Dc Ckt Analysis Ppt

7/28/2019 Dc Ckt Analysis Ppt

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UNIVERSITY OF TECHNOLOGY, SYDNEYUNIVERSITY OF TECHNOLOGY, SYDNEY

FACULTY OF ENGINEERINGFACULTY OF ENGINEERING

48531 Electromechanical Systems48531 Electromechanical Systems

DC and AC Circuit AnalysisDC and AC Circuit Analysis

Topics to cover:

1. DC Circuit Analysis

2. Single Phase AC Circuit Analysis3. Balanced Three Phase AC Circuit Analysis

DC Circuit AnalysisDC Circuit Analysis- Ohm’s Law- Ohm’s Law

Ohm’s law states that the voltage drop across a resistor is equal to theproduct of the current through it and its resistance. That is

V IR= R

+

I

V

DC Circuit AnalysisDC Circuit Analysis-- Kirchhoff’s Kirchhoff’s Voltage Law (KVL)Voltage Law (KVL)

The algebraic sum of all the voltages around a closed path in an electric

circuit is equal to zero. That is

V k k

n

=∑ =

1

0

R I 1

1

I 2

R 2

+ V 1

I 4

R 4

I 3

R 3

+V 3

+

V 2

+

V 4

DC Circuit AnalysisDC Circuit Analysis-- Kirchhoff’s Kirchhoff’s Current Law (KCL)Current Law (KCL)

The algebraic sum of all the currents at any node in an electric circuit is

equal to zero. That is

I k k

n

=∑ =

1

0

R I 1 I 31

I 2

R

R

2

3



DC Circuit AnalysisDC Circuit Analysis-- Thevenin’sThevenin’s TheoremTheorem

A linear circuit containing any number of sources and elements, whenviewed from two nodes (terminals), can be replaced by an equivalent

voltage source (also known as Thevenin’s voltage), V T, in series with an

equivalent resistance, RT (also known as Thevenin’s resistance), where

V T is the open circuit voltage between the two nodes and RT is the ratio

of the open circuit voltage to the short circuit current. If the electric

circuit contains only independent sources, RT can be obtained by looking

at the terminals with the voltage sources replaced by short circuits andthe current sources replaced by open circuits. The open circuit voltage V Tis obtained by removing the load and leaving the terminals open.

DC Circuit AnalysisDC Circuit Analysis-- Thevenin’sThevenin’s TheoremTheorem

R

I

V

RT

+T

R

I DC Circuit

with Voltage

and Current

Sources and

Resistors

L L

where V T is the open circuit voltage and RT is the output resistance

obtained by short circuiting all voltage sources and open circuiting

all current sources in the network of two terminals.

DC Circuit AnalysisDC Circuit Analysis- Maximum Power Transfer- Maximum Power Transfer

The maximum power transfer theorem states that in a dc electric circuit,

maximum power transfer takes place when the load resistance is equal to

Thevenin’s equivalent resistance. That is

R R L T =

DC Circuit AnalysisDC Circuit Analysis- Example- Example

Determine the value of the load resistance RL for maximum power

transfer. What is the maximum power delivered to RL?

100 V + R

L25 V

30 Ω20 Ω

+



DC Circuit AnalysisDC Circuit Analysis- Solution 1: By KVL & KCL- Solution 1: By KVL & KCL

By the Ohm’s law, KVL and KCL, we can write circuit equations as

100 V + R

I

L 25 V

30 Ω20 Ω

L

I 1 I 2

V L +

10020

2530

0− − + − =V V R

V L L

L

L I I I L1 2 0− + = or

Thus,( )

( )V

R

R

R

R

R L

L

L

L

L

L

=+

+ +=

++ +

=+

100

20

25

301

20

1

30

1

300 50

3 2 60

350

5 60

( )

P I RV

R

R

R R

R

R

L L L

L

L

L

L L

L

L

= = = +

=+

2

2 2

2

2

350

5 60

1

350

5 60

DC Circuit AnalysisDC Circuit Analysis- Solution 1: By KVL & KCL ( - Solution 1: By KVL & KCL ( ContCont .) .)

Let

Therefore,

( ) ( )

( )

( )

( ) ( )

dP

dR

R R R

R

R R

R

R

R

L

L

L L L

L

L L

L

L

L

= ×+ − × × + ×

+

= ×+ − × ×

+= ×

−

+=

3505 60 2 5 60 5

5 60

3505 60 2 5

5 60350

60 5

5 600

2

2

4

2

3

2

3

( ) R L= 12 Ω

and

( )( )

( )P R

R L

L

L R L

max.=

+=

=

350

5 60102 08

2

2

12

W

Ω

DC Circuit AnalysisDC Circuit Analysis- Solution 2: By- Solution 2: By Thevenin’sThevenin’s TheoremTheorem

Applying the Thevenin’s theorem, we obtain

100 V + R L 25 V

30Ω20Ω

+

100 V +25 V

30Ω20Ω

V T +

+

R

I

L

L

V T

RT

V L

RT 20Ω 30Ω

I

( )V I T = + = + ×−

+=2 5 3 0 2 5 3 0

1 0 0 2 5

2 0 3 07 0 V

( )

1 1

2 0

1

3 0

1

1 2

1 2

R

R

T

T

= + =

∴ = Ω

( )

( )

( )

P P

I R

V

R RR

L L R R

L L R R

T

T L

L

R R

L T

L T

L T

m ax

.

=

=

=+

= ×+

=

=

=

=

W

2

2

2

2

70 12

12 12

1 02 0 8

AC Circuit AnalysisAC Circuit Analysis

- Instantaneous Expressions- Instantaneous Expressions

In an AC circuit, the voltage, current, and electric power vary with the

time sinusoidally. They can generally be expressed as

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )[ ]

v t V t i t I t

p t v t i t V I t t

V I t

m v m i

m m v i

m m

v i v i

= + = +

= = + +

= − + + +

co s co s

co s co s

co s co s

ω ϕ ω ϕ

ω ϕ ω ϕ

ϕ ϕ ω ϕ ϕ

an d

22

o

ϕ

v(t),

ππ/ 2 π/3 2 π2ω t

i(t), p(t)



AC Circuit AnalysisAC Circuit Analysis- Instantaneous Expressions ( - Instantaneous Expressions ( ContCont .) .)

The average values of the voltage and current are zero, while the average

power

( ) ( ) ( )[ ]( )

P T p t d t T

V I

t d t

V I

T

m m

v i v i

T

m m

v i

= = − + + +

= −

∫ ∫ 1 1

2 2

2

0 0cos co s

co s

ϕ ϕ ω ϕ ϕ

ϕ ϕ

where V V I I m m= =2 2an d

or P VI = cosϕ

ϕ ϕ ϕ = −v i

are the root mean square (rms)

values of the voltage and current defined by

is the phase angle between v(t) and i(t), known as the power

( ) ( )V

T

v t d t V

I

T

i t d t I

T

m

T

m= = = =

∫ ∫

1

2

1

2

2

0

2

0

a n d

factor angle, and cosϕ is known as the power factor.

AC Circuit AnalysisAC Circuit Analysis-- Phasor Phasor Expressions Expressions

By Euler’s identity,

the instantaneous voltage and current can be expressed as

where( ) ( ) ( )e t j t

j t ω ϕ ω ϕ ω ϕ

+ = + + +cos sin j = − 1

( ) ( )[ ] [ ]v t V e Ve em

j t j j t v v= =+Re Re

ω ϕ ϕ ω 2

( ) ( )[ ] [ ]i t I e Ie em

j t j j t i i= =+Re Re

ω ϕ ϕ ω 2

and

Assuming that Re is implied, we can express the voltage and current as

vectors or phasors in terms of the rms values and phase angles:

V = = ∠Ve V j

vvϕ

ϕ I = = ∠ Ie I j

iiϕ

ϕ and

AC Circuit AnalysisAC Circuit Analysis-- Phasor Phasor Expressions ( Expressions ( ContCont .) .)

Ohm’s law for R, L, and C :

Vk

k

n

=∑ =

1

0 and

Component Circuit Symbol Time Domain Phasor Expression

Resistance, R

R

v

i

v Ri= V I= R

Inductance, L

L

v

iv L

di

dt =

V I

I

==

j L

jX L

ω

Capacitance, C

C

v

ii C

dv

dt =

I V

V

=

= −

j C

jX C

ω

KVL and KCL : Ik

k

n

=∑ =

1

0

AC Circuit AnalysisAC Circuit Analysis-- Phasor Phasor Diagrams Diagrams

The relationship between voltage and current phasors can be illustrated

graphically by phasor diagrams, in which phasors are drawn as vectors.

Resistance, V = RI

Inductance, V = jX LI

Capacitance, V = − jX CI

Im - axis

Re - axis

I

O Vϕ = 0o

Im - axis

Re - axis

I

O Vϕ = 90o

Im - axis

Re - axis

I

O V

ϕ =o

90−



AC Circuit AnalysisAC Circuit Analysis- Complex Power- Complex Power

In terms of phasors, we define the complex power as

S V I= = +*

P jQwhere I* is the conjugate of I,

P V I = co s ϕ

Q V I = s in ϕand

are the real power and the reactive

power.

Q = 0 for a resistive circuit,

Q < 0 for a capacitive circuit, and

Q > 0 for an inductive circuit.

Im - axis

Re - axis

S

P

Q

O

ϕ

( )

S VI

= = ×== +

−

−

*

cos sin

Ve IeVIe

VI jVI

j j

j

v i

v i

ϕ ϕ

ϕ ϕ

ϕ ϕ

AC Circuit AnalysisAC Circuit Analysis- Example- Example

Find the current in the circuit shown below. Draw the phasor diagram

and power diagram, and sketch the input voltage and the current in thetime domain.

where v(t) = 14.142 cos 1000t V

1 mH

v(t)

i(t)

µ200 F

3Ω

AC Circuit AnalysisAC Circuit Analysis- Example Solution- Example Solution

Since the angular frequency is 1000 rad/s, we obtain

I3 Ω

= 10V ∠ 0 o V

j1 Ω

j5 Ω−

( ) X L L = = × × =−ω 1000 1 10 13 Ω ( ) X

C C = =

× ×=−

1 1

1000 200 10

56

ω

Ω

and( )Z = + − = + − = − = ∠ − R jX jX j j j L C

o3 1 5 3 4 5 5313. Ω

Therefore,

( )IV

Z= =

∠

∠ −= ∠

14142

20

5 53132 5313

.

..

o

o

o A

or

( ) ( )( ) ( )

i t I t

t

i

o

= += +

2

2 828 1000 5313

cos

. cos .

ω ϕ

A

AC Circuit AnalysisAC Circuit Analysis- Example Solution ( - Example Solution ( ContCont .) .)

The phasor and power diagrams and v(t) & i(t) waveforms

-10

-8

-6

-4

-2

0

2

4

6

8

10

0

v(t) i(t)&

π/2 π/2 π2 1000t

53.13o

V

Im-axis

Re-axis0

jX LI

jX C I

53.13o

I2 53.13

o∠ A2 143.13 o

∠ V

I R6 53.13

o∠ V

10 36.87 o∠ V

Im-axisRe-axis

0

jQ

j16 VAR

P12 W

S= P+jQ j16 VA12

S VI= = ∠ × ∠ −

= −

* .10 0 2 5313

12 16

o o

(VA) j




- Example Solution- Example Solution

0 o∠V = V ph

I

ZY

Z l

By the ∆-Y transform, we obtain ZY = 59− j82 Ω. The balanced 3 phase

circuit can be represented on a per phase basis, as shown below, where

Z Z

V

l Y

o

o

j j j+ = + + − = − = ∠ −= ∠

1 2 59 82 60 80 100 5313866

30

.

and

Ω

∴ =+

=∠ −

= ∠

= =

IV

Z Zl Y

o

o

o

500

100 53135 5313

5313 0 6

..

cos cos . .

A

and (leading)ϕ

I I Ial

o

bl

o

cl

o= ∠ = ∠− = ∠−5 5313 5 6687 5 18687. . .A, and

The line currents and the phase currents of the power supply are


- Example Solution ( - Example Solution ( ContCont .) .)

Therefore, the power dissipated by the load is

P VI source

o= = × × × =3 3 500 5 5313 4500cos cos .ϕ W

P I Rline l= = × × =3 3 5 1 752 2

W

and the power dissipated by the transmission line is

The total power supplied by the power source is

P P Pload source line

= − = − =4500 75 4425 W

The phase currents of the load are

( ) ( )I I

I I

I I

a al

o o o o

b bl

o o

c cl

o o

.= ∠ − = ∠ + = ∠

= ∠ − = ∠ −

= ∠ − = ∠ −

3 30 5 3 5313 30 2 89 8313

3 30 2 89 36 87

3 30 2 89 156 87

. .

. .

. .

A

A

A

Measurement of PowerMeasurement of Power

- in a DC Circuit- in a DC Circuit

P VI =

LoadVs

Is

DC

V

A

RLoad

(a)

(a) for RLoad >> RA and (b) for RLoad comparable to RA, where RA is

the resistance of the ammeter. The electric power dissipated in the

load

LoadVs

Is

DC

V

A

RLoad

(b)


- in a Single Phase AC Circuit- in a Single Phase AC Circuit

0 o

∠Vs = V s

Single

Phase

AC

Load

I s ∠ ϕ= I s i W*

*

(a)

P VI = cos ϕ

0 o

∠Vs = V s

Single

Phase

AC

Load

I s ∠ ϕ= I s i W*

*

(b)

(a) for Z Load >> Z A and (b) for Z Load comparable to Z A, where Z A is

the impedance of the current coil. The electric power dissipated in

the load

where ϕ = −ϕi and cosϕ is the power factor of the load.




- in a Balanced Three Phase AC Circuit- in a Balanced Three Phase AC Circuit

P V I V I ph ph ll l= 3 cos cosϕ = 3 ϕ

Three

Phase

Load

I a

0 o∠Va = V ph

240o∠ −Vc = V ph

120o∠ −Vb V ph=

Neutral

W*

*

W*

*

W*

*

Three

Phase

Load

Ial

0 o∠Va = V ph

120 o∠ −Vb= V ph

240 o

∠ −V phVc =

Ic

Ib

I a

W*

*

W*

*

1

2

( ) ( )

( ) ( )( ) ( )[ ]

P P P V I V I V I V I

V I V I

V I V I V I

a al V I b cl V I

ph ph

o

i

o

ph ph

o

i

o

ph ph

o

i

o

i ph ph

o

i ph ph i

a al b cl= + = + = − − −

= − + − − − += − + + = =

1 2 1 1 1 2 2 2

3 0 30 3 120 270

3 30 30 2 3 30 3

cos cos cos cos

cos cos

cos cos cos cos cos

ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕϕ ϕ ϕ ϕ

Therefore,

Dc Ckt Analysis Ppt

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