกบั แหล่งจ่าย ท าใหค่้า power factor = 1 4 ohm

j16 ohm120 Vrms 60Hz

12 ohm

Example: Final 2/47

Week #11


Example: Final 2/47

Week #11


4 ohm

j16 ohm120 Vrms 60Hz

12 ohm

Example: Final 2/47

Week #11


Ex:

ENE 104

1 µF

7.5 ΩVg

50 µH

5 Ω

Find the average power, the reactive power,

and the apparent power supplied by the voltage source in the circuit if vg = 50 cos 105t V.

Ex:

ENE 104

1 µF

7.5 ΩVg

50 µH

5 Ω

Summary:

Week #11


Hw:

Week #11


Reference:

ENE 104

W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.

Copyright ©2002 McGraw-Hill. All rights reserved.

Circuit Analysis and Electrical Engineering

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