eeu104 01 one phase ac circuit

8/10/2019 EEU104 01 One Phase AC Circuit

1/189

EEU104/3: Electrical Technology

AC CIRCUIT ANALYSIS

Dr Haidi Ibrahim

Semester 2, 2010/2011 Session


2/189

EEU104/3:ElectricalTechnology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number

-- Sinusoidal waveform

Basic AC component

-- Power sources-- Resistance

-- Reactance

-- Impedance

-- Power calculation

Circuit Analysis

Main Reference

Edward Hughes, Hughes Electrical andElectronic Technology, 10th Edition,Pearson Prentice Hall, 2008

Chapter 9: Alternating Voltage and Current

Chapter 10: Single-phase Series Circuits

Chapter 11: Single-phase Parallel Networks

Chapter 12: Power in AC Circuits

Chapter 13: Complex Notation

Chapter 14: Resonance in AC Circuits

Chapter 15: Network Theorem Applied to ACNetworks


3/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Expected Outcomes

After this lesson, the students

should be able to analyze one-phase

AC circuit.

Using complex number and phasor

Sinusoidal and cosine waveform (i.e.

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

Calculate the average and r.m.s values


4/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Complex Number

Complex number consists of real and

imaginary part.

Here, a presents the real part, andjb

presents the imaginary part.

jba

1

12

j

j


5/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Complex Number

If N=a+jb, its complex conjugate is:

Then,

jba*N

22* ba NN


6/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Complex Number

If N1=a+jb, and N2=c+jd

Addition:

Subtraction:

Multiplication:

Division:

dbjcajdcjba 21 NN

dbjcajdcjba 21 NN

adbcjbdacjdcjba 21NN

22*2

*

2

2

1

2

1

dc

adbcjbdac

N

N

N

N

N

N


7/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.1:

If N1= -0.6753 +j0.3425, and

N2= 1.4322 -j0.2155, in within 3minutes, find the followings:

a) N1+N2

b) N1-N

2c) N1N2

d) N1/N2


8/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

TIPS: Casio fx-570MSSTEP 1: Set the calculator into CMPLX mode

STEP 2: Store N1 into memory A

STEP 3: Store N2 into memory B

ON MODE 2

(-) 6 7 5 3 + 3 4 2 5 i. .

SHIFT STO A

4 3 2 2 - 2 1 5 5 i. .

SHIFT STO B

1


9/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

TIPS: Casio fx-570MSSTEP 4: Solve N1 +N2

Use to toggle between real and imaginary.

STEP 5: Use similar operation as STEP 4 to solve N1 -N2, N1 N2,

and N1 /N2.

ALPHA A ALPHA B+ =

SHIFT ReIm


10/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.1:

If N1= -0.6753 +j0.3425, and

N2= 1.4322 -j0.2155, in within 3minutes, find the followings:

a) N1+N2 = 0.7569+j0.1270

b) N1

-N2

= -2.1075+j0.5580

c) N1N2 = -0.8934+j0.6361

d) N1/N2 = -0.4963+j0.1645


11/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

There are three common ways of

representing complex number:

1. Rectangular or Cartesian notation

2. Polar notation

3. Trigonometric notation

jbaA

AA

sincos jA A


12/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

If given: , in Cartesian

space:

jbaA

Im

Re

Horizontal axis for real part, vertical axis for imaginary part


13/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis


space:

jbaA

Im

Re

Treating as vector

aa

jbbjbaA


14/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis


space:

jbaA

Im

Re

Treating as vector

aa

jbbjbaA


15/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Using Pythagoras theorem:

a

bA

22 baA

baA

aA

bA


16/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Using trigonometry theories:

a

bA

a

barctan


17/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Therefore, rectangular notation can

be changed into polar form:

a

bA

Ajba


18/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Using trigonometry theories:

a

bA

cosAa

sinAb


19/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Thus, polar notation can be changed

into trigonometric or rectangular

notation.

a

bA


20/189


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.2:

If A= -3-j4, presents it in polar

notation and trigonometric notation.


21/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

TIPS: Casio fx-570MSSTEP 1: Set the calculator into CMPLX and Deg mode

STEP2: Input the rectangular notation

STEP 3: Convert to polar notation

Use to toggle between magnitude and angle

(-) 3 - 4 i

SHIFT

ON MODE 2 MODE MODE MODE MODE 1

r =

SHIFT ReIm


22/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.2:

If A= -3-j4, presents it in polar

notation and trigonometric notation.

In polar notation:

In trigonometric notation:

8699.1265 A

8699.126sin8699.126cos5 jA


23/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.3:

If A= -3-j4, and B= 75 calculate

C=A+B.


24/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis


STEP2:

Input A in rectangular form

Addition

Input B in polar form

Result

(-) 3 - 4 i

SHIFT


=

+

7 5


25/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.3:


C=A+B.

4694.402229.5

3899.39734.3

jC


26/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.4:


C=A+B.


27/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

TIPS: Casio fx-570MSSTEP 1: Set the calculator into CMPLX and Rad mode

STEP2:

Input A in rectangular form

Addition

Input B in polar form

Result

(-) 3 - 4 i

SHIFT


=

+

7 5


28/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.4:


C=A+B.

6652.17604.10

7125.100144.1

jC

* Unit is important!!!


29/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.5:

If A= 34, and B= 56 calculate

C=A+B.


30/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis


STEP2:

Input A in polar form (degree)

Addition

Convert B to degree

Result

3 4

SHIFT


=

+

5 6

SHIFT

SHIFT DRG 2


31/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.5:


C=A+B.

1512.08835.7

6657.88835.7

1878.17935.7

jC


32/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.6:


C=AB.


33/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.6:


C=AB.

8

18081764)24(

176244

C


34/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.7:


C=A/B.


35/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.7:


C=A/B.

1722

1764)2/4(

1762/44

C


36/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.8:

If A= 44, and B= -5184

calculate C=A+B.


37/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.8:

If A= 44, and B= -5184

calculate C=A+B.

49

4544

1845180144

1845)1(44

184)5(44

C


38/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.9:

If A= 44, and B= -5184

calculate C=(A-1+B-1)-1.


39/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component


-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.9:

If A= 44, and B= -5184

calculate C=(A-1+B-1)-1.

42222.2

calculatorusing

18451

441

184)5(

1

44

1

1

1

C


40/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.20:

If A= 4+3j, and B= 11-12j calculate

C=(A-1+B-1)-1.


41/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.20:

If A= 4+3j, and B= 11-12j calculate

C=(A-1+B-1)-1.

3441.206530.4

calculatorusing1211

1

34

11

jj

C


42/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Sinusoidal Waveform

The cosine or sinusoidal waveform is acommon type of alternating current (ac)and alternating voltage.

-1.5

-1

-0.5

0

0.5

1

1.5


43/189

EEU104/3:

ElectricalTechnology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Sinusoidal Waveform

The cosine waveform is expressed

as:

Ip, Vp,and

Ppare the peak values

(capital letters present constant)

Small letters present values that

change over time

)cos()(

)cos()(

)cos()(

tPtp

tVtv

tIti

p

p

p


44/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Angular velocity, , is given as:

Frequency,f, is given as 1/T, where T

is a period of one cycle.

ff 3602

-1.5

-1

-0.5

0

0.5

1

1.5

T

Peak value


45/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Phase angle, , can be in radian or

degree (thus, unit is important)

Similarly, unit is important to the

angular velocity too.


46/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.11

Find the voltage at t=0.1s, if this ac

voltage is given by:

V)3434cos(75.0)( ttv


47/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.11

Find the voltage at t=0.1s, if this ac

voltage is given by:

V)3434cos(75.0)( ttv

0.4940V)s1.0( v


48/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.12

Find the current at t=1s, if this ac

current is given by:

A)5634cos()( tti


49/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.12

Find the current at t=1s, if this ac

current is given by:

A)5634cos()( tti

0A)s1( i


50/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.13

Find the power at t=1s, t=1.1s and

t=2s if this ac power is given by:

W)3cos(10)( ttp


51/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.13

Find the power at t=1s, t=1.1s and

t=2s if this ac power is given by:

W)3cos(10)( ttp

W8366.2)s2(

-5.7482W)s1.1(

-6.5364W)s1(

p

p

p


52/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.14

For the following waveform,

determine its peak value, angular

velocity, frequency, period, andphase angle.

W)3cos(10)( ttp


53/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Peak value,Pp = 10W

Angular velocity, = 1 rad/s

Frequency,

f = (1 rad/s)/(2 rad) = (1/2)Hz

Period, T= 1/f= 2 s Phase angle, = 3 rad

W)3cos(10)( ttp


54/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.15




V)45sin()( ttv


55/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Peak value, Vp = 1V

Angular velocity, = 45o/s

Frequency,

f = (45o/s)/(360o) = 0.125Hz Period, T= 1/f= 8 s

Phase angle, = -90o

V)9045cos(

V)45sin()(

t

ttv


56/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.16




V)cos(754.0)( ttv


57/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Peak value, Vp = 0.754V

Angular velocity, = o/s

Frequency,

f = ( o/s)/(360o) = 8.7266mHz

Period, T= 1/f= 114.5916 s Phase angle, = rad

V)cos(754.0)( ttv


58/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Exercise 1.17




A)cos(754.0)( ti


59/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

This is not an ac waveform, but a dc

supply

A-0.754

A0)0.754(-1.0

A)cos(754.0)(

ti


60/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

If angular velocity, , decreases:

Frequency,f, decreases, while period,

T, increases

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(0.5t)

cos(t)cos(2t)


61/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance-- Reactance

-- Impedance


Circuit Analysis

If angular velocity, , decreases:

Frequency,f, decreases, while period,

T, increases

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(0.5t)

cos(t)cos(2t)


62/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

If angular velocity, , increases:

Frequency,f, increases, while period, T,

decreases

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(0.5t)

cos(t)cos(2t)


63/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

If angular velocity, , increases:

Frequency,f, increases, while period, T,

decreases

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(0.5t)

cos(t)cos(2t)


64/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak value, (e.g. Vp, Ip, Pp), isproportional to the amplitude of the

waveform.

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

0.5cos(t)

cos(t)

1.5cos(t)


65/189


66/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Negative phase angle : waveformshifted to the right (i.e. lagging).

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(t-90o)

cos(t)

cos(t+90o)


67/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Positive phase angle : waveformshifted to the left (i.e. leading).

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(t-90o)

cos(t)

cos(t+90o)


68/189


69/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630 720

t(degree)

cos(t-90o)

cos(t)

cos(t+90o)

tphase

tphase

3602

phase

T

t


70/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Exercise 1.18:

Find the expression to present the

following voltage waveform. Then,

find the value ofx.

-6

-4

-2

0

2

4

6

0 1 3 5 7

t(seconds)

x


71/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

TIPS: StrategySTEP 1: Find the expression

What are needed? :

1) Peak values2) Angular velocity

3) Phase angle

STEP 2: Find the value ofx

x = v(0)


72/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak value, Vp = 4V

Time for half cycle = 4s

Thus, period T= 8s

Frequency,f= 1/T= 1/(8s) = 0.125Hz

Angular frequency, = 2f= 0.25

From figure, tphase = 7s

75.12

8

72

2

phasephase

T

t

T

t


73/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Therefore, the expression is:

The value ofx:

)V1.754cos(0.25

V)cos()(

t

tVtv p

2.8284V

)V1.7504cos(

)V1.75)0(4cos(0.25

)0(

vx


74/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Positive or negative phase angle?

(i.e. leading, or lagging?)

V360cosV2cos

V)cos()(

o

p

p

p

tV

tVtVtv


75/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

A)75.0cos()(

V)cos(2)(

tti

ttv

-3

-2

-1

0

1

2

3

0 1 3 5 7

t(second)

i(t)

v(t)

v(t) leads i(t) by 0.75 radians, or 135o

i(t) lags v(t) by 0.75 radians, or 135o


76/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

A)25.1cos()(

A)275.0cos()(

V)cos(2)(

tti

tti

ttv

-3

-2

-1

0

1

2

3

0 1 3 5 7

t(second)

i(t)

v(t)

v(t) lags i(t) by 1.25 radians, or 225o

i(t) leads v(t) by 1.25 radians, or 225o


77/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Several ways to present waveform

numerically:

Peak value

Peak-to-peak value

Average value

Root mean square value


78/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak value:

-15

-10

-5

0

5

10

15

Peak value = 10


79/189


80/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak-to-peak value:

-15

-10

-5

0

5

10

15

Peak-to-peak value = 10-(-10) = 20


81/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak-to-peak value:

-10

-5

0

5

10

15

20

Peak-to-peak value = 15-(-5) = 20


82/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Peak-to-peak value: (disadvantage)

-15

-10

-5

0

5

10

15

Both have peak-to-peak value = 20

-10

-5

0

5

10

15

20


83/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Average value:

If given the following sample:

The average is:

}9,4,8{ v

1

3/3

3

)9(48avg

V

*note the notation


84/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis


The average is:

}94,8,9,4,8{ v

1

3/3

)3(2

)9(482

6

)9(48)9(48avg

V


85/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis


The average is:

,...9,4,8)( tv

1

3/3

3

)9(48avg

V

(Repeated)


86/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In general:

Integration = area under the graph

T

dttvT

V0

avg )(1


87/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Average value:

-15

-10

-5

0

5

10

15

Average =0


88/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Average value:

-10

-5

0

5

10

15

20

Average =5


89/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Average value: (disadvantage)

-15

-10

-5

0

5

10

15

0 = No current or voltage?


90/189


91/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Root mean square (rms) value:

T

dttvT

V0

2

rms )(1


92/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

For a full cycle sinusoidal or cosinewaveform:

T

p

T

p

T

p

dtft

VT

V

dtftVT

V

dtftVTV

0

2

rms

0

22

rms

0

2

rms

2

)4cos(11

)2(sin1

)2sin(

1


93/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

p

p

p

Tp

T

p

VV

V

TT

VV

dtftT

VV

dtft

VT

V

7071.02

]0[2

)4cos(12

2

)4cos(11

rms

2

2

rms

0

2

2

rms

0

2

rms


94/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

RMS value:

-15

-10

-5

0

5

10

15

RMS value = 0.7071(10)=7.071


95/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

RMS value:

Irms is normally written asI

Vrms is normally written as V

Power Sources


96/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number

-- Sinusoidal waveforms

Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Power Sources

Power source in ac circuits (i.e.voltage supply or current supply)

Normally, this power source supplies

cosine or sinusoidal waveforms.

For time domain expression:

In complex number notation:

)cos()( tVtv p

sincosrms jV V


97/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In polar form:

The symbols:

Positive cycle is indicated by symbol +

rmsVV

+

-

rmsVV


98/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In polar form:

The symbols:


rmsVV

+

-

)cos()( p tVtv


99/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In polar form:

The symbols:


rmsVV

+

-

rmsII


100/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In polar form:

The symbols:


rmsVV

+

-

)cos()( p tIti


101/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

)32.15cos(1421.14)( tti+

-


102/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Is equivalent to:

+

-

)32.15cos(1421.14)( tti+

-

32.110I


103/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

+

-

3222 V


104/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Is equivalent to:

+

-)32cos(4)( ttv

+

-

3222 V


105/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Is equivalent to:

+

-)32cos(4)( ttv

+

-

3222 V

In case of not given value


106/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:+

-

)327cos(4)( ttv


107/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

+

-

3222 V

+

-

)327cos(4)( ttv


108/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

+

-

3222 V

+

-

)327cos(4)( ttv

+

-

)327cos(4)( ttv


109/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

+

-

3222 V

+

-

)327cos(4)( ttv

+

-

)327cos(4)( ttv


110/189

Reactance


111/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Reactance

In ac circuit, the capacitor andinductor are treated as imaginary

number.

ReactanceX, which is similar toresistance for resistor, is associated

with inductor and capacitor.

Unit for reactance is ohm,


112/189


113/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

2H4

L


114/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

2H4

L

8

)4)(2(

L

j

jLj

jXL

X


115/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

For capacitor:

C

fCj

Cj

jXC

2

1

1

CX


116/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

2F4

C


117/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

2F4

C

125.0

)4)(2(

1

1

C

j

j

Cj

jXC

X

Impedance


118/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Impedance

Impedance, Z, is a complex number,normally presented in rectangular or

polar form.

Impedance


119/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Impedance

Impedance, Z, is a complex number,normally presented in rectangular or

polar form.

Impedance = Resistance + Reactance


120/189


121/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R

4jCX

8j

LX


122/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R

4jCX

8j

LX

9273.05

43

483

j

jj

R CL XXZ


123/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R

4jCX

8j

LX

9273.05

43

483

j

jj

R CL XXZ

3R

4X 5Z

9273.0


124/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R 42 jCX

908LX

9051CX


125/189

EEU104/3:


Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R 42 jCX

908LX

9051CX

1

2

1

11

C

CLR X

XXZ

6148.668374.4

44.492.1

jZ


126/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Example:

Z

3

R 42 jCX

908LX

9051CX

1

2

1

11

C

CLR X

XXZ

6148.668374.4

44.492.1

jZ

92.1R

44.4X 8374.4Z

6148.66


127/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

Units:

Quantity Symbol Unit

Resistance R Reactance X

Impedance Z

Conductance (1/R) G S

Admittance (1/Z) Y S

Power


128/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In resistive circuit:

The current and voltage across resistor,

(i.e. iR(t) and vR(t)) are at the same

phase (i.e. in phase)

)(tvS)(tvR

+

-

+

-

R)(tiR


129/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources


-- Impedance


Circuit Analysis

In resistive circuit:

If

Then,

)(tvS)(tvR

+

-

+

-

)(tiR

V)2sin()( ftVtv pS

V)2sin()( ftVtv pR

A)2sin()(

)(

R

ftV

R

titi

pRR

R


130/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

The instantaneous power dissipated byresistor:

)()()( tvtitp RRR

ftVItp

ftVftItp

ftVftR

Vtp

ppR

ppR

p

p

R

2sin)(

2sin2sin)(

2sin2sin)(

2


131/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.51

1.5

2

2.5

v_R(t)

i_R(t)

p_R(t)


132/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.51

1.5

2

2.5

v_R(t)

i_R(t)

p_R(t)


133/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.51

1.5

2

2.5

v_R(t)

i_R(t)

p_R(t)


134/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

The average power dissipated byresistor:

T

pp

T

pp

T

dtft

VIT

P

dtftVIT

P

dttp

T

P

0

avg

0

2

avg

0

avg

2

)4cos(11

)2(sin1

)(1


135/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

rmsrmsavg

avg

0

avg

0

avg

222

)0()(

2

)4cos(12

2

)4cos(11

VIVIVI

P

T

T

VIP

dtftT

VI

P

dtft

VIT

P

pppp

pp

Tpp

T

pp


136/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

In inductive circuit:

The current and voltage across

inductor, (i.e. iL(t) and vL(t)) are NOT at

the same phase (i.e. out of phase)

)(tvS)(tvL

+

-

+

-

)(tiLL


137/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

In inductive circuit:

If

Then,

V)2cos()( ftVtv pS

V0rmsVSV90 LL X

)(tvS)(tvL

+

-

+

-

)(tiLL


138/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

)(tvS)(tvL

+

-

+

-

)(tiLL

V0rms VSL VV

9090

90

V0

rmsrms

rms

IL

V

L

V

L

LL

X

VI


139/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Im

Re

LI

LV

If seen clockwise, VL leads IL by 90o.


140/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

V0rmsVLV

90rms ILI

V)2cos()( ftVtv pL

A)902cos()( ftIti pL


141/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

The instantaneous power dissipated byinductor:

)()()( tvtitp LLL

ftVI

tp

ftftVItp

ftftftVItp

ftftVItp

pp

L

ppL

ppL

ppL

4sin2

)(

2cos2sin)(

2cos90sin2sin90cos2cos)(

2cos902cos)(


142/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


143/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance



-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


144/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance



-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


145/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

The average power dissipated by inductor:

The inductor takes energy from thesource, and at the other time, return back

the energy. The inductor does notdissipate power.

0)4sin(2

1

)(1

0

avg

0

avg

T

pp

T

dtftVI

TP

dttpT

P


146/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

In capacitive circuit:

The current and voltage across

capacitor, (i.e. iC(t) and vC(t)) are NOT

at the same phase (i.e. out of phase)

)(tvS)(tvC

+

-

+

-

)(tiCC


147/189


148/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

V0rms VSC VV

9090

90)/1(

V0

rmsrms

rms

ICV

C

V

C

C

C

X

V

I

)(tvS)(tvC

+

-

+

-

)(tiC C


149/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

Im

Re

CI

CV

If seen clockwise, VC lags IC by 90o.


150/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

V0rmsVCV

90rmsICI

V)2cos()( ftVtv pC

A)902cos()( ftIti pC


151/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance

-- Impedance


Circuit Analysis

The instantaneous power dissipated bycapacitor:

)()()( tvtitp CCC

ftVI

tp

ftftVItp

ftftftVItp

ftftVItp

pp

C

ppC

ppC

ppC

4sin2

)(

2cos2sin)(

2cos90sin2sin90cos2cos)(

2cos902cos)(


152/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance

-- Reactance-- Impedance


Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


153/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


154/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

-2.5

-2

-1.5-1

-0.5

0

0.5

1

1.5

2

2.5

v_L(t)

i_L(t)

p_L(t)


155/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

The average power dissipated bycapacitor:

The capacitor takes energy from the

source, and at the other time, return backthe energy. The capacitor does notdissipate power.

0)4sin(2

1

)(1

0

avg

0

avg

T

pp

T

dtftVIT

P

dttpT

P

Power in AC Circuit


156/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

Average power: Resistor = (across resistor)

Inductor = 0W

Capacitor = 0W

In AC circuit, there are three kind of

powers, related by:

rmsrmsVIP avg

QPS j


157/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

Power triangle

Related to power triangle!

i.e. takes the angle from the impedance

triangle.

R

XZ

P

QS


158/189

EEU104/3:Electrical

Technology

Dr Haidi Ibrahim

AC CIRCUIT ANALYSIS

Main reference

Expected outcomes

Revisions:

-- Complex number


Basic AC component

-- Power sources

-- Resistance



Circuit Analysis

Power triangle

Related to power triangle!

i.e. takes the angle from the impedance

triangle.

R

XZ

P

QS

eeu104 01 one phase ac circuit

Documents